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Q. Determine the enthalpy of the reaction,
$C _{3} H _{8}(g)+ H _{2}(g) \rightarrow C _{2} H _{6}(g)+ CH _{4}(g)$, at $25^{\circ} C$, using the given heat of combustion values under standard conditions.
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The standard heat of formation of $C _{3} H _{8}(g)$ is $-103\, kJ / mol$.

IIT JEEIIT JEE 1992Thermodynamics

Solution:

First we need to determine heat of combustion of $C _{3} H _{8}$.
$3 C ( gr )+4 H _{2}(g) \longrightarrow C _{3} H _{8}(g) \Delta H_{f}^{\circ}=-103 \,kJ$
$\Rightarrow-103\, kJ =-3 \times 393-4 \times 285.80-\Delta H_{\text{comb }}^{\circ}\left( C _{3} H _{8}\right) $
$\Rightarrow \Delta H_{\text{comb }}^{\circ}\left( C _{3} H _{8}\right)=-2219.20\, kJ$
$\Rightarrow \Delta H_{r}^{\circ}=\Sigma \Delta H_{\text{comb }}^{\circ}$ (reactants) $-\Sigma \Delta H_{\text{comb }}^{\circ}$ (products)
$=-2219.20-285.80+1560+890=-55\, kJ$