Enthalpy Question Bank Set 4

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY - Enthalpy Question Bank - Set 4 Liberty University Question 1 Question Calculate the change in enthalpy for a reaction where 2 moles of methane (CH4 ) are combusted completely with 4 moles of oxygen (O2 ) to form carbon dioxide (CO2 ) and water (H2 O) according to the following reaction: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2 O(l) Given the standard enthalpies of formation (∆Hf◦ ) for the compounds involved are: ∆Hf◦ (CH4 ) = −74.8 kJ/mol ∆Hf◦ (CO2 ) = −393.5 kJ/mol ∆Hf◦ (H2 O) = −285.8 kJ/mol Solution Step 1: Calculate the standard enthalpy change for the reaction using the standard enthalpies of formation of the compounds involved. The standard enthalpy change for the reaction can be calculated using the formula: X X ∆H ◦ = ∆Hf◦ (products) − ∆Hf◦ (reactants) Given: ∆Hf◦ (CH4 ) = −74.8 kJ/mol ∆Hf◦ (CO2 ) = −393.5 kJ/mol ∆Hf◦ (H2 O) = −285.8 kJ/mol The standard enthalpy change for the reaction is:     ∆H ◦ = 1 × ∆Hf◦ (CO2 ) + 2 × ∆Hf◦ (H2 O) − 1 × ∆Hf◦ (CH4 ) + 2 × ∆Hf◦ (O2 ) ∆H ◦ = [1 × (−393.5) + 2 × (−285.8)] − [1 × (−74.8) + 2 × 0] ∆H ◦ = (−393.5 − 571.6) − (−74.8) = −965.1 + 74.8 = −890.3 kJ/mol Therefore, the change in enthalpy for the given reaction is ∆H ◦ = −890.3 kJ/mol. Question 2 Question The enthalpy of formation of ammonia gas (NH3 ) is -45.9 kJ/mol. Calculate the enthalpy change when 3.00 mol of ammonia gas is formed. Solution Step 1: Write the balanced chemical equation for the formation of ammonia gas: N2 (g) + 3H2 (g) → 2NH3 (g) Step 2: Determine the number of moles of ammonia gas being formed: Given that 3.00 mol of ammonia gas is formed. Step 3: Calculate the enthalpy change for the formation of 1 mole of ammonia gas: From the balanced chemical equation, 2 moles of ammonia gas are formed when the enthalpy change is -45.9 kJ. Thus, for the formation of 1 mole of NH3 : −45.9 kJ Enthalpy change = = −22.95 kJ/mol 2 mol Step 4: Calculate the enthalpy change for the formation of 3.00 mol of ammonia gas: Enthalpy change = −22.95 kJ/mol × 3.00 mol = −68.85 kJ Therefore, the enthalpy change when 3.00 mol of ammonia gas is formed is -68.85 kJ. Question 3 Question Given the following reaction: 2H2 (g) + O2 (g) → 2H2 O(l) If the enthalpy change for the reaction is -484 kJ, calculate the enthalpy change when 4 moles of water are formed. 2 Solution Step 1: Calculate the enthalpy change for the formation of 1 mole of water. Given that the enthalpy change for the reaction is -484 kJ and the reaction forms 2 moles of water, the enthalpy change for the formation of 1 mole of water is: −484 kJ/2 = −242 kJ Step 2: Calculate the enthalpy change for the formation of 4 moles of water. Since the enthalpy change for the formation of 1 mole of water is -242 kJ, the enthalpy change for the formation of 4 moles of water is: −242 kJ/mole × 4 moles = −968 kJ Therefore, the enthalpy change when 4 moles of water are formed is -968 kJ. Question 4 Question Calculate the standard enthalpy change (∆H ◦ ) for the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) given the following standard enthalpies of formation: ∆Hf◦ (H2 (g)) = 0 kJ/mol, ∆Hf◦ (O2 (g)) = 0 kJ/mol, ∆Hf◦ (H2 O(l)) = −285.8 kJ/mol. Solution Step 1: Write the balanced chemical equation and identify the standard enthalpies of formation. 2H2 (g) + O2 (g) → 2H2 O(l) Given: ∆Hf◦ (H2 (g)) = 0 kJ/mol ∆Hf◦ (O2 (g)) = 0 kJ/mol ∆Hf◦ (H2 O(l)) = −285.8 kJ/mol Step 2: Calculate the standard enthalpy change (∆H ◦ ) for the reaction using the standard enthalpies of formation. We can use the formula: X X ∆H ◦ = ∆Hf◦ (products) − ∆Hf◦ (reactants) Substitute the values:

∆H ◦ = 2 × ∆Hf◦ (H2 O(l)) − 2 × ∆Hf◦ (H2 (g)) + ∆Hf◦ (O2 (g)) 3 ∆H ◦ = (2 × (−285.8)) − (2 × 0 + 0) ∆H ◦ = −571.6 kJ/mol Answer: The standard enthalpy change for the given reaction is -571.6 kJ/mol. Question 5 Question Calculate the change in enthalpy when 2 moles of water at 100◦ C and 1 atm condenses to liquid water at the same temperature. The enthalpy of vaporization of water at 100◦ C is 40.79 kJ/mol. Solution Step 1: Determine the change in enthalpy from vapor to liquid: Given that the enthalpy of vaporization of water at 100◦ C is 40.79 kJ/mol, the change in enthalpy when 2 moles of water condenses to liquid form can be calculated as follows: ∆H = 40.79 kJ/mol × 2 mol = 81.58 kJ Therefore, the change in enthalpy when 2 moles of water condenses to liquid form is 81.58 kJ. Question 6 Question A reaction is carried out in a bomb calorimeter. The initial temperature of the system is 25°C and the final temperature after the reaction is 45°C. If the heat capacity of the calorimeter is 150 J/°C, and the reaction released 8000 J of heat, calculate the change in enthalpy (∆H) for the reaction. Solution Step 1: Calculate the heat absorbed by the calorimeter Heat absorbed by calorimeter = Heat released by reaction Heat absorbed by calorimeter = 8000 J Step 2: Calculate the temperature change in the calorimeter ∆T = Tf − Ti = 45◦ C − 25◦ C = 20◦ C 4 Step 3: Calculate the heat absorbed by the calorimeter Q = C × ∆T 8000 = 150 × 20 Step 4: Calculate the change in enthalpy for the reaction ∆H = Q − Qcalorimeter ∆H = 8000 − 3000 ∆H = 5000 J Therefore, the change in enthalpy for the reaction is ∆H = 5000 J. Question 7 Question A reaction is carried out in a bomb calorimeter, where the heat capacity of the calorimeter is known to be 120 J/°C. During the reaction, 3.50 grams of a compound are completely burned, raising the temperature of the calorimeter and its contents by 4.50°C. Calculate the enthalpy change (∆H) for the combustion of the compound in kJ/mol. Given: Molar mass of the compound = 102.10 g/mol Solution Step 1: Calculate the heat absorbed by the calorimeter and its contents (qcalorimeter ). qcalorimeter = C × ∆T where C is the heat capacity of the calorimeter and ∆T is the temperature change. qcalorimeter = 120 J/°C × 4.50 °C qcalorimeter = 540 J Step 2: Calculate the molar quantity of the compound burned. Moles of compound = Mass 3.50 g = Molar mass 102.10 g/mol Moles of compound ≈ 0.0343 mol Step 3: Calculate the enthalpy change for the combustion reaction. qcalorimeter ∆H = − moles of compound 540 J 0.0343 mol ∆H ≈ −15765.77 J/mol ≈ −15.77 kJ/mol ∆H = − Therefore, the enthalpy change for the combustion of the compound is approximately -15.77 kJ/mol. 5 Question 8 Question A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50 moles of the reactant are consumed, what is the change in enthalpy per mole of reactant? Solution Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be calculated using the formula: ∆Hper mole = ∆H moles of reactant Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles of reactant is 2.50 mol, we can substitute these values into the formula: ∆Hper mole = −349 kJ 2.50 mol Step 3: Calculate the change in enthalpy per mole of reactant: ∆Hper mole = −349 = −139.6 kJ/mol 2.50 Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol. Question 9 Question Calculate the enthalpy change (∆H) for the reaction: 2H2 (g) + O2 (g) → H2 O(l) given the following information: ˆ ∆Hf◦ for H2 (g) = 0 kJ/mol ˆ ∆Hf◦ for O2 (g) = 0 kJ/mol ˆ ∆Hf◦ for H2 O(l) = -285.8 kJ/mol Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) Step 2: Calculate ∆H for the reaction using the standard enthalpies of formation: X X ∆H = ∆Hf◦ (products) − ∆Hf◦ (reactants) 6 Step 3: Substitute the given values into the equation: ∆H = [2(−285.8 kJ/mol)] − [2(0 kJ/mol) + 0 kJ/mol] Step 4: Solve for ∆H: ∆H = −571.6 kJ/mol Therefore, the enthalpy change for the reaction is ∆H = −571.6 kJ/mol. Question 10 Question Calculate the change in enthalpy (∆H) when 50.0 g of water at 25°C is heated to steam at 100°C. The specific heat capacity of water is 4.18 J/g°C, the molar heat of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to heat the water from 25°C to 100°C. The formula to calculate the heat required is: q = mc∆T Where: - q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Substitute the values: q = (50.0 g)(4.18 J/g°C)(100 − 25)C q = (50.0)(4.18)(75) J Step 2: Calculate the heat required to vaporize the water. The heat required to vaporize the water can be calculated using the formula: q = n · ∆Hvap Where: - n is the number of moles of water, - ∆Hvap is the molar heat of vaporization. First, calculate the number of moles of water: n= 50.0 g 18.015 g/mol Step 3: Convert the heat of vaporization to joules. 40.79 kJ/mol = 40.79 × 103 J/mol 7 Now, calculate the heat required to vaporize the water:   50.0 q= × (40.79 × 103 ) J 18.015 Step 4: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the heat required to raise the temperature and the heat required to vaporize the water: ∆H = qheating + qvaporization Question 11 Question Given the reaction: 2A(g) + 3B(g) → C(g) + 4D(g) ◦ where ∆H = −1258 kJ, calculate the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B. Solution Step 1: Calculate the moles of A and B reacting. Moles of A = 1.5 mol Moles of B = 2.0 mol Step 2: Determine the limiting reactant. Moles of A required for reaction = 22 = 1.0 mol Moles of B required for reaction = 23 = 1.5 mol Since A has fewer moles than required, A is the limiting reactant. Step 3: Calculate the moles of C and D formed. Since 2 moles of A produces 1 mole of C, Moles of C formed = 12 × 1.5 = 0.75 mol Since 2 moles of A produces 4 moles of D, Moles of D formed = 2 × 1.5 = 3.0 mol Step 4: Calculate the enthalpy change for the reaction. ∆H ◦ = −1258 kJ For the reaction of 1 mole of A: = −629 kJ Enthalpy change = −1258 2 For the reaction of 1.5 moles of A: Enthalpy change = −629 × 1.5 = −943.5 kJ Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B is −943.5 kJ . 8 Question 12 Question A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5°C in a closed container water at 25°C. Assuming the heat capacity of the container is negligible and that the density of water is 1 g/cm3 , calculate the enthalpy change for the reaction. Solution Step 1: Calculate the mass of water affected by the heat exchange. Given that the density of water is 1 g/cm3 , the mass of water affected by the heat exchange is: Volume of water = Density × Volume Mass Volume = Density 1g 3 Volume = 3 = 1 cm 1 g/cm The volume of water affected by the heat exchange is 1 cm3 , which is equivalent to 1 gram. Step 2: Calculate the change in temperature of the water. Given that the temperature change is 0.5°C, the heat exchanged is: q = mc∆T q = (1 g)(4.18 J/g°C)(0.5C) = 2.09 J Step 3: Convert the heat exchanged to kilojoules. To convert Joules to kilojoules, divide by 1000: q= 2.09 J = 0.00209 kJ 1000 Step 4: Determine the enthalpy change for the reaction. The enthalpy change for the reaction is equal in magnitude but opposite in sign to the heat exchanged: ∆H = −q = −0.00209 kJ = -2.09 kJ Therefore, the enthalpy change for the reaction is -2.09 kJ. Question 13 Question For a reaction at constant pressure, the enthalpy change (∆H) is given by the equation ∆H = 50 − 2T , where T is the temperature in Kelvin. Determine the temperature at which the reaction becomes exothermic. 