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# At $298\; K, 0.1\; mol$ of methane is burnt in a bomb calorimeter such that $\Delta V=0$. The heat so produced is taken up by a large amount of water so that $\Delta T$ is very small.

$\begin{array}{1 1} (a) \Delta H = \Delta E \\ (b) \Delta H < \Delta E \\ (c) \Delta H > \Delta E \\ (d) \text{given data is not complete} \end{array}$

Answer: $\Delta H < \Delta E$
At 298 K, $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
$\Delta n_g = 1-3 = -2$
$\Delta H = \Delta E + \Delta (PV) = \Delta E + \Delta n_g RT = \Delta E -2RT$
$\therefore \Delta E > \Delta H$