Magnitude of difference between $△ H$ and $△ U$ in $k J$ for the combustion of benzene at 300 K is :
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$

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Solution

Combustion of benzene is represented by following reaction;
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$

The difference in gaseous moles of product and reactant is :
$△ n_{g} = 6 - \frac{15}{2} = - 3 / 2$
And we have relation,
$△ H = △ U + △ n_{g} R T$
so,
$△ H - △ U = △ n_{g} R T$
$△ H - △ U = - \frac{3}{2} \times 8.314 \times 300$
$= 3741.3 J = - 3.741 k J$

Hence the difference between $△ H$ and $△ U$ for the combustion of liquid benzene is 3.74 kJ.

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