Chemistry Homework 1

The average human lung expands by about 0.50 LL during each breath. The air within a piston equipped with a cylinder absorbs 545 JJ of heat and expands from an initial volume of 0.11 LL to a final volume of 0.88 LL against an external pressure of 1.0 barbar. A gas is compressed from an initial volume of 5.35 LL to a final volume of 1.23 LL by an external pressure of 1.00 barbar . During the compression, the gas releases 121 JJ of heat. What is the work associated with decomposition of trinitrotoluene (TNT) upon detonation according to the following reaction at 281.65 KK? C7H5N3O6(s)⟶32N2(g) +52H2(g)+3O2(g)+7C(s) Homework 1 Work = -p x ∆ = -(1.0)x(0.50L)atm = -0.50Lxatm Conversion of 1 Lxatm  J = /101.325 = -50.66 or rounded to -51 W = -p x ∆V ∆V= 0.88-0.11 =.77L =-1 x 0.77 =-77L x atm x 101.325 =-78.02J ∆E= change in internal energy w = 78.02 ∆E = q + w = 545 +(-78.02) = 466.98J =467J q=heat=545J ∆E = q + w q=-121 Calc work done w =p*∆V 1 atm = 1.01*10^5 Pa, and 1 liter = 10^-3 m^3, so: W=(1.01^10^5 N/m^2)*(5.35 - 1.24)*10^-3 m^3 = 415.11J So ∆E = -121 + 415.11 294 But with correct values its 291 1 mole C7H5N3O6 gives 3/2 moles N2 and 5/2 moles H2 and 3 moles O2 gas Total gas moles produced = N2 moles + H2 moles + O2 moles = 1.5 + 2.5 + 3 = 7 work done = - P x dV where dV is change in volume we use ideal gas equation PV = nRT ( since V is nothing but volume occup all gases) work done = - nRT where R = gas constant = 0.08206 liter atm/molK = - 7 mol x 0.08206 liter atm/molK x 281.65 = - 162.8500 atm liter /mol we have 1 atmliter = 101.325 J hence work done = - 162.8500 atm liter /mol x ( 101.325 J/atmliter) = - 16500 J/mol = - 16.500 KJ/mol ( -ve sign indicates expansion work ) (correct answer is -16.39) The change in internal energy If the change in enthalpy is -5074.1 kJ, how much work is done during the for the combustion combustion? of 1.0 mol of octane at a - ∆U = -5084.3 pressure of 1.0 bar is ∆H = -5074.1 -5084.3 kJkJ . ∆H = U +PxV ∆H = ∆U +∆(PxV) ∆H= ∆U +P x ∆V -5074.1 = -5084.3 + P x ∆V Px∆V= -5074.1 + 5084.5 = 10.4 kJ w = p∆V = -10.4 kJ  because you lose the energy in the reaction its lost so its – sig Correct = -10.2 - Determine whether each of the following is exothermic or endothermic and indicate the sign of ΔHΔH. a. dry ice evaporating b. a sparkler burning c. the reaction that occurs in a chemical cold pack often used to ice athletic injuries ΔH + = endothermic ΔH - = exothermic a. Dry ice evaporating to evaporate a dry ice we need to supply heat so heat is absorbed by the syste its endothermic reaction and here, ΔH is having positive value. b. a sparkler burningwhen a sparkler is burning heat is given out to the system sothe reaction is exothermic and so, ΔH value is negative. c. the reaction that occurs in a chemical cold pack often used to ice athletic i the chemical cold pack is used it it absorbs heat so if we consider it a system th absorbs heat and so the reactions endothermic and ΔH value is positive. What mass of natural gas (CH4)(CH4) must burn to emit 264 kJkJ of heat? CH4(g)+2O2(g)→CO2(g) +2H2O(g) ΔrH∘=−802.3kJmol−1 The propane fuel (C3H8) (C3H8) used in gas barbeques burns according to the thermochemical equation: C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)ΔrH∘=−2217kJmol− 1 ΔrH∘=−802.3 Molar mass of CH4 = 16.04g/mol = 264kj x 1molCH4 / molar mass (802.3kJ) x 16.04g CH4/ 1 mol CH4 = 5.28 gCH4 If a pork