Chemistry Answer Note #19

Question 1 Use the following reactions and their equilibrium constants to predict the equilibrium constant for this reaction 2A(s)⇌3D(g) Answer To predict the equilibrium constant for the given reaction, we first need to understand the relationship between equilibrium constants and the individual reactions. According to the Law of Mass Action, the equilibrium constant (K) for a chemical reaction is equal to the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. Based on this, we can express the given reactions as: A(s) ⇌ 12B(g) + C(g), K1= [B]^12 [C] / [A] 3D(g) ⇌ B(g) + 2C(g), K2= [B] [C]^2 / [D]^3 Multiplying the two reactions together, we get: 2A(s) ⇌ 24B(g) + 6C(g) + 6B(g) + 12C(g), K = (K1)x(K2) = [B]^13 [C]^3 / [A][D]^3 Using the given equilibrium constants, K1=0.0334 and K2=2.35, we can plug in the values to the above equation and solve for K: K = [0.0334]^13 [2.35]^3 / [1][1]^3 = 3.17 Therefore, the predicted equilibrium constant for the given reaction, expressed to three significant figures, is 3.17. Question 2 . The conversion of ethene to ethane in the gas phase is found experimentally to be first order reaction with a rate constant of 2.8.10-4 S-1 at 25°C i. If the initial concentration of ethene was 0.125 mol/dm3, calculate its concentration after 12.0 minutes ii. Determine the time taken for the concentration of ethene to decrease from 0.125 mol/dm3to 0.0500 mol/dm3 Answer According to the given information, the conversion of ethene to ethane in the gas phase is a first order reaction with a rate constant of 2.8.10-4 S-1 at 25°C. This means that the rate of the reaction is directly proportional to the concentration of ethene, and the order of the reaction is 1. i. To calculate the concentration of ethene after 12.0 minutes, we can use the first order rate equation: ln([A]t/[A]0) = -kt where [A]t is the concentration of ethene at time t, [A]0 is the initial concentration of ethene, k is the rate constant, and t is the time. Plugging in the given values, we get: ln([A]t/0.125) = -(2.8.10-4)(12.0) Solving for [A]t, we get the concentration of ethene after 12.0 minutes to be 0.0116 mol/dm3. ii. To determine the time taken for the concentration of ethene to decrease from 0.125 mol/dm3 to 0.0500 mol/dm3, we can rearrange the first order rate equation to solve for time: t = ln([A]0/[A]t)/k Plugging in the given values, we get: t = ln(0.125/0.0500)/2.8.10-4 Solving for t, we get the time taken for the concentration of ethene to decrease from 0.125 mol/dm3 to 0.0500 mol/dm3 to be 19.23 minutes. Question 3 .A burring magnesium ribbon was placed in a jar containing carbon (IV) oxide: i. Write an equation for the reaction ii. Explain briefly why the magnesium ribbon burn in the carbon (IV) oxide although the gas does not support combustion Answer Introduction: Burring magnesium ribbon is a widely used demonstration in chemistry laboratories to showcase the reactivity of metals with gases. When placed in a jar containing carbon (IV) oxide, the magnesium ribbon undergoes a vigorous reaction, producing a bright flame and white ash. In this answer, I will provide an equation for the reaction and explain the reason behind the burning of magnesium in carbon (IV) oxide with the help of academic sources. Equation for the reaction: The reaction between burring magnesium ribbon and carbon (IV) oxide can be represented by the following equation: 2Mg(s) + CO2(g) → 2MgO(s) + C(s) Explanation: There are several factors that contribute to the burning of magnesium ribbon in carbon (IV) oxide, even though the gas itself does not support combustion. Some key explanations for this reaction are as follows: 1. High affinity towards oxygen: Magnesium has a high affinity towards oxygen, making it a highly reactive metal. When exposed to oxygen, magnesium readily forms magnesium oxide (MgO), a stable compound. In the above equation, we can see that magnesium reacts with carbon (IV) oxide (CO2) to form magnesium oxide (MgO), thus explaining the white ash produced during the reaction. 2. Thermodynamic instability: The thermodynamic instability of carbon (IV) oxide also contributes to the reaction between magnesium ribbon and CO2. According to the Third Law of Thermodynamics, all substances tend to form more stable compounds. Since magnesium oxide has a higher stability than carbon (IV) oxide, the reaction between them is favored. 3. Endothermic nature of the reaction: The burning of magnesium in carbon (IV) oxide is an endothermic process, meaning it absorbs heat from the surroundings. This heat is provided by the oxidation reaction that occurs between magnesium and carbon (IV) oxide. As a result, the reaction produces a bright flame, making it an excellent demonstration of spontaneous oxidation. Conclusion: In conclusion, the reaction between burring magnesium ribbon and carbon (IV) oxide is a result of several factors that contribute to the high reactivity of magnesium towards oxygen and the thermodynamic instability of carbon (IV) oxide. This burning of magnesium in carbon (IV) oxide, although the gas itself does not support combustion, is an excellent example of a spontaneous exothermic reaction and is widely used as a demonstration in chemistry laboratories. Question 4 . Consider the enthalpy change of formation values given below Compound C2H6O(l) CO2 H2O(l) ΔH◦f/kJ/mol -227.7 -393.5 -286.0 Determine the enthalpy change for the reaction when ethanol is completely burnt in excess oxygen Answer The enthalpy change of a reaction refers to the amount of heat released or absorbed during the course of the reaction. In this case, we are looking at the enthalpy change when ethanol (C2H6O) is completely burnt in excess oxygen. In order to determine the enthalpy change for this reaction, we need to use the enthalpy change of formation values given below: Compound C2H6O(l) CO2 H2O(l) ΔH◦f/kJ/mol -227.7 -393.5 -286.0 In order to determine the enthalpy change of the reaction, we can use the following equation: ΔH◦rxn = ΣnΔH◦f(products) - ΣnΔH◦f(reactants) Where ΔH◦rxn is the enthalpy change of the reaction, ΣnΔH◦f(products) is the sum of the enthalpy change of formation values for the products, and ΣnΔH◦f(reactants) is the sum of the enthalpy change of formation values for the reactants. So, in this case, ΔH◦rxn = [ΔH◦f(CO2) + ΔH◦f(H2O)] - ΔH◦f(C2H6O) Substituting the given values, we get: ΔH◦rxn = [(-393.5) + (-286.0)] - (-227.7) = -393.5 - 286.0 + 227.7 = -452.8 kJ/mol Therefore, the enthalpy change for the reaction when ethanol is completely burnt in excess oxygen is -452.8 kJ/mol. Question 5 Arrange the following substances in increasing boiling point H2O, NaCl, sio2, HF, HCl, CH3CH3, CH2COOH Answer The substances provided can be arranged in increasing boiling point by considering the intermolecular forces present in each substance. The stronger the intermolecular forces, the higher the boiling point. From strongest to weakest, the intermolecular forces are as follows: ion-ion, hydrogen bonding, dipole-dipole, and London dispersion forces. Based on this, the arrangement of the substances in increasing boiling point would be: NaCl < CH3CH3 < CH2COOH < HF < HCl < H2O < SiO2 The substances H2O, HF, and HCl exhibit hydrogen bonding, which is stronger than the dipole-dipole forces present in CH3CH3 and CH2COOH. This is why they have higher boiling points. SiO2 has a higher boiling point due to its network covalent structure and strong London dispersion forces.