9 Solution Step 1: To find the temperature at which the reaction becomes exothermic, set ∆H to zero and solve for T . ∆H = 50 − 2T = 0 50 = 2T T = 25 K Step 2: To determine whether the reaction is exothermic or endothermic at T = 25 K, substitute T = 25 into the equation ∆H = 50 − 2T . ∆H = 50 − 2(25) = 50 − 50 = 0 Since ∆H = 0 at T = 25 K, the reaction becomes exothermic at this temperature. Question 14 Question A gas initially at 300 K and 1 atm pressure is expanded isothermally and reversibly to three times its initial volume. Calculate the change in enthalpy for this process, given that the molar heat capacity at constant pressure, Cp , is 30 J/mol K. Solution Step 1: Calculate the work done during the expansion. Given that the process is isothermal and reversible, the work done is −nRT ln(Vf /Vi ), where n is the number of moles of gas, R is the ideal gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Step 2: Calculate the heat transfer. Since the process is isothermal, the heat transfer is equal to the work done. Therefore, q = −w. Step 3: Calculate the change in enthalpy. The change in enthalpy is given by ∆H = ∆U + P ∆V , where ∆U is the change in internal energy, P is the pressure, and ∆V is the change in volume. For an isothermal process, ∆U = 0, so ∆H = P ∆V . Step 4: Substitute values and calculate. Given: - Initial temperature, Ti = 300 K - Initial pressure, Pi = 1 atm - Final volume, Vf = 3Vi - Molar heat capacity at constant pressure, Cp = 30 J/mol K - Ideal gas constant, R = 8.314 J/mol K From the ideal gas law, P V = nRT , we can find n and R. Then, we can calculate the change in enthalpy using the formula ∆H = P ∆V . Finally, substitute the known values to find the change in enthalpy. 10 Question 15 Question Given the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) with the following enthalpy changes: ∆Hf◦ (H2 O(l)) = −285.8 kJ/mol ∆Hf◦ (O2 (g)) = 0 kJ/mol ∆Hf◦ (H2 (g)) = 0 kJ/mol calculate the standard enthalpy change for the reaction. Solution Step 1: Calculate the standard enthalpy change for the reaction using the standard enthalpy of formation values provided. X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants)

∆H ◦ = 2 · ∆Hf◦ (H2 O(l)) − 2 · ∆Hf◦ (H2 (g)) + ∆Hf◦ (O2 (g)) ∆H ◦ = 2(−285.8 kJ/mol) − 2(0 kJ/mol) − 0 kJ/mol ∆H ◦ = −571.6 kJ/mol Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol . Question 16 Question Given the following reaction: 2A(g) + 3B(g) −→ C(g) + D(g) If the standard enthalpy of formation for each substance is as follows: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol Calculate the standard enthalpy change, ∆H ◦ , for the reaction. 11 Solution To calculate the standard enthalpy change for the reaction, we can use Hess’s Law, which states that the total enthalpy change for a reaction is the same regardless of the number of steps in the reaction or the pathway taken. Step 1: Write the given reaction in terms of standard enthalpies of formation. The given reaction is: 2A(g) + 3B(g) −→ C(g) + D(g) The standard enthalpy change for this reaction can be expressed as: X X ∆H ◦ = ∆Hf◦ (products) − ∆Hf◦ (reactants) Step 2: Calculate the standard enthalpy change for the reaction. Given that: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol The standard enthalpy change can be calculated as follows: ∆H ◦ = [(∆Hf◦ (C) + ∆Hf◦ (D)) − (∆Hf◦ (2A) + ∆Hf◦ (3B))] ∆H ◦ = [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))] ∆H ◦ = [−900 kJ/mol − (−200 kJ/mol + −900 kJ/mol)] ∆H ◦ = −900 kJ/mol + 1100 kJ/mol ∆H ◦ = 200 kJ/mol Question 17 Question A reaction has an enthalpy change of ∆H = −285.8 kJ. If 4.96 moles of the reactant undergo the reaction, what is the total heat absorbed or released during the reaction? 12 Solution Step 1: Determine the total heat absorbed or released per mole of the reactant. Given: ∆H = −285.8 kJ We know that the total heat absorbed or released during the reaction can be calculated as: Total heat = ∆H × moles of reactant Step 2: Calculate the total heat absorbed or released. Plugging in the values: Total heat = −285.8 kJ × 4.96 moles Total heat = −1416.688 kJ Therefore, the total heat absorbed or released during the reaction is -1416.688 kJ. Question 18 Question Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦ C is converted to steam at 100◦ C. The specific heat capacity of water is 4.18 J/(g·°C), the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to raise the temperature of the water from 25◦ C to 100◦ C using the formula: q1 = m × c × ∆T where - m is the mass of water (35.0 g), - c is the specific heat capacity of water (4.18 J/(g·°C)), - ∆T is the change in temperature (100◦ C - 25◦ C). Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing by 1000. Step 3: Calculate the moles of water in 35.0 g using the molar mass of water. Step 4: Calculate the heat required to convert the water to steam at 100◦ C using the formula: q2 = n × ∆Hvap where - n is the number of moles of water calculated in Step 3, - ∆Hvap is the molar enthalpy of vaporization of water (40.79 kJ/mol). Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat change, ∆H. 13 ∆H = q1 + q2 Question 19 Question Given the balanced chemical equation for the combustion of ethane: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) Calculate the standard enthalpy change (∆H ◦ ) for this reaction. Given the following standard enthalpy of formation values: ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol ∆Hf◦ (CO2 ) = −393.51 kJ/mol ∆Hf◦ (H2 O) = −285.83 kJ/mol Solution Step 1: Calculate the standard enthalpy of formation for the products using the given values. ∆Hf◦ (2CO2 ) = 2(−393.51 kJ/mol) = −787.02 kJ/mol ∆Hf◦ (3H2 O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol Step 2: Calculate the standard enthalpy of formation for the reactants using the given value. ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol Step 3: Calculate the net standard enthalpy change for the reaction using the formula: X X ∆H ◦ = ( ∆Hf◦ products) − ( ∆Hf◦ reactants) ∆H ◦ = [(−787.02 kJ/mol) + (−857.49 kJ/mol)] − (−84.68 kJ/mol) ∆H ◦ = −1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol Therefore, the standard enthalpy change (∆H ◦ ) for the combustion of ethane is -1559.83 kJ/mol. 14 Question 20 Question Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2 ) react according to the following balanced chemical equation: N2 (g) + 3H2 (g) → 2N H3 (g) Given the enthalpies of formation (∆Hf◦ ) for N2 (g), H2 (g), and N H3 (g) are 0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively. Solution Step 1: Write the given balanced chemical equation and the enthalpy change for the reaction. N2 (g) + 3H2 (g) → 2N H3 (g) The enthalpy change for the reaction is given by: ∆H = Σ∆Hf◦ (products) − Σ∆Hf◦ (reactants) Step 2: Calculate the enthalpy change for the reaction. The enthalpies of formation are: ∆Hf◦ (N H3 ) = −46.2 kJ/mol ∆Hf◦ (N2 ) = 0 kJ/mol ∆Hf◦ (H2 ) = 0 kJ/mol Substitute the values into the equation for ∆H:

∆H = 2 × ∆Hf◦ (N H3 ) − ∆Hf◦ (N2 ) + 3 × ∆Hf◦ (H2 ) ∆H = (2 × −46.2 kJ/mol) − (0 + 3 × 0 kJ/mol) ∆H = −92.4 kJ/mol Step 3: Calculate the change in enthalpy when 5.00 moles of N2 react. Since the enthalpy change is given per mole, we need to calculate the total enthalpy change for 5 moles of N2 : Change in enthalpy = moles × ∆H Change in enthalpy = 5.00 moles × −92.4 kJ/mol Change in enthalpy = −462 kJ Therefore, the change in enthalpy when 5.00 moles of N2 react is -462 kJ. 15 Question 21 Question A reaction is thought to have a certain enthalpy change. By performing a series of measurements, it is found that the reaction has an enthalpy change of −289.2 kJ when carried out at 25°C. Assuming that the heat capacities of all substances are constant over the temperature range in question, calculate the enthalpy change for the reaction at 55°C. Solution Step 1: Determine the heat capacity change (∆Cp ) of the reaction. Z 55C 25C ∆Hreaction = ∆Hreaction + ∆Cp dT 25C Z 55C ∆Hreaction = −289.2 kJ + ∆Cp dT 25C Step 2: Since the heat capacities are constant, we can consider ∆Cp as a constant value, say Cp . Z 55C ∆Hreaction = −289.2 kJ + Cp dT 25C ∆Hreaction = −289.2 kJ + Cp · (55 − 25) K ∆Hreaction = −289.2 kJ + 30Cp Step 3: Given that the enthalpy change of the reaction at 25°C is -289.2 kJ: −289.2 = −289.2 + 30Cp 0 = 30Cp =⇒ Cp = 0 Step 4: The assumption that the heat capacities are constant is incorrect. We must therefore consider the correct dependence of heat capacity on temperature for the substances involved. Question 22 Question A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of 298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for the reaction. Given: Gas constant, R = 8.314 J/(mol·K) 16 Solution Step 1: Calculate the work done by the gas during the reaction. The work done by the gas is given by the equation: W = −Pext ∆V Where: Pext = external pressure = 1.00 atm = 1.013 × 105 Pa ∆V = change in volume = 10.0 L Convert 10.0 L to m³: 10.0L = 10.0 × 10−3 m3 = 0.0100m3 Now, calculate the work done: W = −(1.013 × 105 P a)(0.0100m3 ) = −1013J Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed by the calorimeter is given by the equation: qcal = C∆T Where: C = heat capacity of the calorimeter = 750 J/K ∆T = change in temperature = 316 K - 298 K = 18 K Now, calculate the heat absorbed by the calorimeter: qcal = (750J/K)(18K) = 13500J Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction is given by the equation: qrxn = qcal − W Now, calculate the heat of the reaction: qrxn = 13500J − (−1013J) = 14513J Step 4: Calculate the change in enthalpy. Since ∆H = qrxn under constant pressure conditions, the change in enthalpy for the reaction is ∆H = 14513 J. Therefore, the change in enthalpy for the reaction is 14513 J. Question 23 Question A reaction is known to have an enthalpy change of ∆H = −250 kJ/mol. If the reaction is exothermic, in which direction will the enthalpy change (∆H) be when the reaction is reversed? 17 Solution 1. When a reaction is reversed, the sign of the enthalpy change is also reversed. Therefore, if the original reaction has ∆H = −250 kJ/mol, the reversed reaction will have ∆H = +250 kJ/mol. 2. This change in sign occurs because when a reaction is reversed, the products become the reactants and vice versa. As a result, the direction of heat flow (exothermic or endothermic) also changes. Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol. Question 24 Question An ideal gas undergoes a process at constant pressure. The initial temperature of the gas is 300 K and the final temperature is 400 K. Calculate the change in enthalpy of the gas during the process. Solution Step 1: Recall the definition of enthalpy in terms of internal energy and pressure: ∆H = ∆U + P ∆V Step 2: Since the process is at constant pressure, ∆U = Q − W where Q is the heat added to the system and W is the work done by the gas on its surroundings. Step 3: We can rewrite the expression for enthalpy change as: ∆H = (Q − W ) + P ∆V Step 4: The work done by the gas can be expressed as W = P ∆V . Substituting this into the equation gives: ∆H = Q Step 5: To calculate the heat added to the system, we can use the formula: Q = nCp ∆T where n is the number of moles of gas and Cp is the molar heat capacity at constant pressure. Step 6: First, we need to find the number of moles of the gas. We can use the ideal gas law: P V = nRT Solving for n, we have: n= PV RT 18 Step 7: Substituting n into the formula for heat gives:   PV Q= Cp ∆T RT Step 8: Substituting the given values P = constant, V = constant, R = constant, T1 = 300 K, T2 = 400 K and Cp = specific heat at constant pressure into the equation gives the change in enthalpy. Question 25 Question A reaction between hydrogen gas and oxygen gas to form water vapor has an enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much heat is absorbed or released? Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(g) Step 2: Determine the moles of oxygen gas reacting, based on the stoichiometry of the balanced equation. Since the balanced equation states that 1 mole of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas required will be half of the moles of hydrogen gas: 1 × 8.00 moles = 4.00 moles 2 Step 3: Calculate the heat absorbed or released using the enthalpy change and the moles of the limiting reactant. Given enthalpy change: ∆H = −483.6 kJ Since the equation is for 2 moles of hydrogen gas reacting, the heat change for 1 mole of hydrogen gas will be half of the given value: moles of O2 = 483.6 kJ = −241.8 kJ/mole 2 Therefore, the heat change for 8.00 moles of hydrogen gas will be: Heat change for 1 mole of H2 = − Heat change = −241.8 kJ/mole × 8.00 moles = −1934.4 kJ So, 1934.4 kJ of heat is released in this reaction. Question 26 Question A reaction has an enthalpy change of ∆H = −185 kJ/mol. If 2.50 moles of the reaction takes place, calculate the total enthalpy change for the reaction. 19 Solution Step 1: Determine the total enthalpy change for the reaction per mole. Given: ∆H = −185 kJ/mol Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles of the reaction takes place. Total Enthalpy Change = ∆H × Number of moles Total Enthalpy Change = −185 kJ/mol × 2.50 mol Total Enthalpy Change = −185 kJ/mol × 2.50 = −462.5 kJ Therefore, the total enthalpy change for the reaction when 2.50 moles of the reaction takes place is −462.5 kJ. Question 27 Question Calculate the change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure. Given that the heat of vaporization of water is 40.79 kJ/mol. Solution Step 1: Determine the change in enthalpy during the phase change from vapor to liquid. For 2 moles of water vapor being condensed to liquid water, the change in enthalpy can be calculated using the heat of vaporization formula: Change in Enthalpy = moles × heat of vaporization Change in Enthalpy = 2 moles × 40.79 kJ/mol Change in Enthalpy = 81.58 kJ Therefore, the change in enthalpy during the phase change is 81.58 kJ. Step 2: Account for the temperature change. Since the temperature remains constant during the phase change, there is no additional change in enthalpy due to the temperature change. Step 3: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the change in enthalpy during the phase change and any changes due to temperature: Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change) Total Change in Enthalpy = 81.58 kJ + 0 kJ Total Change in Enthalpy = 81.58 kJ 20 Therefore, the total change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure is 81.58 kJ. Question 28 Question A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at constant temperature to a final volume of 25 L. If the enthalpy change during the process is 1500 J, calculate the final pressure of the gas. Solution Step 1: Calculate the initial number of moles of gas using the ideal gas law, P V = nRT . PV n= RT Given P = 10 atm, V = 5 L, R = 0.0821 L atm mol−1 K−1 , and T is constant for the process. Substitute the given values to find n. n= 50 (10 atm)(5 L) 610 = = 0.0821T T (0.0821 L atm mol−1 K−1 )(T ) Step 2: Calculate the final pressure of the gas using the enthalpy change and the number of moles. ∆H = nCp ∆T Given ∆H = 1500 J and n = 610/T , where T is the gas temperature for the process, and Cp is the molar heat capacity at constant pressure. Since the process is at constant temperature, ∆T = 0 and nCp ∆T = 0. Thus, the enthalpy change during the process does not depend on the molar heat capacity at constant pressure. So, ∆H = nCp ∆T = 0. Since ∆H = nCp ∆T = 0, nCp ∆T = 1500 J implies nCp = 1500 J. Substitute n = 610/T into nCp = 1500 J and simplify. 610Cp = 1500 T Step 3: Calculate the final pressure. P V = nRT At the final state, Pfinal Vfinal = nRT . Given Vfinal = 25 L, n = 610/T , and R = 0.0821 L atm mol−1 K−1 . 21 Substitute the values to find the final pressure. Pfinal · 25 = 610 · 0.0821 · T T 25Pfinal = 50.31Pfinal = 50.31 25 Therefore, the final pressure of the gas is 2.012 atm. Question 29 Question A certain reaction has an enthalpy change of ∆H = −128 kJ/mol at 25°C. If the reaction is exothermic, calculate the standard enthalpy change at 298 K. Solution Step 1: Calculate the change in enthalpy due to the change in temperature from 25°C to 298 K. Z 298 ∆H298 = ∆H25 + Cp dT 25 where Cp is the molar heat capacity at constant pressure. Step 2: Use the heat capacity values and the relationship Cp = calculate the integral.  Z 298 Z 298  dH dT Cp dT = dT p 25 25 Z 298 = 25  128 298 − 25 dH dT p
to  dT Step 3: Perform the integration and calculate the new enthalpy change at 298 K.   298 = 128 × ln kJ/mol 25 ∆H298 = −126.39 kJ/mol Therefore, the standard enthalpy change at 298 K is ∆H298 = −126.39 kJ/mol. 22 Question 30 Question Calculate the change in enthalpy for the reaction 2H2 (g) + O2 (g) → 2H2 O(l) given the following bond energies: H − H = 436 kJ/mol, O = O = 498 kJ/mol, H − O = 463 kJ/mol. Solution Step 1: Calculate the total energy required to break the bonds in the reactants. Step 2: Calculate the total energy released when the new bonds are formed in the products. Step 3: Find the difference between the energy required to break the bonds in the reactants and the energy released when the new bonds are formed in the products. Step 1: The total energy required to break the bonds in the reactants is: 2 × (H − H) + 1 × (O = O) = 2 × 436 + 1 × 498 = 1370 kJ/mol Step 2: The total energy released when the new bonds are formed in the products is: 2 × (H − O) = 2 × 463 = 926 kJ/mol Step 3: The change in enthalpy for the reaction is given by: ∆H = (Energy required to break the bonds)−(Energy released when new bonds are formed) ∆H = 1370 − 926 = 444 kJ/mol Therefore, the change in enthalpy for the reaction is 444 kJ/mol. Question 31 Question Given the following reaction involving the combustion of ethane gas: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) If the standard enthalpies of formation are −84.7 kJ/mol for CO2 and −285.8 kJ/mol for H2 O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the standard enthalpy of formation for ethane. 23 Solution Step 1: Calculate the standard enthalpy change of the reaction using the given standard enthalpies of formation. X X ◦ ◦ ◦ ∆Hrxn = ν∆Hf,products − ν∆Hf,reactants Given data: ∆Hf◦ (CO2 ) = −84.7 kJ/mol ∆Hf◦ (H2 O) = −285.8 kJ/mol ◦ ∆Hrxn = −1558.0 kJ/mol ◦ ∆Hrxn = 2(−84.7) + 3(−285.8) − (−1558.0) ◦ ∆Hrxn = −169.4 kJ/mol Step 2: Write the standard enthalpy change of the reaction using the enthalpy of formation of ethane. C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) ◦ ∆Hrxn = 2∆Hf◦ (CO2 ) + 3∆Hf◦ (H2 O) − ∆Hf◦ (C2 H6 ) Step 3: Solve for the standard enthalpy of formation of ethane. −169.4 = 2(−84.7) + 3(−285.8) − ∆Hf◦ (C2 H6 ) −169.4 = −169.4 − ∆Hf◦ (C2 H6 ) ∆Hf◦ (C2 H6 ) = 0 kJ/mol Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol. Question 32 Question A sample of an ideal gas undergoes a process where the pressure changes from 2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and final temperatures are 300 K and 500 K respectively, calculate the change in enthalpy of the gas during this process. 24 Solution Step 1: Calculate the initial and final internal energies of the gas using the first law of thermodynamics: ∆U = Q − W where ∆U is the change in internal energy, Q is the heat added to the gas, and W is the work done by the gas. Given that the gas undergoes an isobaric process, the work done can be calculated as: W = −Pext · ∆V Since the gas is ideal, we can use the ideal gas law to express Pext :   Vf W = −Pext · ∆V = −nRT ln Vi Substitute the given values:     2 1 W = −nRT ln = −nRT ln = −nRT ln(0.25) 8 4 ∆U = Q − W = Q + nRT ln(0.25) ∆U = nCV ∆T where CV is the molar heat capacity at constant volume. Since the gas is ideal, CV = f2 R where f is the degrees of freedom (f=3 for monoatomic gas). Substitute and rearrange to solve for Q: f Q = ∆U − nR∆T ln(0.25) = n( R)∆T − nR∆T ln(0.25) 2 3 f − ln(0.25)) = nR∆T ( − ln(0.25)) 2 2 Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation: Q = nR∆T ( ∆H = ∆U + ∆(P V ) Given that the process is isobaric, substitute the change in internal energy and pressure-volume work: ∆H = Q + W + P ∆V Substitute the expressions for Q and W calculated earlier: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T ln(0.25) + Pext ∆V 2 25 Given that P ∆V = nR∆T for an isobaric process: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T + nR∆T 2 5 ∆H = nR∆T ( − ln(0.25)) 2 Now, substitute the given values for n, R, and ∆T to find the change in enthalpy. Question 33 Question Calculate the standard enthalpy change for the reaction below at 298 K: 2CH4 (g) + 4O2 (g) → 2CO2 (g) + 4H2 O(l) Given the standard enthalpies of formation (∆Hf◦ ) for CH4 (g), CO2 (g), and H2 O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. Solution Step 1: Calculate the standard enthalpy change (∆H ◦ ) for the reaction using the standard enthalpies of formation: X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants) where n and m are the stoichiometric coefficients of the products and reactants, respectively. Step 2: Substitute the given values and calculate: ∆H ◦ = 2×∆Hf◦ (CO2 (g))+4×∆Hf◦ (H2 O(l))−[2×∆Hf◦ (CH4 (g))+4×∆Hf◦ (O2 (g))] Step 3: Plug in the values and calculate: ∆H ◦ = 2 × (−393.5) + 4 × (−285.8) − [2 × (−74.8) + 4 × 0] ∆H ◦ = −787.0 + (−1143.2) − (−149.6) ∆H ◦ = −787.0 − 1143.2 + 149.6 ∆H ◦ = −1780.6 kJ/mol Therefore, the standard enthalpy change for the given reaction at 298 K is -1780.6 kJ/mol. 26 Question 34 Question Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C. Given that the heat capacity of steam is 2.08 J/g·K, the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water is 40.79 kJ/mol. Solution Step 1: Calculate the heat required to cool the steam to 100◦ C to water at 100◦ C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol + 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K * 36 g/mol = 74.88 J/mol·K. The change in temperature is 100◦ C - 25◦ C = 75◦ C = 75 K. Therefore, the heat required to cool 10.0 moles of steam to water at 100◦ C is: 10.0 mol × 75 K × 74.88 J/mol · K = 56160 J. Step 2: Calculate the heat required to condense the water at 100◦ C to water at 25◦ C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol. Therefore, the heat required to condense 10.0 moles of steam at 100◦ C to water at 100◦ C is: 10.0 mol × 40790 J/mol = 407900 J. Step 3: Add the heats from Step 1 and Step 2 to find the total change in enthalpy. ∆H = 56160 J + 407900 J = 464060 J. Therefore, the change in enthalpy when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C is 464060 J. Question 35 Question An experiment involves the mixing of 150 g of water at 25°C with 300 g of water at 85°C in a thermally insulated container. Calculate the final temperature of the mixture. Assume specific heat capacity of water is 4.18 J/g°C and neglect any heat losses to the surroundings. Solution Step 1: Calculate the heat lost by the hot water and the heat gained by the cold water. The heat lost by the hot