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

Obtaining the Gibbs Free Energy Equation

The formula for Gibbs free energy is defined as follows: \[\Delta G = \Delta H - T\Delta S\]where \(\Delta G\) = Gibbs free energy change, \(\Delta H\) = Enthalpy change, \(\Delta S\) = Entropy change,and \(T\) = Temperature in Kelvin.

02

Deriving an expression relating lnK, ΔHo, and ΔSo

Recall that at equilibrium, \(\Delta G = 0\), yielding:\[0 = \Delta H - T\Delta S\]Solving for \(\Delta S\) gives us \(\Delta S = \Delta H / T\).We also know that \(\Delta G = -RT\ln(K)\), where \(R\) is the ideal gas constant and \(K\) is the equilibrium constant. At equilibrium, this becomes \(0 = -RT\ln(K)\), yielding \(\ln(K) = -\Delta H / (RT)\), substituting \(\Delta S\) into the equation we get \(\ln(K) = -\Delta S\).So, the expression relating \(\ln K\), \(\Delta H^{\circ}\), and \(\Delta S^{\circ}\) is \(\ln(K) = \Delta S - \Delta H / T\).

03

Determine the effect of temperature on lnK when ΔH > 0

In the derived expression, \(\ln(K) = \Delta S - \Delta H / T\), when \(\Delta H\) is greater than 0 (endothermic reaction) and the temperature increases, the term \(-\Delta H / T\) decreases. Therefore, \(\ln K\) will increase, meaning that the equilibrium constant increases when the temperature is increased for reactions with a positive enthalpy.

04

Determine the effect of temperature on lnK when ΔH < 0

In the derived expression, \(\ln(K) = \Delta S - \Delta H / T\), when \(\Delta H\) is less than 0 (exothermic reaction) and the temperature increases, the term \(-\Delta H / T\) increases, leading to a decrease in \(\ln K\). Therefore, the equilibrium constant decreases when the temperature is increased for reactions with a negative enthalpy.

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