roast must absorb 1.7×103 kJ1.7×103 kJ to fully cook, and if only 14 % the heat produced by the barbeque is actually absorbed by the roast, what ma of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast - heat required by pork roast = 1.7*10^3 KJ Let heat produced be x KJ then 14 % of x equals 1.7*10^3 KJ 14*x/100 = 1.7*10^3 KJ x = 1.21*10^4 KJ Determine the mass of CO2CO2 produced by burning enough carbon (in the form of charcoal) to produce 4.80×102 kJkJ of heat. C(s) +O2(g)→CO2(g)ΔrH∘=−393.5 kJmol−1 When 1 mol of a fuel burns at constant pressure, it produces 3458 kJ of heat and does 15 kJkJ of work. This heat must be produced by reaction From reaction, for 3 mol of CO2. heat produced = 2217 KJ So, for 1.21*10^4 KJ, mol of CO2 required = 3*1.21*10^4 / 2217 = 16.43 mol Molar mass of CO2, MM = 1*MM(C) + 2*MM(O) = 1*12.01 + 2*16.0 = 44.01 g/mol use: mass of CO2, m = number of mol * molar mass = 16.43 mol * 44.01 g/mol = 723 g Answer: 720g C(s)+O2(g)→CO2(g) ΔrH∘= −393.5kJ mol−1 One mole of CO2 produces from one mole of C and gives -393.5 kJ heat. Now x moles of CO2 gives (4.80×102 kJkJ ) 480 kJ of heat. x moles of CO2 = 480/393.5 = 1.143 moles of CO2 Mass of 1.143 moles of CO2 = 1.219822109 =1.20982211x10^0 × 44 = 5.32321728x10^1 =53.231728 =53.23 Answer= 53.7g Homework 2 Heat of comb (q=-3458kJ/mol) Work done = 15 kJ/mol ΔE = ΔU ΔU = q-w = -(3458) – 15 = - 3473 To solve for ΔH ΔH = 3473 – 3458 kJ To solve for ΔE, you need to add -(3458)+(-15)/ 1mol = -3473kJ (it's negative du the fact that you're producing heat) To solve for ΔH, you need to multiply -3473 by the pressure (in this case, it's as we're at standard pressure) so: -3473*1= -3473 kJ Determine whether each of the following is exothermic or endothermic and indicate the sign of ΔHΔH. a. natural gas burning on a stove b. isopropyl alcohol evaporating from skin c. water condensing from steam What mass of natural gas (CH4) must burn to emit 261 kJ of heat? CH4(g)+2O2(g)→CO2(g) +2H2O(g)ΔrH∘=−802.3kJmol −1 A silver block, initially at 55.3 ∘C∘C, is submerged into 100.0 gg of water at 24.6 ∘C∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.5 ∘C∘C. A silver block, initially at 55.3∘C, is submerged into 100.0 g of water at 24.9∘C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.4∘C. The specific heat capacities for water and silver are Cs,water=4.18J/ (g⋅∘C) and Cs,silver=0.235J/ (g⋅∘C 1.What is the mass of the silver block? ΔH + = endothermic a) ΔH – b) ΔH + c) ΔH - ΔH - = exothermic ΔrH∘=−802.3kJmol−1 kJ released = 261 kJ molar mass of CH4 = 16.04 g/mol 261 x 1/802.3 x 16.04/1 = 5.218048112 = 5.21 (correct: 5.22g) H2O: Ti = 24.6 Tf = 26.5 ∆T= 1.9 Mass H2O= 100 g/ml Silver: Ti = 55.3 Mass of silver? c = 0.235 g x °C 100g H2O x 4.184 x 1.9 = 794.96 Mass x 0.235 x 55.3 -26.5 M = 794.96 / .233 x (55.3 – 26.5) = 117.4 (586288) Correct: 120 g Two substances, AA and BB, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance AA is 6.38 gg and its initial temperature is 20.0 ∘C∘C . The mass of substance BB is 25.8 gg and its initial temperature is 52.4 ∘C∘C . The final temperature of both substances at thermal equilibrium is 46.7 ∘C∘C . If the specific heat capacity of substance B is 1.17 Jg−1∘C−1Jg−1∘C−1, what is specific heat capacity of substance A? ma = 6.38 Ta = 20 mb = 25.8 tb = 52.4 Tb = 46.7 cpb = 1.17 cpa=? Qa = -Qb ma*cpa*(Tf-Ta) = -mb*cpb*(Tf-Tb) 6.19*cpa*(46.7-20.0) = -25.8*1.17*(46.7-52.4) Cpa = -25.8*1.17*(46.7-52.4) / ((46.7-20.0) *6.38) Cpa = 1.01 (0063048) J/gC Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.19 g and its initial temperature is 20.7 ∘C. The mass of substance B is 25.8 g and its initial temperature is 52.4 ∘C. The final temperature of both substances at thermal equilibrium is 46.9 ∘C. If the specific heat capacity of substance B is 1.17 J/g⋅∘C, what is the specific heat capacity of substance A? Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8C10H8). When 1.400 gg of naphthalene burns in a bomb calorimeter, the temperature rises from 26.431 ∘C∘C to 30.338 ∘C∘C. Find ΔrU for the combustion of naphthalene. The heat capacity of the calorime determined in separate experiment, is 5.112 kJ∘C−1kJ∘C−1. - Given: mass of Mothball = 1.400g Initial Temp: 26.431 Final temp: 30.338 Molar mass: 128.1705 g/mol n = 1.400/128.1705 = 0.01092295029 Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.325 g of naphthalene burns in a bomb calorimeter, the temperature rises from 25.493 ∘C to 30.720 ∘C. Find Δr for the combustion of naphthalene. The heat capacity of the calorimeter, determined in separate experiment, is 5.112 kJ∘C−1. (-2585 kJ/mol) 5.112 x (30.338 – 26.431) 5.112 x 3.907 = 19.972584 Now heat / mols 19.972584 / 0.01092295029 = -1828.497198 =-1828.5 kJ mol-1 (correct :P) Find ΔrHΔrH for the combustion of naphthalene at 298 KK. Express your answer using four significant figures. - ∆H = ∆Y + nxRxT ∆n = 10-2 = -2 mol R= 0.008314 kJ/molxK ∆H = ∆U +∆nxRxT = 1828.5 + (-4.955144) = -1833.455144 (correct :P) HELP: Find Δr for the combustion of naphthalene at 298K The human body obtains 925 kJkJ of energy from a candy bar. If this energy were used to vaporize water at 100.0 ∘C∘C, how much water (in could be vaporized? (Assume the density of water is 1.00 gmL−1gmL−1.) - Q= m x Hvap Hvap = 2257 kJ/kg Q = 925 925 = m x 2257 m = 0.41 = 410g Density = 1g/mL Volume = mass / density = 410/1g/mL = 410m: = 0.410 L: A 95.0 mLmL sample of water How much heat (in kJkJ) is required to vaporize it? (Assume a density of is heated to its boiling point. 1.00 g/mLg/mL.) - Hvap = 2260 J/g Mass = desity x volume = 95g x 2260 J/g =214700 J =214.7kJ Suppose that 0.91 gg of If the heat released during condensation goes only to warming the iron block, water condenses on a the final temperature (in Celsius) of the iron block? (Assume a constant enthal 85.0 gg block of iron that is vaporization for water of 44.0 kJmol−1kJmol−1, CFeCFe= 0.449 Jg−1∘C−1Jg−1∘ - G: initially at 20 ∘C Moles of water = mass of water/molar mass of water = 0.91/18.02 = 0.05049944506 mol Heat released by water condensing = moles of water x enthalpy of vaporization = 0.05049944506 x 44.0 x 103 = 2221.975583 J Let T be the final temperature of the iron block Heat absorbed by iron block = mass of iron block x specific heat of iron x temperature change How much energy is released when 42.5 gg of water freezes? How much heat (in kilojoules) is evolved in converting 1.00 mol of steam at 160.0 ∘C∘C to ice at -45.0 ∘C∘C? The heat capacity of steam is 2.01 Jg−1∘C−1Jg−1∘C−1 and of that ice is 2.09 Jg−1∘C− = 85.0 x 0.450 x (T - 20) heat absorbed = heat released 85.0 x 0.450 x (T - 20) = 2221.975583 2221.975583 = (85)(0.45)(t−20) 2221.975583 = ((85)(0.45))(t)+((85)(0.45))(−20)(Distribute) 2221.975583 = 38.25t+−765 2221.975583 = 38.25t−765 38.25t−765=2221.975583 38.25t−765+765=2221.975583+765 38.25t=2986.975583 