For the chemical equilibrium, CaCO_ 3(s) leftharpoons CaO_ (s) +CO_...

For the reaction, $K_p=p_{\mathrm{CO}_2}$ $\text{ and } K_c= \left[\mathrm{CO}_2\right]\left(\because\left[\mathrm{CaCO}_{3}\right]=1\right. and [ \mathrm{CaO}]=1$ for solids $]$ According to Arrhenius equation we have K=A $e^{-\DeltaH_r^{{}^{\circ}} / R T} Taking logarithm, we have \log K_p=\log A-\frac{\Delta H_r^{{}^{\circ}}}{R T(2.303)}This is an equation of straight line. When \log K_p$ is plotted against $1 / T$. we get a straight line. The intercept of this line $=\log A$, slope $=-\Delta H_r^{{}^{\circ}} / 2.303 R$ Knowing the value of slope from the plot and universal gas constant $R, \Delta H_r^{{}^{\circ}}$ can be calculated. (Equation of straight line : $Y=m x+C$. Here, $$\begin{aligned}& \log K_p=-\frac{\Delta H_r^{{}^{\circ}}}{2.303 R}\left(\frac{1}{T}\right)+\logA\\ & \begin{array}{llll}Y & m & x & C \end{array} \\ & \end{aligned}$$