Question
If $$\Delta E$$ is the heat of reaction for $${C_2}{H_5}OH\left( l \right) + 3{O_2}\left( g \right) \to $$ $$2C{O_2}\left( g \right) + 3{H_2}O\left( l \right)$$ at constant volume, the $$\Delta H$$ ( heat of reaction at constant pressure ), then the correct relation is
A.
$$\Delta H = \Delta E + RT$$
B.
$$\Delta H = \Delta E - RT$$
C.
$$\Delta H = \Delta E - 2\,RT$$
D.
$$\Delta H = \Delta E + 2\,RT$$
Answer :
$$\Delta H = \Delta E - RT$$
Solution :
We know that, $$\,\Delta H = \Delta E + \Delta {n_g}RT$$
where, $$\,\Delta {n_g} = $$ total number of moles of gaseous product $$-$$ total number of moles of gaseous reactant
$$\eqalign{
& = 2 - 3 = - 1 \cr
& {\text{So,}}\,\Delta H = \Delta E - RT \cr} $$