38.25 T / 38.25 = 2986.975583 / 38.25 t=78.090865 First, let us explain what happens when water freeze: Water goes form liquid state to solid phase in the "freezing" point So Q = dH dH = Hfinal - Hinitial = Hsolid - Hliquid dHfreezing = Hsolid - Hliquid = 334.77 J/g so m = 42.5 g of water Q = m*HL = 42.5*334.77 J/g Q = 14,227.725 J Q = 14.227 kJ are released that is, water is losing energy in order to go solid state Specific = 2.09 ∆Hf = 6.020 kJ = heat (ci) 6020.0 j/mol Specific = 4.180 ∆Hv = 40.70 kj/mol heat of = 40700 j/mol water (Cw) Specific = 2.010 heat of steam (Cs) Mass of = 18.02 ice (m) Molar = 18.015 mass H2o Mols of Mass / 18.02 / 18.015 = 1.00 ice molar mol mass Initial 160°C temp T2 Final Temp T1 Q1 Q2 Q3 Q4 Q5 Q evolved Calculate the amount of heat required to completely sublime 78.0 gg of solid dry ice (CO2) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJmol−1kJmol−1. Zinc metal reacts with hydrochloric acid according to the following balanced -45 °C q = m x c x ∆T = m x Cs (160 -100) = 18.02 x 2.01 x 60 = 2172.609 = 2.172609 kJ = mass of water (when it is ice) x ∆H = 18.02 x 2172.609 = 39139.55114 J = 39.13955114 kJ = water from 100°C  0.0°C = m x Cw x ∆T = 18.02 x 4.180 x 100 =7530.27 J = 7.53027kJ =18.02 x 334 =6012 = 6.012 Ice (0°C )  -45°C = 18.02 x 2.09 x 45 = 1694.781 = 1.694781 = q1 + q2 + q3 + q4 + q5 = 2.172609 + 39.13955114 + 7.53027 + 6.012 + 1.694781 = 56.5421114 kJ (correct answer is 58.1) Molar mass of CO2 = 44 g/mol Moles of CO2 = 78 / 44 = 1.772 Heat required = (32.3 x 1.772) = 57.2590 (correct) equation. Zn(s)+2HCl(aq)→ZnCl2(aq) +H2(g)Zn(s) +2HCl(aq)→ZnCl2(aq)+H2(g) When 0.117 gg of Zn(s)Zn(s) is combined with enough HClHCl to make 53.0 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.0 ∘C∘C to 24.2 ∘C∘C . Find ΔrHΔrH for this reaction as written. (Use 1.0 gmL−1gmL−1 for the density of the solution and 4.184 Jg−1∘C−1Jg−1∘C−1 as the specific heat capacity.) Hess's law states that "the heat released or absorbed in a chemical process is the same whether the process takes place in one or in several steps." It is important to recall the following rules: 1. When two reactions are added, their enthalpy values are added. 2. When a reaction is reversed, the sign of its enthalpy value changes. 3. When the coefficients of a reaction are multiplied by a factor, the enthalpy value is multiplied by that same factor. Calculate the enthalpyΔfH∘A=33.2 of the x 2 = 66.4 kJ mol−1 reaction 2NO(g)+O2(g)→2NO2(g) ΔfH∘B=90.2 = -180.4 kJ mol−1 given the following reactions and enthalpies of formation: i) 12N2(g) +O2(g)→NO2(g), ΔfH∘A=33.2 kJ mol−1 ii) 12N2(g) +12O2(g)→NO(g), ΔfH∘B=90.2 kJ mol−1 Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: i) B2O3(s) +3H2O(g)→3O2(g) +B2H6(g), ΔrH∘A=+2035 kJ mol−1 ii) 2B(s)+3H2(g)→B2H6(g), ΔrH∘B=+36 kJ mol−1 iii) H2(g)+12O2(g)→H2O(l), ΔrH∘C=−285 kJ mol−1 iv) H2O(l)→H2O(g), ΔrH∘D=+44 kJ mol−1 Chemical energy is released or absorbed from reactions in various forms. The most easily measurable form of energy comes in the form of heat, or enthalpy. The enthalpy of a reaction can be calculated from the heats of formation of the substances involved in the reaction: ΔrH∘=Σ npΔfH∘(products)−Σ nrΔfH∘(reactants) where n represents the stoichiometric coefficients. -180.4 + 66.4 = -114 kJ mol−1 Final rxn 2NO(g)+O2(g)→2NO2(g) ΔfH∘ = -114 kJ mol−1 BUT they want it in J/ mol -114000 Δf H ∘ ∘ kJ⋅mol−1 ⋅ ⇌ ∘ kJ ⋅mol−1 