water is given by: qhot = mhot · c · ∆T 27 Question 8 Question A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50 moles of the reactant are consumed, what is the change in enthalpy per mole of reactant? Solution Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be calculated using the formula: ∆Hper mole = ∆H moles of reactant Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles of reactant is 2.50 mol, we can substitute these values into the formula: ∆Hper mole = −349 kJ 2.50 mol Step 3: Calculate the change in enthalpy per mole of reactant: ∆Hper mole = −349 = −139.6 kJ/mol 2.50 Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol. Question 9 Question Calculate the enthalpy change (∆H) for the reaction: 2H2 (g) + O2 (g) → H2 O(l) given the following information: ˆ ∆Hf◦ for H2 (g) = 0 kJ/mol ˆ ∆Hf◦ for O2 (g) = 0 kJ/mol ˆ ∆Hf◦ for H2 O(l) = -285.8 kJ/mol Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) Step 2: Calculate ∆H for the reaction using the standard enthalpies of formation: X X ∆H = ∆Hf◦ (products) − ∆Hf◦ (reactants) 6 Step 3: Substitute the given values into the equation: ∆H = [2(−285.8 kJ/mol)] − [2(0 kJ/mol) + 0 kJ/mol] Step 4: Solve for ∆H: ∆H = −571.6 kJ/mol Therefore, the enthalpy change for the reaction is ∆H = −571.6 kJ/mol. Question 10 Question Calculate the change in enthalpy (∆H) when 50.0 g of water at 25°C is heated to steam at 100°C. The specific heat capacity of water is 4.18 J/g°C, the molar heat of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to heat the water from 25°C to 100°C. The formula to calculate the heat required is: q = mc∆T Where: - q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Substitute the values: q = (50.0 g)(4.18 J/g°C)(100 − 25)C q = (50.0)(4.18)(75) J Step 2: Calculate the heat required to vaporize the water. The heat required to vaporize the water can be calculated using the formula: q = n · ∆Hvap Where: - n is the number of moles of water, - ∆Hvap is the molar heat of vaporization. First, calculate the number of moles of water: n= 50.0 g 18.015 g/mol Step 3: Convert the heat of vaporization to joules. 40.79 kJ/mol = 40.79 × 103 J/mol 7 Now, calculate the heat required to vaporize the water:   50.0 q= × (40.79 × 103 ) J 18.015 Step 4: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the heat required to raise the temperature and the heat required to vaporize the water: ∆H = qheating + qvaporization Question 11 Question Given the reaction: 2A(g) + 3B(g) → C(g) + 4D(g) ◦ where ∆H = −1258 kJ, calculate the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B. Solution Step 1: Calculate the moles of A and B reacting. Moles of A = 1.5 mol Moles of B = 2.0 mol Step 2: Determine the limiting reactant. Moles of A required for reaction = 22 = 1.0 mol Moles of B required for reaction = 23 = 1.5 mol Since A has fewer moles than required, A is the limiting reactant. Step 3: Calculate the moles of C and D formed. Since 2 moles of A produces 1 mole of C, Moles of C formed = 12 × 1.5 = 0.75 mol Since 2 moles of A produces 4 moles of D, Moles of D formed = 2 × 1.5 = 3.0 mol Step 4: Calculate the enthalpy change for the reaction. ∆H ◦ = −1258 kJ For the reaction of 1 mole of A: = −629 kJ Enthalpy change = −1258 2 For the reaction of 1.5 moles of A: Enthalpy change = −629 × 1.5 = −943.5 kJ Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B is −943.5 kJ . 8 Question 12 Question A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5°C in a closed container water at 25°C. Assuming the heat capacity of the container is negligible and that the density of water is 1 g/cm3 , calculate the enthalpy change for the reaction. Solution Step 1: Calculate the mass of water affected by the heat exchange. Given that the density of water is 1 g/cm3 , the mass of water affected by the heat exchange is: Volume of water = Density × Volume Mass Volume = Density 1g 3 Volume = 3 = 1 cm 1 g/cm The volume of water affected by the heat exchange is 1 cm3 , which is equivalent to 1 gram. Step 2: Calculate the change in temperature of the water. Given that the temperature change is 0.5°C, the heat exchanged is: q = mc∆T q = (1 g)(4.18 J/g°C)(0.5C) = 2.09 J Step 3: Convert the heat exchanged to kilojoules. To convert Joules to kilojoules, divide by 1000: q= 2.09 J = 0.00209 kJ 1000 Step 4: Determine the enthalpy change for the reaction. The enthalpy change for the reaction is equal in magnitude but opposite in sign to the heat exchanged: ∆H = −q = −0.00209 kJ = -2.09 kJ Therefore, the enthalpy change for the reaction is -2.09 kJ. Question 13 Question For a reaction at constant pressure, the enthalpy change (∆H) is given by the equation ∆H = 50 − 2T , where T is the temperature in Kelvin. Determine the temperature at which the reaction becomes exothermic. 9 Solution Step 1: To find the temperature at which the reaction becomes exothermic, set ∆H to zero and solve for T . ∆H = 50 − 2T = 0 50 = 2T T = 25 K Step 2: To determine whether the reaction is exothermic or endothermic at T = 25 K, substitute T = 25 into the equation ∆H = 50 − 2T . ∆H = 50 − 2(25) = 50 − 50 = 0 Since ∆H = 0 at T = 25 K, the reaction becomes exothermic at this temperature. Question 14 Question A gas initially at 300 K and 1 atm pressure is expanded isothermally and reversibly to three times its initial volume. Calculate the change in enthalpy for this process, given that the molar heat capacity at constant pressure, Cp , is 30 J/mol K. Solution Step 1: Calculate the work done during the expansion. Given that the process is isothermal and reversible, the work done is −nRT ln(Vf /Vi ), where n is the number of moles of gas, R is the ideal gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Step 2: Calculate the heat transfer. Since the process is isothermal, the heat transfer is equal to the work done. Therefore, q = −w. Step 3: Calculate the change in enthalpy. The change in enthalpy is given by ∆H = ∆U + P ∆V , where ∆U is the change in internal energy, P is the pressure, and ∆V is the change in volume. For an isothermal process, ∆U = 0, so ∆H = P ∆V . Step 4: Substitute values and calculate. Given: - Initial temperature, Ti = 300 K - Initial pressure, Pi = 1 atm - Final volume, Vf = 3Vi - Molar heat capacity at constant pressure, Cp = 30 J/mol K - Ideal gas constant, R = 8.314 J/mol K From the ideal gas law, P V = nRT , we can find n and R. Then, we can calculate the change in enthalpy using the formula ∆H = P ∆V . Finally, substitute the known values to find the change in enthalpy. 10 Question 15 Question Given the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) with the following enthalpy changes: ∆Hf◦ (H2 O(l)) = −285.8 kJ/mol ∆Hf◦ (O2 (g)) = 0 kJ/mol ∆Hf◦ (H2 (g)) = 0 kJ/mol calculate the standard enthalpy change for the reaction. Solution Step 1: Calculate the standard enthalpy change for the reaction using the standard enthalpy of formation values provided. X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants)

∆H ◦ = 2 · ∆Hf◦ (H2 O(l)) − 2 · ∆Hf◦ (H2 (g)) + ∆Hf◦ (O2 (g)) ∆H ◦ = 2(−285.8 kJ/mol) − 2(0 kJ/mol) − 0 kJ/mol ∆H ◦ = −571.6 kJ/mol Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol . Question 16 Question Given the following reaction: 2A(g) + 3B(g) −→ C(g) + D(g) If the standard enthalpy of formation for each substance is as follows: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol Calculate the standard enthalpy change, ∆H ◦ , for the reaction. 11 Solution To calculate the standard enthalpy change for the reaction, we can use Hess’s Law, which states that the total enthalpy change for a reaction is the same regardless of the number of steps in the reaction or the pathway taken. Step 1: Write the given reaction in terms of standard enthalpies of formation. The given reaction is: 2A(g) + 3B(g) −→ C(g) + D(g) The standard enthalpy change for this reaction can be expressed as: X X ∆H ◦ = ∆Hf◦ (products) − ∆Hf◦ (reactants) Step 2: Calculate the standard enthalpy change for the reaction. Given that: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol The standard enthalpy change can be calculated as follows: ∆H ◦ = [(∆Hf◦ (C) + ∆Hf◦ (D)) − (∆Hf◦ (2A) + ∆Hf◦ (3B))] ∆H ◦ = [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))] ∆H ◦ = [−900 kJ/mol − (−200 kJ/mol + −900 kJ/mol)] ∆H ◦ = −900 kJ/mol + 1100 kJ/mol ∆H ◦ = 200 kJ/mol Question 17 Question A reaction has an enthalpy change of ∆H = −285.8 kJ. If 4.96 moles of the reactant undergo the reaction, what is the total heat absorbed or released during the reaction? 12 Solution Step 1: Determine the total heat absorbed or released per mole of the reactant. Given: ∆H = −285.8 kJ We know that the total heat absorbed or released during the reaction can be calculated as: Total heat = ∆H × moles of reactant Step 2: Calculate the total heat absorbed or released. Plugging in the values: Total heat = −285.8 kJ × 4.96 moles Total heat = −1416.688 kJ Therefore, the total heat absorbed or released during the reaction is -1416.688 kJ. Question 18 Question Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦ C is converted to steam at 100◦ C. The specific heat capacity of water is 4.18 J/(g·°C), the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to raise the temperature of the water from 25◦ C to 100◦ C using the formula: q1 = m × c × ∆T where - m is the mass of water (35.0 g), - c is the specific heat capacity of water (4.18 J/(g·°C)), - ∆T is the change in temperature (100◦ C - 25◦ C). Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing by 1000. Step 3: Calculate the moles of water in 35.0 g using the molar mass of water. Step 4: Calculate the heat required to convert the water to steam at 100◦ C using the formula: q2 = n × ∆Hvap where - n is the number of moles of water calculated in Step 3, - ∆Hvap is the molar enthalpy of vaporization of water (40.79 kJ/mol). Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat change, ∆H. 13 ∆H = q1 + q2 Question 19 Question Given the balanced chemical equation for the combustion of ethane: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) Calculate the standard enthalpy change (∆H ◦ ) for this reaction. Given the following standard enthalpy of formation values: ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol ∆Hf◦ (CO2 ) = −393.51 kJ/mol ∆Hf◦ (H2 O) = −285.83 kJ/mol Solution Step 1: Calculate the standard enthalpy of formation for the products using the given values. ∆Hf◦ (2CO2 ) = 2(−393.51 kJ/mol) = −787.02 kJ/mol ∆Hf◦ (3H2 O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol Step 2: Calculate the standard enthalpy of formation for the reactants using the given value. ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol Step 3: Calculate the net standard enthalpy change for the reaction using the formula: X X ∆H ◦ = ( ∆Hf◦ products) − ( ∆Hf◦ reactants) ∆H ◦ = [(−787.02 kJ/mol) + (−857.49 kJ/mol)] − (−84.68 kJ/mol) ∆H ◦ = −1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol Therefore, the standard enthalpy change (∆H ◦ ) for the combustion of ethane is -1559.83 kJ/mol. 14 Question 20 Question Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2 ) react according to the following balanced chemical equation: N2 (g) + 3H2 (g) → 2N H3 (g) Given the enthalpies of formation (∆Hf◦ ) for N2 (g), H2 (g), and N H3 (g) are 0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively. Solution Step 1: Write the given balanced chemical equation and the enthalpy change for the reaction. N2 (g) + 3H2 (g) → 2N H3 (g) The enthalpy change for the reaction is given by: ∆H = Σ∆Hf◦ (products) − Σ∆Hf◦ (reactants) Step 2: Calculate the enthalpy change for the reaction. The enthalpies of formation are: ∆Hf◦ (N H3 ) = −46.2 kJ/mol ∆Hf◦ (N2 ) = 0 kJ/mol ∆Hf◦ (H2 ) = 0 kJ/mol Substitute the values into the equation for ∆H:

∆H = 2 × ∆Hf◦ (N H3 ) − ∆Hf◦ (N2 ) + 3 × ∆Hf◦ (H2 ) ∆H = (2 × −46.2 kJ/mol) − (0 + 3 × 0 kJ/mol) ∆H = −92.4 kJ/mol Step 3: Calculate the change in enthalpy when 5.00 moles of N2 react. Since the enthalpy change is given per mole, we need to calculate the total enthalpy change for 5 moles of N2 : Change in enthalpy = moles × ∆H Change in enthalpy = 5.00 moles × −92.4 kJ/mol Change in enthalpy = −462 kJ Therefore, the change in enthalpy when 5.00 moles of N2 react is -462 kJ. 15 Question 21 Question A reaction is thought to have a certain enthalpy change. By performing a series of measurements, it is found that the reaction has an enthalpy change of −289.2 kJ when carried out at 25°C. Assuming that the heat capacities of all substances are constant over the temperature range in question, calculate the enthalpy change for the reaction at 55°C. Solution Step 1: Determine the heat capacity change (∆Cp ) of the reaction. Z 55C 25C ∆Hreaction = ∆Hreaction + ∆Cp dT 25C Z 55C ∆Hreaction = −289.2 kJ + ∆Cp dT 25C Step 2: Since the heat capacities are constant, we can consider ∆Cp as a constant value, say Cp . Z 55C ∆Hreaction = −289.2 kJ + Cp dT 25C ∆Hreaction = −289.2 kJ + Cp · (55 − 25) K ∆Hreaction = −289.2 kJ + 30Cp Step 3: Given that the enthalpy change of the reaction at 25°C is -289.2 kJ: −289.2 = −289.2 + 30Cp 0 = 30Cp =⇒ Cp = 0 Step 4: The assumption that the heat capacities are constant is incorrect. We must therefore consider the correct dependence of heat capacity on temperature for the substances involved. Question 22 Question A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of 298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for the reaction. Given: Gas constant, R = 8.314 J/(mol·K) 16 Solution Step 1: Calculate the work done by the gas during the reaction. The work done by the gas is given by the equation: W = −Pext ∆V Where: Pext = external pressure = 1.00 atm = 1.013 × 105 Pa ∆V = change in volume = 10.0 L Convert 10.0 L to m³: 10.0L = 10.0 × 10−3 m3 = 0.0100m3 Now, calculate the work done: W = −(1.013 × 105 P a)(0.0100m3 ) = −1013J Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed by the calorimeter is given by the equation: qcal = C∆T Where: C = heat capacity of the calorimeter = 750 J/K ∆T = change in temperature = 316 K - 298 K = 18 K Now, calculate the heat absorbed by the calorimeter: qcal = (750J/K)(18K) = 13500J Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction is given by the equation: qrxn = qcal − W Now, calculate the heat of the reaction: qrxn = 13500J − (−1013J) = 14513J Step 4: Calculate the change in enthalpy. Since ∆H = qrxn under constant pressure conditions, the change in enthalpy for the reaction is ∆H = 14513 J. Therefore, the change in enthalpy for the reaction is 14513 J. Question 23 Question A reaction is known to have an enthalpy change of ∆H = −250 kJ/mol. If the reaction is exothermic, in which direction will the enthalpy change (∆H) be when the reaction is reversed? 17 Solution 1. When a reaction is reversed, the sign of the enthalpy change is also reversed. Therefore, if the original reaction has ∆H = −250 kJ/mol, the reversed reaction will have ∆H = +250 kJ/mol. 2. This change in sign occurs because when a reaction is reversed, the products become the reactants and vice versa. As a result, the direction of heat flow (exothermic or endothermic) also changes. Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol. Question 24 Question An ideal gas undergoes a process at constant pressure. The initial temperature of the gas is 300 K and the final temperature is 400 K. Calculate the change in enthalpy of the gas during the process. Solution Step 1: Recall the definition of enthalpy in terms of internal energy and pressure: ∆H = ∆U + P ∆V Step 2: Since the process is at constant pressure, ∆U = Q − W where Q is the heat added to the system and W is the work done by the gas on its surroundings. Step 3: We can rewrite the expression for enthalpy change as: ∆H = (Q − W ) + P ∆V Step 4: The work done by the gas can be expressed as W = P ∆V . Substituting this into the equation gives: ∆H = Q Step 5: To calculate the heat added to the system, we can use the formula: Q = nCp ∆T where n is the number of moles of gas and Cp is the molar heat capacity at constant pressure. Step 6: First, we need to find the number of moles of the gas. We can use the ideal gas law: P V = nRT Solving for n, we have: n= PV RT 18 Step 7: Substituting n into the formula for heat gives:   PV Q= Cp ∆T RT Step 8: Substituting the given values P = constant, V = constant, R = constant, T1 = 300 K, T2 = 400 K and Cp = specific heat at constant pressure into the equation gives the change in enthalpy. Question 25 Question A reaction between hydrogen gas and oxygen gas to form water vapor has an enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much heat is absorbed or released? Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(g) Step 2: Determine the moles of oxygen gas reacting, based on the stoichiometry of the balanced equation. Since the balanced equation states that 1 mole of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas required will be half of the moles of hydrogen gas: 1 × 8.00 moles = 4.00 moles 2 Step 3: Calculate the heat absorbed or released using the enthalpy change and the moles of the limiting reactant. Given enthalpy change: ∆H = −483.6 kJ Since the equation is for 2 moles of hydrogen gas reacting, the heat change for 1 mole of hydrogen gas will be half of the given value: moles of O2 = 483.6 kJ = −241.8 kJ/mole 2 Therefore, the heat change for 8.00 moles of hydrogen gas will be: Heat change for 1 mole of H2 = − Heat change = −241.8 kJ/mole × 8.00 moles = −1934.4 kJ So, 1934.4 kJ of heat is released in this reaction. Question 26 Question A reaction has an enthalpy change of ∆H = −185 kJ/mol. If 2.50 moles of the reaction takes place, calculate the total enthalpy change for the reaction. 19 Solution Step 1: Determine the total enthalpy change for the reaction per mole. Given: ∆H = −185 kJ/mol Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles of the reaction takes place. Total Enthalpy Change = ∆H × Number of moles Total Enthalpy Change = −185 kJ/mol × 2.50 mol Total Enthalpy Change = −185 kJ/mol × 2.50 = −462.5 kJ Therefore, the total enthalpy change for the reaction when 2.50 moles of the reaction takes place is −462.5 kJ. Question 27 Question Calculate the change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure. Given that the heat of vaporization of water is 40.79 kJ/mol. Solution Step 1: Determine the change in enthalpy during the phase change from vapor to liquid. For 2 moles of water vapor being condensed to liquid water, the change in enthalpy can be calculated using the heat of vaporization formula: Change in Enthalpy = moles × heat of vaporization Change in Enthalpy = 2 moles × 40.79 kJ/mol Change in Enthalpy = 81.58 kJ Therefore, the change in enthalpy during the phase change is 81.58 kJ. Step 2: Account for the temperature change. Since the temperature remains constant during the phase change, there is no additional change in enthalpy due to the temperature change. Step 3: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the change in enthalpy during the phase change and any changes due to temperature: Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change) Total Change in Enthalpy = 81.58 kJ + 0 kJ Total Change in Enthalpy = 81.58 kJ 20 Therefore, the total change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure is 81.58 kJ. Question 28 Question A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at constant temperature to a final volume of 25 L. If the enthalpy change during the process is 1500 J, calculate the final pressure of the gas. Solution Step 1: Calculate the initial number of moles of gas using the ideal gas law, P V = nRT . PV n= RT Given P = 10 atm, V = 5 L, R = 0.0821 L atm mol−1 K−1 , and T is constant for the process. Substitute the given values to find n. n= 50 (10 atm)(5 L) 610 = = 0.0821T T (0.0821 L atm mol−1 K−1 )(T ) Step 2: Calculate the final pressure of the gas using the enthalpy change and the number of moles. ∆H = nCp ∆T Given ∆H = 1500 J and n = 610/T , where T is the gas temperature for the process, and Cp is the molar heat capacity at constant pressure. Since the process is at constant temperature, ∆T = 0 and nCp ∆T = 0. Thus, the enthalpy change during the process does not depend on the molar heat capacity at constant pressure. So, ∆H = nCp ∆T = 0. Since ∆H = nCp ∆T = 0, nCp ∆T = 1500 J implies nCp = 1500 J. Substitute n = 610/T into nCp = 1500 J and simplify. 610Cp = 1500 T Step 3: Calculate the final pressure. P V = nRT At the final state, Pfinal Vfinal = nRT . Given Vfinal = 25 L, n = 610/T , and R = 0.0821 L atm mol−1 K−1 . 21 Substitute the values to find the final pressure. Pfinal · 25 = 610 · 0.0821 · T T 25Pfinal = 50.31Pfinal = 50.31 25 Therefore, the final pressure of the gas is 2.012 atm. Question 29 Question A certain reaction has an enthalpy change of ∆H = −128 kJ/mol at 25°C. If the reaction is exothermic, calculate the standard enthalpy change at 298 K. Solution Step 1: Calculate the change in enthalpy due to the change in temperature from 25°C to 298 K. Z 298 ∆H298 = ∆H25 + Cp dT 25 where Cp is the molar heat capacity at constant pressure. Step 2: Use the heat capacity values and the relationship Cp = calculate the integral.  Z 298 Z 298  dH dT Cp dT = dT p 25 25 Z 298 = 25  128 298 − 25 dH dT p
