Question

$K$At $25.0^{o}C$for the reaction

The values of $\Delta H^o$and $\Delta S^o$are $-58.03kJ/mol$and $-176.6J{K}^{-1}mo{l}^{-1}$, respectively. Calculate the value of$K$ at $25.0^{o}C$. Assuming role="math" localid="1658203617877" $\Delta H^o$and$\Delta S^o$ are temperature-independent, estimate the value of $K$at $100.0^{o}C$.

Short answer

The value of$K$ at ${25.0}^{\circ }C$.and$\Delta H^o$ and $\Delta H^o$are temperature – independent, $K$at the value of$100.0^{o}C$ at is $K_2=7.96\times {10}^{-2}$.

Step-by-step solution

Calculation for

The relation between standard Gibbs energy, entropy and enthalpy for a system is given below.

$\begin{array}{c}\Delta G^o=\Delta H^o-T\Delta S^o\\ =-58.03kJ/mol-(298K\times (-176.6J/Kmol))\\ =-5.40kJ/mol\end{array}$

Calculation for equilibrium constant K

$\begin{array}{c}\Delta G_R^0=-RTlnK\\ K=e^{-\frac{\Delta G_R^0}{RT}}\\ K=e^{-\frac{-5.40kJ/mol}{0.008314kJ/mol-K\times 298K}}\\ K=8.84\end{array}$

Calculation of K at T=100oC

Based of equation

$\mathrm{l}n\frac{K_2}{K_1}=\frac{\Delta H^o}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$\ln \frac{K_2}{8.84}=\frac{-58.03kJ/mol}{0.008314kJ/molK}(\frac{1}{(25+273)K}-\frac{1}{(100+273)K})$

$\ln \frac{K_2}{8.84}=\frac{-58.03kJ/mol}{0.008314kJ/molK}(\frac{1}{\left(298\right)K}-\frac{1}{\left(373\right)K})$

$ln\frac{K_2}{8.84}=-4.709$

$\begin{array}{l}{K}_2=e^{4.709}\times 8.84\\ K_2=7.96\times {10}^{-2}\end{array}$

Conclusion

Therefore, the value of$K$ at$25.0^{o}C$ .and $\Delta H^o$and $\Delta S^o$are temperature -independent, at the value of $K$at$100.0^{o}C$ is $K_2=7.96\times {10}^{-2}$.