For the reaction given in Part A, how much heat is absorbed when 3.00 mol of AA re ∆H rxn = 1 mol A = 1 mol x 287/ 2 mol Energy from 3 mols (287/ 2 mol) x 3 mol = 430.5 kJ ∘ ∘ The standard heat of formation, ΔfH∘, is defined as the enthalpy change for the formation of one mole of substance from its constituent elements in their standard states. Thus, elements in their standard states have ΔfH∘=0. Heat of formation values can be used to calculate the enthalpy change of any reaction. Consider, for example, the reaction -   - 2NO(g)+O2(g)⇌2NO2(g) with heat of formation values given by the following table: Substance (kJ mol−1) NO(g) 90.2 O2(g) 0 NO2(g) 33.2 Then the heat of formation for the overall reaction is ΔrH∘===∑npΔfH∘(products)2(3 3.2)−114 kJ mol−1− −∑nrΔfH∘(reactants)[2(90.2)+0] Ozone, O3, is destroyed when ClO levels are high. ClO is likely formed from the decomposition of chlorofluorocarbons (CFCs) in sunlight. Standard enthalpies of formation for selected substances are given in the table below. Substance (kJ mol−1) ClO(g) 101.0 ClO2(g) 102.0 O(g) 247.5 O2(g) 0 O3(g) 142.3 ∘ ΔfH∘ ∆H = [(3 x -393.5)+(4x-241.8)] – [(-1180.5)+(-967.2)] – [-104.7] =-2043 kJ⋅mol−1kJ⋅mol−1 - Calculate the standard enthalpy change of the reaction - Calculate the standard enthalpy change of the reaction ΔfH∘ - - What is the enthalpy for the overall reaction? ∆H = ∆H1 + ∆H2 ΔrH∘ -389.8 kJ m = ol Use standard enthalpies of formation to calculate ΔrH∘ΔrH∘ for the following reaction. 1. C2H5OH(l) +3O2(g)→2CO2(g) +3H2O(g) 2. SO2(g) +1/2O2(g)→SO3(g) 3. C(s) +H2O(g)→CO(g) +H2(g) 4. N2O4(g) +4H2(g)→N2(g) +4H2O(g) For all reactions: 1. 2. 3. 4. -1235.5 -98.89 131.293 -976.44 In chemical reactions, heat is converted into chemical energy (the potential energy stored in chemical bonds) or vice versa. Bond energy is the energy required to break one mole of the bond in the gas phase. Since it takes energy to break a bond, bond energies are always positive. Conversely, energy is released when bonds are formed. Thus, the enthalpy change for a reaction can be approximated from ΔrH=∑(ΔHbreaking) +∑(ΔHforming) where H represents bond energies for the breaking (positive bond energy) or forming (negative bond energy) of a bond and Hr represents the overall enthalpy for the reaction. Calculate the bond energy for all the bonds in a mole of a) methane, CH4 4 mol x 414 = 1656 kJ/mol b) O2 molecules? 1 mol x 498 = 498 kJ/mol c) H2O 2 mol x 464 = 928 kJ Becomes (-) because the bond formation energy = and opposite to the bond dissociati energy = - 928 kJ d) CO2 2 mol x 799 = 1598 kJ Bond formation energy = and opposite to bond dissociation so the energy would be its = -1598 Calculate the approximate enthalpy change, ΔrHΔrH, for the combustion of methane: CH4+2O2→2H2O+CO2 ΔrH = = = ΔHCH4+2×ΔHO2+2×ΔHH2O+ΔHCO2 1656+2×498+2×(−928)+(−1598) −802 kJ mol−1 Use the table to answer questions about bond energies. Bond Bond energy (kJ mol−1 ) C−C 347 C=C 611 C−H 414 C−O 360 C=O 799 O−O 142 O=O 498 H−O 464 Consider benzene in the gas phase, C6H6(g)C6H6(g). - What is the standard enthalpy change for the reaction that represents breaking all bonds in gaseous benzene, C6H6(g)C6H6(g)? 6 6 o Use the heat of formation, ΔfH∘ΔfH∘, values below to answer the questions. Substance ΔfH∘ΔfH∘ (kJ mol−1kJ mol−1 ) o - reaction f o f o f o 6 6 reaction What is the chemical equation for the reaction that corresponds to breaking just th