\(CaCO_{3(g)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}\)
\(K_p = P_{CO_2}\) 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,
\(\log {K_p} = - \frac{{\Delta H_r^o}}{{2.303R}}\left( {\frac{1}{T}} \right) + \log A\)
(ઉપયોગ $R =8.31\, J\, K ^{-1}\, mol ^{-1} ; \log 2=0.3010$. In $10=$ $2.3, \log 3=0.477$ )
($R = 0.082\, L\, atm\, mol^{-1}\, K^{-1}$, મોલર દળ $S = 32\, g\, mol^{-1}$, મોલર દળ $N = 14\, g\, mol^{-1}$)