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  4. The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C6H6(.l.)+15O2(.g.) arrow 12CO2(.g.)+6H2O(.l.) at 25° C in kJ is

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Q. The difference between heats of reaction at constant pressure and constant volume for the reaction,
$2C_{6}H_{6}\left(\right.l\left.\right)+15O_{2}\left(\right.g\left.\right) \rightarrow 12CO_{2}\left(\right.g\left.\right)+6H_{2}O\left(\right.l\left.\right)$ at $25^\circ C$ in kJ is

NTA AbhyasNTA Abhyas 2022

Solution:

$\Delta H-\Delta E=\Delta n_{g}RT$
$\Delta n_{g}=12-15=-3$
So $\Delta H -\Delta E =-3 \times 8.314 J mol ^{-1} K ^{-1} \times 298 K$ $=-7.43 kJ mol ^{-1}$