\(CaCO_{3(g)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}\)
\(K_p = P_{CO_2}\) 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,
\(\log {K_p} = - \frac{{\Delta H_r^o}}{{2.303R}}\left( {\frac{1}{T}} \right) + \log A\)
$A + B$ $\rightleftharpoons$ $C + D$ $+$ ઉષ્મા
સંતુલન સુધી પહોંચી ગઈ છે. પ્રક્રિયાને આગળ વધારવા માટે શું કરી શકાય?
$(i)\,CO(g)+ H_2O(g) \rightleftharpoons CO_2(g)+H_2(g)\,;\,K_1$
$(ii)\,CH_4(g)+H_2O(g) \rightleftharpoons CO(g)+3H_2(g)\,;\,K_2$
$(iii)\,CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g)+ 4H_2(g)\,;\,K_3$
$400\, K$ $\Delta G ^{\circ}=+25.2\, kJ mol ^{-1}$. એ $2 A ( g ) \rightleftharpoons A _{2}( g )$
આ પ્રકિયા માટે સંતુલન અચળાંક $K _{ C }$$...... \times 10^{-2}$
$\left.\log _{10} 2=0.30,1\, atm =1\, bar \right]$
$[$ antilog $(-0.3)=0.501]$