$CaCO_{3(g)} \rightleftharpoons  CaO_{(s)} + CO_{2(g)}$

$K_p = P_{CO_2}$   and $K_C  = [CO_2]$

  $(\because [CaCO_3] = 1$ and $[CaO] = 1$ for solids$)$

According to Arrhenius equation we have

$K = A{e^{ - \Delta H{^\circ _r}/RT}}$

 Taking logarithm, we have

$\log {K_p} = \log\, A - \frac{{\Delta H_r^o}}{{RT(2.303)}}$

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 $= -\Delta H^°_r / 2.303 \,R$

 Knowing the value of slope from the plot and universal gas constant $R$, $∆H^°_r$ can be calculated.

 (Equation of straight line : $Y = mx + C$. Here,

$\log {K_p} =  - \frac{{\Delta H_r^o}}{{2.303R}}\left( {\frac{1}{T}} \right) + \log A$