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Chapter 1: Problem 59

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

Short Answer

Expert verified
Answer: (D) The value of \(ΔG°\) would remain the same because it doesn't change with variations in concentration of the reactants, only the equilibrium position shifts, but the equilibrium constant \(K\) and hence \(ΔG°\) remain the same.

Step by step solution

01

Understanding Gibbs Free Energy and Equilibrium

First note, that the Gibbs Free Energy relationship with the equilibrium constant is given by the formula \(ΔG° = -RTlnK\), where \(ΔG°\) is the standard Gibbs free energy change, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant. Examining this equation tells us that at any given temperature \(T\), and \(R\) being a constant, \(ΔG°\) can only change if \(K\) changes.
02

Reactant Concentration and Equilibrium

The equilibrium constant \(K\) is determined by the specific reaction and the temperature, it is not influenced by the concentrations of the reactants or the products. In this exercise, the concentration of nitrogen (\(N_{2}\)) is supposed to be increased. However, this increase doesn't lead to a shift in the equilibrium constant \(K\), it only leads to a shift in the position of the equilibrium according to Le Chatelier's principle, by moving the equilibrium to the right.
03

Relation with Gibbs Free Energy

Since we established that an increase in the concentration of nitrogen will not change the equilibrium constant \(K\), it means that it also doesn't change the standard Gibbs free energy change \(ΔG°\). Any changes in reactant or product concentrations may shift the equilibrium position, but they do not alter the value of ∆G° for the reaction.

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Most popular questions from this chapter

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Which gas has the strongest IMFs? (A) He (B) Ne (C) NO (D) All gases have identical IMFs.
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