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Chapter 19: Problem 79

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

Short Answer

Expert verified
(a) To calculate the standard Gibbs free energy change (ΔG⁰), first find the standard enthalpy change (ΔH⁰) and standard entropy change (ΔS⁰) using the data in Appendix C: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) and ΔS⁰(reaction) = Σ ΔS⁰(products) - Σ ΔS⁰(reactants). Then, use the equation ΔG⁰ = ΔH⁰ - TΔS⁰ to find the standard Gibbs free energy change. (b) To calculate ΔG at 298 K given the partial pressures, first find the reaction quotient (Q) using the formula Q = (P(N₂O₄))/(P(NO₂)²) with the given partial pressures of NO₂ and N₂O₄ as 0.40 atm and 1.60 atm, respectively. Next, use the equation ΔG = ΔG⁰ + RT ln(Q) to find ΔG, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K).

Step by step solution

01

Find the standard enthalpy change and standard entropy change

Look up the standard enthalpy change (ΔH⁰) and the standard entropy change (ΔS⁰) for the reaction in Appendix C. We can find the values for the individual species and use the equation ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) and ΔS⁰(reaction) = Σ ΔS⁰(products) - Σ ΔS⁰(reactants).
02

Calculate the standard Gibbs free energy change

Use the equation ΔG⁰ = ΔH⁰ - TΔS⁰ to find the standard Gibbs free energy change. (b) Calculate ΔG at 298K given the partial pressures.
03

Find the reaction quotient (Q)

The reaction quotient Q is defined as the ratio of the product of partial pressures of products to the product of partial pressures of reactants. In this case, Q = (P(N₂O₄))/(P(NO₂)²). Given the partial pressures of NO₂ and N₂O₄ as 0.40 atm and 1.60 atm respectively, calculate the value of Q.
04

Calculate ΔG using the reaction quotient

Use the equation ΔG = ΔG⁰ + RT ln(Q) to find ΔG. R is the gas constant (8.314 J/mol K), T is the temperature (298 K), and Q is the reaction quotient calculated in step 3. Following these steps above will give you the answers for ΔG⁰ and ΔG at 298 K for the given reaction and partial pressures.

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Most popular questions from this chapter

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ},\) as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Use Appendix \(\mathrm{C}\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g)\), \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q)\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol} \mathrm{P}_{2}(g)\), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values.

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K\). (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s,\) graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)
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