Question

Consider the reaction

$\mathbf{2}{\mathbf{POCl}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\to \mathbf{2}{\mathbf{PCl}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

1. Calculate $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$for this reaction. The $\mathit{\Delta }{\mathit{G}}_{\mathit{R}}^{\mathit{0}}$values for $\mathbf{2}\mathit{P}\mathit{O}\mathit{C}{\mathit{l}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$and $\mathbf{2}\mathit{P}\mathit{C}{\mathit{l}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$are $\mathbf{\backslash (}\mathbf{}\mathbf{-}\mathbf{}\mathbf{502}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{\backslash )}$and $\mathbf{\backslash (}\mathbf{}\mathbf{-}\mathbf{}\mathbf{270}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{\backslash )}$, respectively.
2. Is this reaction spontaneous under standard conditions at 298K?
3. The value of $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$for this reaction is 179J/K . At what temperatures is this reaction spontaneous at standard conditions? Assume that $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$ and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathit{}$do not depend on temperature.

Short answer

$\Delta G_r°=464kJ/\text{mol}$, non-spontaneous and temperature more than $T =2890K$are the solutions.

Step-by-step solution

 Calculation for ∆G0 for a

In the calculation,

$$\begin{gathered} 2POC{l}_{3\left(g\right)}\to 2PC{l}_{3\left(g\right)}+O_{2\left(g\right)} \\ \Delta G_R°=2\left[\Delta G_f°\left(PC{l}_{3\left(g\right)}\right)+1\Delta G_f°\left(O_{2\left(g\right)}\right)\right]-\left[1\Delta G_f°\left(POC{l}_{3\left(g\right)}\right)\right] \\ =\left[2\text{mol}-270\text{kJ/mol + 1mol}\times \text{0kJ/mol}\right]-\left[2\text{mol}\times -502\text{kJ/mol}\right] \\ \Delta G_R°=464KJ \end{gathered}$$

 Calculation for b

The chemical reaction is an example of a non-spontaneous reaction

 Calculation for C

$\Delta T° \text{For} \text{the} \text{reaction} T>2890K$

From part a$\Delta G_R^0=53kJ\mathit{,}\mathit{}\Delta H_{}^{\mathit{0}}=100kJ$

$$\begin{gathered} T=(25+273)K \\ =298K \end{gathered}$$

In the calculation,

$$\begin{gathered} \Delta H^o=\Delta G_{}^0+T\Delta S^o \\ =464kJ+\left(298K\right)(179kJ/K) \\ =517.34kJ \\ \Delta H_{}^{\mathit{0}}=T\Delta S_{}^{\mathit{0}} \\ =\frac{\mathit{\Delta }{\mathit{H}}_{}^{\mathit{0}}}{\mathit{\Delta }{\mathit{S}}_{}^{\mathit{0}}} \\ =\frac{\mathit{100}}{\mathit{0}\mathit{.}\mathit{1577}} \\ \Delta H_{}^{\mathit{0}}=634K \end{gathered}$$

Equilibrium at $\Delta G°=0$

$$\begin{gathered} \Delta H_{}^0=T\Delta S^o \\ T =\frac{\Delta H^o}{\Delta S_{}^0} \\ =\frac{517.34kJ}{0.179kJ/K} \\ T =2890K \end{gathered}$$

Conclusion

Therefore, $\Delta G_r°=464kJ/\text{mol}$, non-spontaneous and temperature more than$T=2890K$ are the solutions.