Chemistry - Chemical Equilibrium Question with Solution | TestHub

For the reaction, $$K_p=p_{{\mathrm{CO}}_2}\text{ and }{K}_c=\left[{\mathrm{CO}}_2\right]$$ $(\because \left[{\mathrm{CaCO}}_3\right]=1$ and $[\mathrm{CaO}]=1$ for solids $]$ According to Arrhenius equation we have $$K=A{e}^{-\Delta H_r^{\circ }/RT}$$ Taking logarithm, we have $$\log K_p=\log A-\frac{\Delta H_r^{\circ }}{RT(2.303)}$$ This is an equation of straight line. When $\log K_p$ is plotted against $1/T$. we get a straight line. The intercept of this line $=\log A$, slope $=-\Delta H_r^{\circ }/2.303R$ Knowing the value of slope from the plot and universal gas constant $R,\Delta H_r^{\circ }$ can be calculated. (Equation of straight line : $Y=mx+C$. Here, $$\begin{array}{rl}& \log K_p=-\frac{\Delta H_r^{\circ }}{2.303R}\left(\frac{1}{T}\right)+\log A\\ & \begin{array}{llll}Y& m& x& C\end{array}\\ & \end{array}$$