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? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

Step by step solution

01

(a) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = -844 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = -165 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = -844\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(-165 \, \mathrm{J/K}) $$ Now we'll calculate the value of delta G: $$ \Delta G^{\circ} = -764820 \, \mathrm{J} = -764.82 \, \mathrm{kJ} $$

02

(a) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} < 0\), the reaction is spontaneous at \(298 \mathrm{~K}\). Therefore, there is no need to calculate the temperature at which the reaction becomes spontaneous.

03

(b) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = 572 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = 179 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 572\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(179 \, \mathrm{J/K}) $$ Calculating the value of delta G: $$ \Delta G^{\circ} = 161662 \, \mathrm{J} = 161.662 \, \mathrm{kJ} $$

04

(b) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} > 0\), the reaction is not spontaneous at \(298 \mathrm{~K}\). We need to find the temperature at which the reaction becomes spontaneous.

05

(b) Find the temperature for spontaneous reaction

To find the temperature at which \(\Delta G^{\circ} = 0\), we will plug the value of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) into the formula, and solve for T: $$ 0 = 572\times10^3 \, \mathrm{J} - T(179 \, \mathrm{J/K}) $$ Now solve for T: $$ T = \frac{572\times10^3 \, \mathrm{J}}{179 \, \mathrm{J/K}} \approx 3196 \, \mathrm{K} $$ So, the reaction becomes spontaneous at a temperature of approximately \(3196 \mathrm{~K}\).

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