Recall that \(\Delta G^{\circ}\) can be written as a function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\). Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not temperature dependent and answer each of the following: a) Derive an expression relating \(\ln K, \Delta H^{\circ}\), and \(\Delta S^{\circ} .\) That is, derive an expression that looks like \(\ln K=\) some function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ} .\) The temperature, \(T\), should appear only once in this equation. b) How are the equilibrium constants for reactions with \(\Delta H^{\circ}>0\) affected by an increase in temperature? c) How are equilibrium constants for reactions with \(\Delta H^{\circ}<0\) affected by an increase in temperature?

Step by step solution

01

Derive the required expression

Begin with the standard Gibbs free energy equation: \( \Delta G^\circ = -RTlnK \). Using the relation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \), we can arrange the equations and substitute to obtain \( \ln K = - \frac{\Delta H^\circ}{R} +\frac{\Delta S^\circ}{R} \).

02

Discuss effect of temperature increase for \(\Delta H^{\circ}>0\)

In this case, the reaction is endothermic and the change in enthalpy is \(>\) 0. An increase in temperature will increase \( S \) which consequently increases the value of \( \ln K \). Therefore, for endothermic reactions, an increase in temperature tends to increase the equilibrium constant.

03

Discuss effect of temperature increase for \(\Delta H^{\circ}

Here, the reaction is exothermic, and the change in enthalpy is \(<\) 0. An increase in temperature will decrease \( S \) which in turn decreases the value of \( \ln K \). Therefore, for exothermic reactions, an increase in temperature tends to decrease the equilibrium constant.

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