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If 2 moles of $\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})$ are completely burnt $4100 \mathrm{~kJ}$ of heat is liberated. If $\Delta H^{\circ}$ for $\mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(l)$ are -410 and $-285 \mathrm{~kJ}$ per mole respectively then the heat of formation of $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ is
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Verified Answer
The correct answer is:
$-375 \mathrm{~kJ}$
$\begin{array}{r}
2 \mathrm{C}_2 \mathrm{H}_6(g)+7 \mathrm{O}_2(g) \longrightarrow 4 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H=-4100 \mathrm{~kJ}
\end{array}$
Let, $\Delta H_f^{\circ}\left(\mathrm{C}_2 \mathrm{H}_6\right)=x$,
$\begin{aligned}
\Delta H & =\Sigma H_{f(\text { products })}^{\circ}-\Sigma H_{f_{(\text {reactants })}^{\circ}}^{\circ} \\
-4100 & =[4(-410)+6(-285)]-[2 x+0] \\
x & =-375 \mathrm{~kJ}
\end{aligned}$
2 \mathrm{C}_2 \mathrm{H}_6(g)+7 \mathrm{O}_2(g) \longrightarrow 4 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H=-4100 \mathrm{~kJ}
\end{array}$
Let, $\Delta H_f^{\circ}\left(\mathrm{C}_2 \mathrm{H}_6\right)=x$,
$\begin{aligned}
\Delta H & =\Sigma H_{f(\text { products })}^{\circ}-\Sigma H_{f_{(\text {reactants })}^{\circ}}^{\circ} \\
-4100 & =[4(-410)+6(-285)]-[2 x+0] \\
x & =-375 \mathrm{~kJ}
\end{aligned}$
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