to  dT Step 3: Perform the integration and calculate the new enthalpy change at 298 K.   298 = 128 × ln kJ/mol 25 ∆H298 = −126.39 kJ/mol Therefore, the standard enthalpy change at 298 K is ∆H298 = −126.39 kJ/mol. 22 Question 30 Question Calculate the change in enthalpy for the reaction 2H2 (g) + O2 (g) → 2H2 O(l) given the following bond energies: H − H = 436 kJ/mol, O = O = 498 kJ/mol, H − O = 463 kJ/mol. Solution Step 1: Calculate the total energy required to break the bonds in the reactants. Step 2: Calculate the total energy released when the new bonds are formed in the products. Step 3: Find the difference between the energy required to break the bonds in the reactants and the energy released when the new bonds are formed in the products. Step 1: The total energy required to break the bonds in the reactants is: 2 × (H − H) + 1 × (O = O) = 2 × 436 + 1 × 498 = 1370 kJ/mol Step 2: The total energy released when the new bonds are formed in the products is: 2 × (H − O) = 2 × 463 = 926 kJ/mol Step 3: The change in enthalpy for the reaction is given by: ∆H = (Energy required to break the bonds)−(Energy released when new bonds are formed) ∆H = 1370 − 926 = 444 kJ/mol Therefore, the change in enthalpy for the reaction is 444 kJ/mol. Question 31 Question Given the following reaction involving the combustion of ethane gas: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) If the standard enthalpies of formation are −84.7 kJ/mol for CO2 and −285.8 kJ/mol for H2 O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the standard enthalpy of formation for ethane. 23 Solution Step 1: Calculate the standard enthalpy change of the reaction using the given standard enthalpies of formation. X X ◦ ◦ ◦ ∆Hrxn = ν∆Hf,products − ν∆Hf,reactants Given data: ∆Hf◦ (CO2 ) = −84.7 kJ/mol ∆Hf◦ (H2 O) = −285.8 kJ/mol ◦ ∆Hrxn = −1558.0 kJ/mol ◦ ∆Hrxn = 2(−84.7) + 3(−285.8) − (−1558.0) ◦ ∆Hrxn = −169.4 kJ/mol Step 2: Write the standard enthalpy change of the reaction using the enthalpy of formation of ethane. C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) ◦ ∆Hrxn = 2∆Hf◦ (CO2 ) + 3∆Hf◦ (H2 O) − ∆Hf◦ (C2 H6 ) Step 3: Solve for the standard enthalpy of formation of ethane. −169.4 = 2(−84.7) + 3(−285.8) − ∆Hf◦ (C2 H6 ) −169.4 = −169.4 − ∆Hf◦ (C2 H6 ) ∆Hf◦ (C2 H6 ) = 0 kJ/mol Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol. Question 32 Question A sample of an ideal gas undergoes a process where the pressure changes from 2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and final temperatures are 300 K and 500 K respectively, calculate the change in enthalpy of the gas during this process. 24 Solution Step 1: Calculate the initial and final internal energies of the gas using the first law of thermodynamics: ∆U = Q − W where ∆U is the change in internal energy, Q is the heat added to the gas, and W is the work done by the gas. Given that the gas undergoes an isobaric process, the work done can be calculated as: W = −Pext · ∆V Since the gas is ideal, we can use the ideal gas law to express Pext :   Vf W = −Pext · ∆V = −nRT ln Vi Substitute the given values:     2 1 W = −nRT ln = −nRT ln = −nRT ln(0.25) 8 4 ∆U = Q − W = Q + nRT ln(0.25) ∆U = nCV ∆T where CV is the molar heat capacity at constant volume. Since the gas is ideal, CV = f2 R where f is the degrees of freedom (f=3 for monoatomic gas). Substitute and rearrange to solve for Q: f Q = ∆U − nR∆T ln(0.25) = n( R)∆T − nR∆T ln(0.25) 2 3 f − ln(0.25)) = nR∆T ( − ln(0.25)) 2 2 Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation: Q = nR∆T ( ∆H = ∆U + ∆(P V ) Given that the process is isobaric, substitute the change in internal energy and pressure-volume work: ∆H = Q + W + P ∆V Substitute the expressions for Q and W calculated earlier: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T ln(0.25) + Pext ∆V 2 25 Given that P ∆V = nR∆T for an isobaric process: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T + nR∆T 2 5 ∆H = nR∆T ( − ln(0.25)) 2 Now, substitute the given values for n, R, and ∆T to find the change in enthalpy. Question 33 Question Calculate the standard enthalpy change for the reaction below at 298 K: 2CH4 (g) + 4O2 (g) → 2CO2 (g) + 4H2 O(l) Given the standard enthalpies of formation (∆Hf◦ ) for CH4 (g), CO2 (g), and H2 O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. Solution Step 1: Calculate the standard enthalpy change (∆H ◦ ) for the reaction using the standard enthalpies of formation: X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants) where n and m are the stoichiometric coefficients of the products and reactants, respectively. Step 2: Substitute the given values and calculate: ∆H ◦ = 2×∆Hf◦ (CO2 (g))+4×∆Hf◦ (H2 O(l))−[2×∆Hf◦ (CH4 (g))+4×∆Hf◦ (O2 (g))] Step 3: Plug in the values and calculate: ∆H ◦ = 2 × (−393.5) + 4 × (−285.8) − [2 × (−74.8) + 4 × 0] ∆H ◦ = −787.0 + (−1143.2) − (−149.6) ∆H ◦ = −787.0 − 1143.2 + 149.6 ∆H ◦ = −1780.6 kJ/mol Therefore, the standard enthalpy change for the given reaction at 298 K is -1780.6 kJ/mol. 26 Question 34 Question Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C. Given that the heat capacity of steam is 2.08 J/g·K, the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water is 40.79 kJ/mol. Solution Step 1: Calculate the heat required to cool the steam to 100◦ C to water at 100◦ C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol + 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K * 36 g/mol = 74.88 J/mol·K. The change in temperature is 100◦ C - 25◦ C = 75◦ C = 75 K. Therefore, the heat required to cool 10.0 moles of steam to water at 100◦ C is: 10.0 mol × 75 K × 74.88 J/mol · K = 56160 J. Step 2: Calculate the heat required to condense the water at 100◦ C to water at 25◦ C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol. Therefore, the heat required to condense 10.0 moles of steam at 100◦ C to water at 100◦ C is: 10.0 mol × 40790 J/mol = 407900 J. Step 3: Add the heats from Step 1 and Step 2 to find the total change in enthalpy. ∆H = 56160 J + 407900 J = 464060 J. Therefore, the change in enthalpy when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C is 464060 J. Question 35 Question An experiment involves the mixing of 150 g of water at 25°C with 300 g of water at 85°C in a thermally insulated container. Calculate the final temperature of the mixture. Assume specific heat capacity of water is 4.18 J/g°C and neglect any heat losses to the surroundings. Solution Step 1: Calculate the heat lost by the hot water and the heat gained by the cold water. The heat lost by the hot water is given by: qhot = mhot · c · ∆T 27 Question 8 Question A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50 moles of the reactant are consumed, what is the change in enthalpy per mole of reactant? Solution Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be calculated using the formula: ∆Hper mole = ∆H moles of reactant Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles of reactant is 2.50 mol, we can substitute these values into the formula: ∆Hper mole = −349 kJ 2.50 mol Step 3: Calculate the change in enthalpy per mole of reactant: ∆Hper mole = −349 = −139.6 kJ/mol 2.50 Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol. Question 9 Question Calculate the enthalpy change (∆H) for the reaction: 2H2 (g) + O2 (g) → H2 O(l) given the following information: ˆ ∆Hf◦ for H2 (g) = 0 kJ/mol ˆ ∆Hf◦ for O2 (g) = 0 kJ/mol ˆ ∆Hf◦ for H2 O(l) = -285.8 kJ/mol Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) Step 2: Calculate ∆H for the reaction using the standard enthalpies of formation: X X ∆H = ∆Hf◦ (products) − ∆Hf◦ (reactants) 6 Step 3: Substitute the given values into the equation: ∆H = [2(−285.8 kJ/mol)] − [2(0 kJ/mol) + 0 kJ/mol] Step 4: Solve for ∆H: ∆H = −571.6 kJ/mol Therefore, the enthalpy change for the reaction is ∆H = −571.6 kJ/mol. Question 10 Question Calculate the change in enthalpy (∆H) when 50.0 g of water at 25°C is heated to steam at 100°C. The specific heat capacity of water is 4.18 J/g°C, the molar heat of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to heat the water from 25°C to 100°C. The formula to calculate the heat required is: q = mc∆T Where: - q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Substitute the values: q = (50.0 g)(4.18 J/g°C)(100 − 25)C q = (50.0)(4.18)(75) J Step 2: Calculate the heat required to vaporize the water. The heat required to vaporize the water can be calculated using the formula: q = n · ∆Hvap Where: - n is the number of moles of water, - ∆Hvap is the molar heat of vaporization. First, calculate the number of moles of water: n= 50.0 g 18.015 g/mol Step 3: Convert the heat of vaporization to joules. 40.79 kJ/mol = 40.79 × 103 J/mol 7 Now, calculate the heat required to vaporize the water:   50.0 q= × (40.79 × 103 ) J 18.015 Step 4: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the heat required to raise the temperature and the heat required to vaporize the water: ∆H = qheating + qvaporization Question 11 Question Given the reaction: 2A(g) + 3B(g) → C(g) + 4D(g) ◦ where ∆H = −1258 kJ, calculate the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B. Solution Step 1: Calculate the moles of A and B reacting. Moles of A = 1.5 mol Moles of B = 2.0 mol Step 2: Determine the limiting reactant. Moles of A required for reaction = 22 = 1.0 mol Moles of B required for reaction = 23 = 1.5 mol Since A has fewer moles than required, A is the limiting reactant. Step 3: Calculate the moles of C and D formed. Since 2 moles of A produces 1 mole of C, Moles of C formed = 12 × 1.5 = 0.75 mol Since 2 moles of A produces 4 moles of D, Moles of D formed = 2 × 1.5 = 3.0 mol Step 4: Calculate the enthalpy change for the reaction. ∆H ◦ = −1258 kJ For the reaction of 1 mole of A: = −629 kJ Enthalpy change = −1258 2 For the reaction of 1.5 moles of A: Enthalpy change = −629 × 1.5 = −943.5 kJ Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B is −943.5 kJ . 8 Question 12 Question A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5°C in a closed container water at 25°C. Assuming the heat capacity of the container is negligible and that the density of water is 1 g/cm3 , calculate the enthalpy change for the reaction. Solution Step 1: Calculate the mass of water affected by the heat exchange. Given that the density of water is 1 g/cm3 , the mass of water affected by the heat exchange is: Volume of water = Density × Volume Mass Volume = Density 1g 3 Volume = 3 = 1 cm 1 g/cm The volume of water affected by the heat exchange is 1 cm3 , which is equivalent to 1 gram. Step 2: Calculate the change in temperature of the water. Given that the temperature change is 0.5°C, the heat exchanged is: q = mc∆T q = (1 g)(4.18 J/g°C)(0.5C) = 2.09 J Step 3: Convert the heat exchanged to kilojoules. To convert Joules to kilojoules, divide by 1000: q= 2.09 J = 0.00209 kJ 1000 Step 4: Determine the enthalpy change for the reaction. The enthalpy change for the reaction is equal in magnitude but opposite in sign to the heat exchanged: ∆H = −q = −0.00209 kJ = -2.09 kJ Therefore, the enthalpy change for the reaction is -2.09 kJ. Question 13 Question For a reaction at constant pressure, the enthalpy change (∆H) is given by the equation ∆H = 50 − 2T , where T is the temperature in Kelvin. Determine the temperature at which the reaction becomes exothermic. 9 Solution Step 1: To find the temperature at which the reaction becomes exothermic, set ∆H to zero and solve for T . ∆H = 50 − 2T = 0 50 = 2T T = 25 K Step 2: To determine whether the reaction is exothermic or endothermic at T = 25 K, substitute T = 25 into the equation ∆H = 50 − 2T . ∆H = 50 − 2(25) = 50 − 50 = 0 Since ∆H = 0 at T = 25 K, the reaction becomes exothermic at this temperature. Question 14 Question A gas initially at 300 K and 1 atm pressure is expanded isothermally and reversibly to three times its initial volume. Calculate the change in enthalpy for this process, given that the molar heat capacity at constant pressure, Cp , is 30 J/mol K. Solution Step 1: Calculate the work done during the expansion. Given that the process is isothermal and reversible, the work done is −nRT ln(Vf /Vi ), where n is the number of moles of gas, R is the ideal gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Step 2: Calculate the heat transfer. Since the process is isothermal, the heat transfer is equal to the work done. Therefore, q = −w. Step 3: Calculate the change in enthalpy. The change in enthalpy is given by ∆H = ∆U + P ∆V , where ∆U is the change in internal energy, P is the pressure, and ∆V is the change in volume. For an isothermal process, ∆U = 0, so ∆H = P ∆V . Step 4: Substitute values and calculate. Given: - Initial temperature, Ti = 300 K - Initial pressure, Pi = 1 atm - Final volume, Vf = 3Vi - Molar heat capacity at constant pressure, Cp = 30 J/mol K - Ideal gas constant, R = 8.314 J/mol K From the ideal gas law, P V = nRT , we can find n and R. Then, we can calculate the change in enthalpy using the formula ∆H = P ∆V . Finally, substitute the known values to find the change in enthalpy. 10 Question 15 Question Given the reaction: 2H2 (g) + O2 (g) → 2H2 O(l) with the following enthalpy changes: ∆Hf◦ (H2 O(l)) = −285.8 kJ/mol ∆Hf◦ (O2 (g)) = 0 kJ/mol ∆Hf◦ (H2 (g)) = 0 kJ/mol calculate the standard enthalpy change for the reaction. Solution Step 1: Calculate the standard enthalpy change for the reaction using the standard enthalpy of formation values provided. X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants)

