Question

The equilibrium constant K for the reaction

$2C{l}_{\left(g\right)}\rightleftarrows C{l}_{2\left(g\right)}$

was measured as a function of temperature (in Kelvins). A graph of In( ) versus $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$ for this reaction gives a straight line with a slope of $1.352\times {10}^{4}K$ and y intercept of $-14.51$. Determine the value of $\Delta H^o$and $\Delta S^o$for this reaction. (See Exercise 88)

Short answer

The solution for $\Delta H^o$and $\Delta S^o$ is$-112.4kJ/mol$ and $-120.6kJ/mol$ respectively

Step-by-step solution

Description of Solution on Slope

$2C{l}_{\left(g\right)}\rightleftarrows C{l}_{2\left(g\right)}$

According to the plot ofIn($K$) Versus $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$

$\text{slope\hspace{0.17em}}=\frac{{\text{-ΔH}}^{\text{o}}}{\text{R}}$

$\begin{array}{c}\frac{-\Delta H^{\circ }}{R}=1.352\times {10}^{4}K\\ \Delta H^{\circ }=-1.352\times {10}^{4}K\times 0.008314kJ/\text{mol}-K\\ \Delta H^{\circ }=-112.4kJ/\text{mol}\end{array}$

Description of Solution on Intercept

$\text{intercept}=\frac{{\text{ΔS}}^{\text{o}}}{\text{R}}$

$\begin{array}{c}\frac{\Delta S^{\circ }}{R}=-14.51\\ \Delta S^{\circ }=-14.51\times 0.008314kJ/\text{mol}-K\\ \Delta S^{\circ }=-120.6kJ/\text{mol}\end{array}$

Conclusion

The solution for$\Delta H^o$ and $\Delta S^o$ is $-112.4kJ/mol$and $-120.6kJ/mol$ respectively.