The molar enthalpies of combustion of isobutane and $n$ -butane are $-2870 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-2878 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ respectively at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$. Calculate $\Delta \mathrm{H}^{\circ}$ for the conversion of 1 mole of n-butane to 1 mole…

Question

The molar enthalpies of combustion of isobutane and $n$ -butane are $-2870 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-2878 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ respectively at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$. Calculate $\Delta \mathrm{H}^{\circ}$ for the conversion of 1 mole of n-butane to 1 mole of isobutane

Aicharya · Accepted Answer

Isobutane + oxygen $ightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$\Delta \mathrm{H}=-2870 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots \ldots$ (i)
$$
\begin{aligned}
\text { n-butane }+\text { oxygen } & ightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \\
\Delta \mathrm{H} &=-2878 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots . \text { (ii) }
\end{aligned}
$$
(ii) $-(\mathrm{i}) ; \mathrm{n}$ -butane $-$ Isobutane,
$$
\begin{aligned}
\Delta \mathrm{H}=&(-2878+2870) \\
&=-8 \mathrm{~kJ} \mathrm{~mol}^{-1} .
\end{aligned}
$$ ,

The molar enthalpies of combustion of isobutane and $n$ -butane are $-2870 \mathrm{~kJ} \mathrm{~mol}^{-1}$…

Solution