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Chapter 19: Problem 75

Indicate whether \(\Delta G\) increases, decreases, or stays the same for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(\mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$ (c)

Short Answer

Expert verified
In summary, for an increase in the partial pressure of O₂: - In reaction (a), ∆G increases. - In reaction (b), ∆G decreases. - In reaction (c), ∆G decreases.

Step by step solution

01

A) Analyze Reaction (a)

The reaction in this problem is: \[ \mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{O}_{2}(g) \] As the partial pressure of O₂ is increased, the only species in the reaction affected is O₂ as it is the only species in gaseous form. The reaction quotient Q for this reaction is: \[ Q = \frac{P_{\mathrm{O}_2}}{1} \] As the partial pressure of O₂ is increased, Q also increases.
02

B) Determine the direction of Reaction (a)

For a reaction to be spontaneous, the Gibbs free energy change must be negative: \[ \Delta G < 0 \] If Q is less than the equilibrium constant K (Q < K), the forward reaction is spontaneous (favoring products), and ∆G will be negative. As ∆G becomes less negative, it means it's increasing. Since Q is increasing, we infer that, under the given conditions, ∆G will increase.
03

C) Analyze Reaction (b)

The reaction in this problem is: \[ 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \] The reaction quotient Q for this reaction is: \[ Q = \frac{P_{\mathrm{SO_3}}^2}{P_{\mathrm{SO_2}}^2 \times P_{\mathrm{O_2}}} \] As the partial pressure of O₂ is increased, the denominator of Q increases.
04

D) Determine the direction of Reaction (b)

If the denominator of Q increases, Q decreases. If Q < K, the forward reaction is spontaneous (favoring products), and ∆G will be negative. Since Q is decreasing, we infer that, under the given conditions, ∆G will decrease.
05

E) Analyze Reaction (c)

The reaction in this problem is: \[ 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \] The reaction quotient Q for this reaction is: \[ Q = \frac{1}{P_{\mathrm{H_2}}^2 \times P_{\mathrm{O_2}}} \] As the partial pressure of O₂ is increased, the denominator of Q increases.
06

F) Determine the direction of Reaction (c)

If the denominator of Q increases, Q decreases. If Q < K, the forward reaction is spontaneous (favoring products), and ∆G will be negative. Since Q is decreasing, we infer that, under the given conditions, ∆G will decrease.
07

G) Conclusion:

In summary, for an increase in the partial pressure of O₂: - In reaction (a), ∆G increases. - In reaction (b), ∆G decreases. - In reaction (c), ∆G decreases.

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Most popular questions from this chapter

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix \(D\). (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b}\). (b) By using the value of \(K_{b}\), calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when $\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}$ and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

Consider the reaction $\mathrm{CH}_{4}(\mathrm{~g})+4 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+$ \(4 \mathrm{HCl}(g) .\). (a) Using data from Appendix C, calculate $\Delta G^{\circ}\( at \)298 \mathrm{~K} .(\mathbf{b})\( Calculate \)\Delta G\( at \)298 \mathrm{~K}\( if the reaction mixture consists of \)50.7 \mathrm{kPa}$ of \(\mathrm{CH}_{4}(g), 25.3 \mathrm{kPa}\) of $\mathrm{Cl}_{2}(g), 10.13 \mathrm{kPa}$ of \(\mathrm{CCl}_{4}(\mathrm{~g})\) and \(15.2 \mathrm{kPa}\) of \(\mathrm{HCl}(\mathrm{g})\)

Indicate whether each statement is true or false. (a) The third law of thermodynamics says that the entropy of a perfect, pure crystal at absolute zero increases with the mass of the crystal. (b) "Translational motion" of molecules refers to their change in spatial location as a function of time. (c) "Rotational" and "vibrational" motions contribute to the entropy in atomic gases like He and Xe. (d) The larger the number of atoms in a molecule, the more degrees of freedom of rotational and vibrational motion it likely has.

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) or \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), (b) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (c) \(1 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(g)\) or $2 \mathrm{~mol} \mathrm{NO}_{2}(g)$ (d) \(\mathrm{HCl}(g)\) or \(\mathrm{HCl}(a q) .\) Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond $)=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /$ $\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\( a n d \)\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /$ mol-K. All but \(S\) n have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?
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