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Chapter 18: Problem 83

Derive the following equation $$ \Delta G=R T \ln (Q / K) $$ where \(Q\) is the reaction quotient and describe how you would use it to predict the spontaneity of a reaction.

Short Answer

Expert verified
\(\Delta G\) can be expressed as \(\Delta G=RT \ln (Q / K)\). \(\Delta G<0\) indicates a spontaneous reaction, \(\Delta G>0\) indicates a non-spontaneous reaction, and \(\Delta G=0\) indicates the reaction is at equilibrium.

Step by step solution

01

Starting Point

Start with the definition of \(\Delta G\). \(\Delta G=\Delta G^{\circ} + RT \ln Q\), where \(\Delta G^{\circ}\) is the standard Gibbs free energy change, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(Q\) is the reaction quotient.
02

Express standard Gibbs free energy change

\(\Delta G^{\circ}\) can also be expressed in terms of the equilibrium constant as \(\Delta G^{\circ}=-RT \ln K\). Substitute this back into the first equation to obtain \(\Delta G = -RT \ln K + RT \ln Q\).
03

Combine the logarithms

Using the property of logarithms \(\ln a - \ln b = \ln (a/b)\), the equation can be simplified to \(\Delta G=RT \ln (Q / K)\)
04

Interpretation

The sign of \(\Delta G\) can be used to predict whether a reaction will be spontaneous. If \(\Delta G\) is negative, the reaction proceeds spontaneously in the forward direction. If \(\Delta G\) is positive, the reaction is non-spontaneous and proceeds spontaneously in the reverse direction. If \(\Delta G\) is zero, the reaction is at equilibrium.

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Most popular questions from this chapter

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{f}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$. (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\).

A carbon monoxide (CO) crystal is found to have entropy greater than zero at absolute zero of temperature. Give two possible explanations for this observation.

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)
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