$\Delta H = -3129\, KJ$
$\Delta H = \Delta H_f (product) - \Delta H_f (reactants)$
$= [4 × \Delta H_f(CO_2) + 6×\Delta H_f(H_2O)] =[2× \Delta H_f(C_2H_6) + 7 \Delta H_f(O_2)]$
$- 3129 = [4 × (-395) + 6 ×(-286)] - [2× \Delta H_f(C_2H_6) + 7 ×0]$
$2×\Delta Hf(C_2H_6) = - 167 \,kJ$
$\Delta \,{H_f}({C_2}{H_6})\,\, = \,\, - \frac{{167}}{2}\,\, = \,\, - \,83.5\,\,KJ$
$A + B \rightleftharpoons C+D$
$27^{\circ}\,C$ પર પ્રમાણિત મુક્ત ઊર્જા ફેરફાર $\left(\Delta_{ r } G ^0\right)$ $(-)$ $............kJ mol ^{-1}$ છે. (નજીકની પૂર્ણાંક)
(આપેલ : $R =8.3\,Jk ^{-1}\, mol ^{-1}$ અને In $10=2.3$ )