From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at 298 \(\mathrm{K}\) . If the reaction is not spontaneous under standard conditions at 298 \(\mathrm{K}\) , at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l}{\text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)} \\ {\Delta H^{\circ}=-844 \mathrm{kk} ; \Delta S^{\circ}=-165 \mathrm{J} / \mathrm{K}} \\\ {\text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)} \\ {\Delta H^{\circ}=572 \mathrm{kJ} ; \Delta S^{\circ}=179 \mathrm{J} / \mathrm{K}}\end{array} $$

Step by step solution

01

Calculate \(\Delta G^{\circ}\) at 298 K

Using the given values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and the equation for Gibbs free energy change: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} = -844 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (-165\,\mathrm{J} /\mathrm{K})) \] Convert the entropy change in J/K to kJ/K before performing the calculation: \[ \Delta G^{\circ} = -844 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (-0.165\,\mathrm{kJ} /\mathrm{K})) = -844\,\mathrm{kJ} + 49.17\,\mathrm{kJ} = -794.83\,\mathrm{kJ} \]

02

Determine, if the reaction is spontaneous at 298 K

Since the Gibbs free energy change is negative (\(\Delta G^{\circ} = -794.83\,\mathrm{kJ}\)) at 298 K, the reaction is spontaneous at this temperature. ##Reaction (b)##

03

Calculate \(\Delta G^{\circ}\) at 298 K

Using the given values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and the Gibb's free energy change equation: \[ \Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}= 572 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (179\,\mathrm{J}/\mathrm{K})) \] Convert the entropy change in J/K to kJ/K before performing the calculation: \[ \Delta G^{\circ} = 572 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (0.179\,\mathrm{kJ}/\mathrm{K}))= 572\,\mathrm{kJ} - 53.34\,\mathrm{kJ} = 518.66\,\mathrm{kJ} \]

04

Determine if the reaction is spontaneous at 298 K

Since the Gibbs free energy change is positive (\(\Delta G^{\circ}= 518.66\,\mathrm{kJ}\)) at 298 K, the reaction is not spontaneous at this temperature.

05

Find the temperature at which the reaction becomes spontaneous

To find the temperature at which the reaction becomes spontaneous, we're looking for the temperature at which \(\Delta G^{\circ} = 0\). So, set the equation for Gibbs free energy change equal to 0 and solve for T: \[ 0 = \Delta H^{\circ} - T \Delta S^{\circ} \Longrightarrow T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{572\,\mathrm{kJ}}{0.179\,\mathrm{kJ}/\mathrm{K}} \approx 3197\,\mathrm{K} \] The reaction (b) will become spontaneous at approximately 3197 K.

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