$CaCO_{3(g)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$K_p = P_{CO_2}$ 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,
$\log {K_p} = - \frac{{\Delta H_r^o}}{{2.303R}}\left( {\frac{1}{T}} \right) + \log A$
$CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right)$
${A_2}(g)\, + \,{B_2}(g)\,\overset {{K_1}} \leftrightarrows \,2AB(g)\,\,\,......(1)$
$6AB\,(g)\,\,\overset {{K_2}} \leftrightarrows \,\,3{A_2}(g)\, + \,3{B_2}(g)......(2)$
તો $K_1$ અને $K_2$ વચ્ચેનો સંબંધ શું થાય?
$N_{2}=3.0 \times 10^{-3} M$
$O_{2}=4.2 \times 10^{-3} M$
અને $N O=2.8 \times 10^{-3} M$
આપેલ પ્રક્રિયા માટે બંધ કરેલા વાસણમાં $800 \,K$ અને $1$ $atm$ દબાણે $K_{p}$ ......... $atm$ હશે ?
$N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N O(g)$
(અહીં : $SrCO_{3(s)} \rightleftharpoons SrO_{(s)}+ CO_{2(g)} \,, K_p=1.6\,atm$)