∆H ◦ = 2 · ∆Hf◦ (H2 O(l)) − 2 · ∆Hf◦ (H2 (g)) + ∆Hf◦ (O2 (g)) ∆H ◦ = 2(−285.8 kJ/mol) − 2(0 kJ/mol) − 0 kJ/mol ∆H ◦ = −571.6 kJ/mol Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol . Question 16 Question Given the following reaction: 2A(g) + 3B(g) −→ C(g) + D(g) If the standard enthalpy of formation for each substance is as follows: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol Calculate the standard enthalpy change, ∆H ◦ , for the reaction. 11 Solution To calculate the standard enthalpy change for the reaction, we can use Hess’s Law, which states that the total enthalpy change for a reaction is the same regardless of the number of steps in the reaction or the pathway taken. Step 1: Write the given reaction in terms of standard enthalpies of formation. The given reaction is: 2A(g) + 3B(g) −→ C(g) + D(g) The standard enthalpy change for this reaction can be expressed as: X X ∆H ◦ = ∆Hf◦ (products) − ∆Hf◦ (reactants) Step 2: Calculate the standard enthalpy change for the reaction. Given that: ∆Hf◦ (A) = −200 kJ/mol ∆Hf◦ (B) = −300 kJ/mol ∆Hf◦ (C) = −400 kJ/mol ∆Hf◦ (D) = −500 kJ/mol The standard enthalpy change can be calculated as follows: ∆H ◦ = [(∆Hf◦ (C) + ∆Hf◦ (D)) − (∆Hf◦ (2A) + ∆Hf◦ (3B))] ∆H ◦ = [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))] ∆H ◦ = [−900 kJ/mol − (−200 kJ/mol + −900 kJ/mol)] ∆H ◦ = −900 kJ/mol + 1100 kJ/mol ∆H ◦ = 200 kJ/mol Question 17 Question A reaction has an enthalpy change of ∆H = −285.8 kJ. If 4.96 moles of the reactant undergo the reaction, what is the total heat absorbed or released during the reaction? 12 Solution Step 1: Determine the total heat absorbed or released per mole of the reactant. Given: ∆H = −285.8 kJ We know that the total heat absorbed or released during the reaction can be calculated as: Total heat = ∆H × moles of reactant Step 2: Calculate the total heat absorbed or released. Plugging in the values: Total heat = −285.8 kJ × 4.96 moles Total heat = −1416.688 kJ Therefore, the total heat absorbed or released during the reaction is -1416.688 kJ. Question 18 Question Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦ C is converted to steam at 100◦ C. The specific heat capacity of water is 4.18 J/(g·°C), the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass of water is 18.015 g/mol. Solution Step 1: Calculate the heat required to raise the temperature of the water from 25◦ C to 100◦ C using the formula: q1 = m × c × ∆T where - m is the mass of water (35.0 g), - c is the specific heat capacity of water (4.18 J/(g·°C)), - ∆T is the change in temperature (100◦ C - 25◦ C). Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing by 1000. Step 3: Calculate the moles of water in 35.0 g using the molar mass of water. Step 4: Calculate the heat required to convert the water to steam at 100◦ C using the formula: q2 = n × ∆Hvap where - n is the number of moles of water calculated in Step 3, - ∆Hvap is the molar enthalpy of vaporization of water (40.79 kJ/mol). Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat change, ∆H. 13 ∆H = q1 + q2 Question 19 Question Given the balanced chemical equation for the combustion of ethane: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) Calculate the standard enthalpy change (∆H ◦ ) for this reaction. Given the following standard enthalpy of formation values: ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol ∆Hf◦ (CO2 ) = −393.51 kJ/mol ∆Hf◦ (H2 O) = −285.83 kJ/mol Solution Step 1: Calculate the standard enthalpy of formation for the products using the given values. ∆Hf◦ (2CO2 ) = 2(−393.51 kJ/mol) = −787.02 kJ/mol ∆Hf◦ (3H2 O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol Step 2: Calculate the standard enthalpy of formation for the reactants using the given value. ∆Hf◦ (C2 H6 ) = −84.68 kJ/mol Step 3: Calculate the net standard enthalpy change for the reaction using the formula: X X ∆H ◦ = ( ∆Hf◦ products) − ( ∆Hf◦ reactants) ∆H ◦ = [(−787.02 kJ/mol) + (−857.49 kJ/mol)] − (−84.68 kJ/mol) ∆H ◦ = −1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol Therefore, the standard enthalpy change (∆H ◦ ) for the combustion of ethane is -1559.83 kJ/mol. 14 Question 20 Question Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2 ) react according to the following balanced chemical equation: N2 (g) + 3H2 (g) → 2N H3 (g) Given the enthalpies of formation (∆Hf◦ ) for N2 (g), H2 (g), and N H3 (g) are 0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively. Solution Step 1: Write the given balanced chemical equation and the enthalpy change for the reaction. N2 (g) + 3H2 (g) → 2N H3 (g) The enthalpy change for the reaction is given by: ∆H = Σ∆Hf◦ (products) − Σ∆Hf◦ (reactants) Step 2: Calculate the enthalpy change for the reaction. The enthalpies of formation are: ∆Hf◦ (N H3 ) = −46.2 kJ/mol ∆Hf◦ (N2 ) = 0 kJ/mol ∆Hf◦ (H2 ) = 0 kJ/mol Substitute the values into the equation for ∆H:

∆H = 2 × ∆Hf◦ (N H3 ) − ∆Hf◦ (N2 ) + 3 × ∆Hf◦ (H2 ) ∆H = (2 × −46.2 kJ/mol) − (0 + 3 × 0 kJ/mol) ∆H = −92.4 kJ/mol Step 3: Calculate the change in enthalpy when 5.00 moles of N2 react. Since the enthalpy change is given per mole, we need to calculate the total enthalpy change for 5 moles of N2 : Change in enthalpy = moles × ∆H Change in enthalpy = 5.00 moles × −92.4 kJ/mol Change in enthalpy = −462 kJ Therefore, the change in enthalpy when 5.00 moles of N2 react is -462 kJ. 15 Question 21 Question A reaction is thought to have a certain enthalpy change. By performing a series of measurements, it is found that the reaction has an enthalpy change of −289.2 kJ when carried out at 25°C. Assuming that the heat capacities of all substances are constant over the temperature range in question, calculate the enthalpy change for the reaction at 55°C. Solution Step 1: Determine the heat capacity change (∆Cp ) of the reaction. Z 55C 25C ∆Hreaction = ∆Hreaction + ∆Cp dT 25C Z 55C ∆Hreaction = −289.2 kJ + ∆Cp dT 25C Step 2: Since the heat capacities are constant, we can consider ∆Cp as a constant value, say Cp . Z 55C ∆Hreaction = −289.2 kJ + Cp dT 25C ∆Hreaction = −289.2 kJ + Cp · (55 − 25) K ∆Hreaction = −289.2 kJ + 30Cp Step 3: Given that the enthalpy change of the reaction at 25°C is -289.2 kJ: −289.2 = −289.2 + 30Cp 0 = 30Cp =⇒ Cp = 0 Step 4: The assumption that the heat capacities are constant is incorrect. We must therefore consider the correct dependence of heat capacity on temperature for the substances involved. Question 22 Question A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of 298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for the reaction. Given: Gas constant, R = 8.314 J/(mol·K) 16 Solution Step 1: Calculate the work done by the gas during the reaction. The work done by the gas is given by the equation: W = −Pext ∆V Where: Pext = external pressure = 1.00 atm = 1.013 × 105 Pa ∆V = change in volume = 10.0 L Convert 10.0 L to m³: 10.0L = 10.0 × 10−3 m3 = 0.0100m3 Now, calculate the work done: W = −(1.013 × 105 P a)(0.0100m3 ) = −1013J Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed by the calorimeter is given by the equation: qcal = C∆T Where: C = heat capacity of the calorimeter = 750 J/K ∆T = change in temperature = 316 K - 298 K = 18 K Now, calculate the heat absorbed by the calorimeter: qcal = (750J/K)(18K) = 13500J Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction is given by the equation: qrxn = qcal − W Now, calculate the heat of the reaction: qrxn = 13500J − (−1013J) = 14513J Step 4: Calculate the change in enthalpy. Since ∆H = qrxn under constant pressure conditions, the change in enthalpy for the reaction is ∆H = 14513 J. Therefore, the change in enthalpy for the reaction is 14513 J. Question 23 Question A reaction is known to have an enthalpy change of ∆H = −250 kJ/mol. If the reaction is exothermic, in which direction will the enthalpy change (∆H) be when the reaction is reversed? 17 Solution 1. When a reaction is reversed, the sign of the enthalpy change is also reversed. Therefore, if the original reaction has ∆H = −250 kJ/mol, the reversed reaction will have ∆H = +250 kJ/mol. 2. This change in sign occurs because when a reaction is reversed, the products become the reactants and vice versa. As a result, the direction of heat flow (exothermic or endothermic) also changes. Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol. Question 24 Question An ideal gas undergoes a process at constant pressure. The initial temperature of the gas is 300 K and the final temperature is 400 K. Calculate the change in enthalpy of the gas during the process. Solution Step 1: Recall the definition of enthalpy in terms of internal energy and pressure: ∆H = ∆U + P ∆V Step 2: Since the process is at constant pressure, ∆U = Q − W where Q is the heat added to the system and W is the work done by the gas on its surroundings. Step 3: We can rewrite the expression for enthalpy change as: ∆H = (Q − W ) + P ∆V Step 4: The work done by the gas can be expressed as W = P ∆V . Substituting this into the equation gives: ∆H = Q Step 5: To calculate the heat added to the system, we can use the formula: Q = nCp ∆T where n is the number of moles of gas and Cp is the molar heat capacity at constant pressure. Step 6: First, we need to find the number of moles of the gas. We can use the ideal gas law: P V = nRT Solving for n, we have: n= PV RT 18 Step 7: Substituting n into the formula for heat gives:   PV Q= Cp ∆T RT Step 8: Substituting the given values P = constant, V = constant, R = constant, T1 = 300 K, T2 = 400 K and Cp = specific heat at constant pressure into the equation gives the change in enthalpy. Question 25 Question A reaction between hydrogen gas and oxygen gas to form water vapor has an enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much heat is absorbed or released? Solution Step 1: Write the balanced chemical equation for the reaction: 2H2 (g) + O2 (g) → 2H2 O(g) Step 2: Determine the moles of oxygen gas reacting, based on the stoichiometry of the balanced equation. Since the balanced equation states that 1 mole of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas required will be half of the moles of hydrogen gas: 1 × 8.00 moles = 4.00 moles 2 Step 3: Calculate the heat absorbed or released using the enthalpy change and the moles of the limiting reactant. Given enthalpy change: ∆H = −483.6 kJ Since the equation is for 2 moles of hydrogen gas reacting, the heat change for 1 mole of hydrogen gas will be half of the given value: moles of O2 = 483.6 kJ = −241.8 kJ/mole 2 Therefore, the heat change for 8.00 moles of hydrogen gas will be: Heat change for 1 mole of H2 = − Heat change = −241.8 kJ/mole × 8.00 moles = −1934.4 kJ So, 1934.4 kJ of heat is released in this reaction. Question 26 Question A reaction has an enthalpy change of ∆H = −185 kJ/mol. If 2.50 moles of the reaction takes place, calculate the total enthalpy change for the reaction. 