Chemistry › Thermodynamics (C) › Laws of Thermochemistry and Enthalpy Change
Heat of combustion of $\mathrm{C}_{(\mathrm{s})}, \mathrm{H}_{2(\mathrm{~g})}$ and $\mathrm{C}_{2}…
Heat of combustion of $\mathrm{C}_{(\mathrm{s})}, \mathrm{H}_{2(\mathrm{~g})}$ and $\mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}$ are $-x_{1},-x_{2}$ and $-x_{3}$ respectively. Hence heat of formation of $\mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}$ is
$-x_{1}-x_{2}+x_{3}$ $-2 x_{1}-3 x_{2}+x_{3}$ $x_{1}+x_{2}-x_{3}$ $-x_{3}+2 x_{1}+3 x_{2}$
Solution
i. $\quad \mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} \quad \Delta \mathrm{H}_{1}=-\mathrm{x}_{1}$
ii. $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \Delta \mathrm{H}_{2}=-\mathrm{x}_{2}$
iii. $\mathrm{C}_{2} \mathrm{H}_{6}+\frac{7}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \quad \Delta \mathrm{H}_{3}=-\mathrm{x}_{3}$
Multiply equation (i) by $2+$ multiply equation (ii) by 3 - equation (iii)
$\begin{array}{l}
\therefore 2 \mathrm{C}+3 \mathrm{H}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6} \\
\therefore \Delta \mathrm{H}=2 \Delta \mathrm{H}_{1}+3 \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{3} \\
\therefore \Delta \mathrm{H}=-2 \mathrm{x}_{1}-3 \mathrm{x}_{2}+\mathrm{x}_{3}
\end{array}$
Asked in: MHT CET 2020 (14 Oct Shift 1)
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