Question

The relationship between \(\Delta G^{\circ}\) and \(T\) is linear. Draw the graph of a reaction with the properties described below. You need only label the point at which \(\Delta G^{\circ}=0\). The reaction is exothermic. \(\Delta n_{\mathrm{g}}<0\left(\Delta n_{\mathrm{g}}=\right.\) moles of gas products \(-\) moles of gas reactants). At \(300 \mathrm{~K},\) the system is at equilibrium and \(K=1\).

Short answer

The temperature at which ΔG°=0 for the given exothermic reaction with Δng<0 is 300 K. This is because, at equilibrium with K=1, ΔG° becomes 0 according to the equation ΔG° = -RT ln K.

Step-by-step solution

Recall the mathematical relationship between \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and \(T\)

The relationship between Gibbs free energy change, enthalpy change, entropy change, and temperature is given by the following equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] We know that the reaction is exothermic, which means \(\Delta H^{\circ} < 0\). As \(\Delta n_{\mathrm{g}}<0\), the entropy change must be negative, which means \(\Delta S^{\circ} < 0\).

Determine \(\Delta G^{\circ}\) when \(K=1\)

At equilibrium, we can use the relationship between Gibbs free energy change and the equilibrium constant K: \[\Delta G^{\circ} = -RT \ln K\] Since at \(300\mathrm{~K}\), the system is at equilibrium and \(K=1\), we can substitute these values into the equation: \[\Delta G^{\circ}_{300} = - (8.314)(300) \ln1\] As \(\ln1 = 0\), we get \(\Delta G^{\circ}_{300} = 0\).

Draw the graph

Plot the Temperature \(T\) on the x-axis and Gibbs free energy change \(\Delta G^{\circ}\) on the y-axis. The graph of the reaction will be linear. Since we know that \(\Delta G^{\circ}_{300}=0\), we can label this point on the graph as \((300,0)\). Now, considering that the reaction is exothermic and \(\Delta n_{\mathrm{g}}<0\), the line should have a positive slope. In summary, the graph is a straight line with positive slope, having a point \((300,0)\) where \(\Delta G^{\circ}=0\).