19 Solution Step 1: Determine the total enthalpy change for the reaction per mole. Given: ∆H = −185 kJ/mol Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles of the reaction takes place. Total Enthalpy Change = ∆H × Number of moles Total Enthalpy Change = −185 kJ/mol × 2.50 mol Total Enthalpy Change = −185 kJ/mol × 2.50 = −462.5 kJ Therefore, the total enthalpy change for the reaction when 2.50 moles of the reaction takes place is −462.5 kJ. Question 27 Question Calculate the change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure. Given that the heat of vaporization of water is 40.79 kJ/mol. Solution Step 1: Determine the change in enthalpy during the phase change from vapor to liquid. For 2 moles of water vapor being condensed to liquid water, the change in enthalpy can be calculated using the heat of vaporization formula: Change in Enthalpy = moles × heat of vaporization Change in Enthalpy = 2 moles × 40.79 kJ/mol Change in Enthalpy = 81.58 kJ Therefore, the change in enthalpy during the phase change is 81.58 kJ. Step 2: Account for the temperature change. Since the temperature remains constant during the phase change, there is no additional change in enthalpy due to the temperature change. Step 3: Calculate the total change in enthalpy. The total change in enthalpy is the sum of the change in enthalpy during the phase change and any changes due to temperature: Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change) Total Change in Enthalpy = 81.58 kJ + 0 kJ Total Change in Enthalpy = 81.58 kJ 20 Therefore, the total change in enthalpy when 2 moles of water vapor at 100°C and 1 atm pressure is condensed to liquid water at the same temperature and pressure is 81.58 kJ. Question 28 Question A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at constant temperature to a final volume of 25 L. If the enthalpy change during the process is 1500 J, calculate the final pressure of the gas. Solution Step 1: Calculate the initial number of moles of gas using the ideal gas law, P V = nRT . PV n= RT Given P = 10 atm, V = 5 L, R = 0.0821 L atm mol−1 K−1 , and T is constant for the process. Substitute the given values to find n. n= 50 (10 atm)(5 L) 610 = = 0.0821T T (0.0821 L atm mol−1 K−1 )(T ) Step 2: Calculate the final pressure of the gas using the enthalpy change and the number of moles. ∆H = nCp ∆T Given ∆H = 1500 J and n = 610/T , where T is the gas temperature for the process, and Cp is the molar heat capacity at constant pressure. Since the process is at constant temperature, ∆T = 0 and nCp ∆T = 0. Thus, the enthalpy change during the process does not depend on the molar heat capacity at constant pressure. So, ∆H = nCp ∆T = 0. Since ∆H = nCp ∆T = 0, nCp ∆T = 1500 J implies nCp = 1500 J. Substitute n = 610/T into nCp = 1500 J and simplify. 610Cp = 1500 T Step 3: Calculate the final pressure. P V = nRT At the final state, Pfinal Vfinal = nRT . Given Vfinal = 25 L, n = 610/T , and R = 0.0821 L atm mol−1 K−1 . 21 Substitute the values to find the final pressure. Pfinal · 25 = 610 · 0.0821 · T T 25Pfinal = 50.31Pfinal = 50.31 25 Therefore, the final pressure of the gas is 2.012 atm. Question 29 Question A certain reaction has an enthalpy change of ∆H = −128 kJ/mol at 25°C. If the reaction is exothermic, calculate the standard enthalpy change at 298 K. Solution Step 1: Calculate the change in enthalpy due to the change in temperature from 25°C to 298 K. Z 298 ∆H298 = ∆H25 + Cp dT 25 where Cp is the molar heat capacity at constant pressure. Step 2: Use the heat capacity values and the relationship Cp = calculate the integral.  Z 298 Z 298  dH dT Cp dT = dT p 25 25 Z 298 = 25  128 298 − 25 dH dT p
to  dT Step 3: Perform the integration and calculate the new enthalpy change at 298 K.   298 = 128 × ln kJ/mol 25 ∆H298 = −126.39 kJ/mol Therefore, the standard enthalpy change at 298 K is ∆H298 = −126.39 kJ/mol. 22 Question 30 Question Calculate the change in enthalpy for the reaction 2H2 (g) + O2 (g) → 2H2 O(l) given the following bond energies: H − H = 436 kJ/mol, O = O = 498 kJ/mol, H − O = 463 kJ/mol. Solution Step 1: Calculate the total energy required to break the bonds in the reactants. Step 2: Calculate the total energy released when the new bonds are formed in the products. Step 3: Find the difference between the energy required to break the bonds in the reactants and the energy released when the new bonds are formed in the products. Step 1: The total energy required to break the bonds in the reactants is: 2 × (H − H) + 1 × (O = O) = 2 × 436 + 1 × 498 = 1370 kJ/mol Step 2: The total energy released when the new bonds are formed in the products is: 2 × (H − O) = 2 × 463 = 926 kJ/mol Step 3: The change in enthalpy for the reaction is given by: ∆H = (Energy required to break the bonds)−(Energy released when new bonds are formed) ∆H = 1370 − 926 = 444 kJ/mol Therefore, the change in enthalpy for the reaction is 444 kJ/mol. Question 31 Question Given the following reaction involving the combustion of ethane gas: C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) If the standard enthalpies of formation are −84.7 kJ/mol for CO2 and −285.8 kJ/mol for H2 O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the standard enthalpy of formation for ethane. 23 Solution Step 1: Calculate the standard enthalpy change of the reaction using the given standard enthalpies of formation. X X ◦ ◦ ◦ ∆Hrxn = ν∆Hf,products − ν∆Hf,reactants Given data: ∆Hf◦ (CO2 ) = −84.7 kJ/mol ∆Hf◦ (H2 O) = −285.8 kJ/mol ◦ ∆Hrxn = −1558.0 kJ/mol ◦ ∆Hrxn = 2(−84.7) + 3(−285.8) − (−1558.0) ◦ ∆Hrxn = −169.4 kJ/mol Step 2: Write the standard enthalpy change of the reaction using the enthalpy of formation of ethane. C2 H6 (g) + 7/2O2 (g) → 2CO2 (g) + 3H2 O(l) ◦ ∆Hrxn = 2∆Hf◦ (CO2 ) + 3∆Hf◦ (H2 O) − ∆Hf◦ (C2 H6 ) Step 3: Solve for the standard enthalpy of formation of ethane. −169.4 = 2(−84.7) + 3(−285.8) − ∆Hf◦ (C2 H6 ) −169.4 = −169.4 − ∆Hf◦ (C2 H6 ) ∆Hf◦ (C2 H6 ) = 0 kJ/mol Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol. Question 32 Question A sample of an ideal gas undergoes a process where the pressure changes from 2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and final temperatures are 300 K and 500 K respectively, calculate the change in enthalpy of the gas during this process. 24 Solution Step 1: Calculate the initial and final internal energies of the gas using the first law of thermodynamics: ∆U = Q − W where ∆U is the change in internal energy, Q is the heat added to the gas, and W is the work done by the gas. Given that the gas undergoes an isobaric process, the work done can be calculated as: W = −Pext · ∆V Since the gas is ideal, we can use the ideal gas law to express Pext :   Vf W = −Pext · ∆V = −nRT ln Vi Substitute the given values:     2 1 W = −nRT ln = −nRT ln = −nRT ln(0.25) 8 4 ∆U = Q − W = Q + nRT ln(0.25) ∆U = nCV ∆T where CV is the molar heat capacity at constant volume. Since the gas is ideal, CV = f2 R where f is the degrees of freedom (f=3 for monoatomic gas). Substitute and rearrange to solve for Q: f Q = ∆U − nR∆T ln(0.25) = n( R)∆T − nR∆T ln(0.25) 2 3 f − ln(0.25)) = nR∆T ( − ln(0.25)) 2 2 Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation: Q = nR∆T ( ∆H = ∆U + ∆(P V ) Given that the process is isobaric, substitute the change in internal energy and pressure-volume work: ∆H = Q + W + P ∆V Substitute the expressions for Q and W calculated earlier: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T ln(0.25) + Pext ∆V 2 25 Given that P ∆V = nR∆T for an isobaric process: 3 ∆H = nR∆T ( − ln(0.25)) + nR∆T + nR∆T 2 5 ∆H = nR∆T ( − ln(0.25)) 2 Now, substitute the given values for n, R, and ∆T to find the change in enthalpy. Question 33 Question Calculate the standard enthalpy change for the reaction below at 298 K: 2CH4 (g) + 4O2 (g) → 2CO2 (g) + 4H2 O(l) Given the standard enthalpies of formation (∆Hf◦ ) for CH4 (g), CO2 (g), and H2 O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. Solution Step 1: Calculate the standard enthalpy change (∆H ◦ ) for the reaction using the standard enthalpies of formation: X X ∆H ◦ = n∆Hf◦ (products) − m∆Hf◦ (reactants) where n and m are the stoichiometric coefficients of the products and reactants, respectively. Step 2: Substitute the given values and calculate: ∆H ◦ = 2×∆Hf◦ (CO2 (g))+4×∆Hf◦ (H2 O(l))−[2×∆Hf◦ (CH4 (g))+4×∆Hf◦ (O2 (g))] Step 3: Plug in the values and calculate: ∆H ◦ = 2 × (−393.5) + 4 × (−285.8) − [2 × (−74.8) + 4 × 0] ∆H ◦ = −787.0 + (−1143.2) − (−149.6) ∆H ◦ = −787.0 − 1143.2 + 149.6 ∆H ◦ = −1780.6 kJ/mol Therefore, the standard enthalpy change for the given reaction at 298 K is -1780.6 kJ/mol. 26 Question 34 Question Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C. Given that the heat capacity of steam is 2.08 J/g·K, the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water is 40.79 kJ/mol. Solution Step 1: Calculate the heat required to cool the steam to 100◦ C to water at 100◦ C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol + 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K * 36 g/mol = 74.88 J/mol·K. The change in temperature is 100◦ C - 25◦ C = 75◦ C = 75 K. Therefore, the heat required to cool 10.0 moles of steam to water at 100◦ C is: 10.0 mol × 75 K × 74.88 J/mol · K = 56160 J. Step 2: Calculate the heat required to condense the water at 100◦ C to water at 25◦ C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol. Therefore, the heat required to condense 10.0 moles of steam at 100◦ C to water at 100◦ C is: 10.0 mol × 40790 J/mol = 407900 J. Step 3: Add the heats from Step 1 and Step 2 to find the total change in enthalpy. ∆H = 56160 J + 407900 J = 464060 J. Therefore, the change in enthalpy when 10.0 moles of steam at 100◦ C is cooled to water at 25◦ C is 464060 J. Question 35 Question An experiment involves the mixing of 150 g of water at 25°C with 300 g of water at 85°C in a thermally insulated container. Calculate the final temperature of the mixture. Assume specific heat capacity of water is 4.18 J/g°C and neglect any heat losses to the surroundings. Solution Step 1: Calculate the heat lost by the hot water and the heat gained by the cold water. The heat lost by the hot water is given by: qhot = mhot · c · ∆T 27 where mhot = 300 g is the mass of hot water, c = 4.18 J/g°C is the specific heat capacity of water, and ∆T = 85 − Tfinal is the decrease in temperature of the hot water. The heat gained by the cold water is given by: qcold = mcold · c · ∆T where mcold = 150 g is the mass of cold water and ∆T = Tfinal − 25 is the increase in temperature of the cold water. Since the system is thermally isolated, the heat lost by the hot water is equal to the heat gained by the cold water: qhot = qcold Step 2: Set up and solve the equation based on the heat lost and gained. mhot · c · (85 − Tfinal ) = mcold · c · (Tfinal − 25) Substitute in the given values: 300 · 4.18 · (85 − Tfinal ) = 150 · 4.18 · (Tfinal − 25) Step 3: Solve for the final temperature (Tfinal ). Simplify the equation: Expand: 1260 · (85 − Tfinal ) = 630 · (Tfinal − 25) 107100 − 1260Tfinal = 630Tfinal − 15750 Combine like terms: 107100 + 15750 = 630Tfinal + 1260Tfinal 107100 + 15750 ≈ 70.37C Solve for Tfinal : Tfinal = 1890 Therefore, the final temperature of the mixture is approximately 70.37C. 28