Question
Assume each reaction is carried out in an open container. For which reaction will $$\Delta H = \Delta E?$$
A.
$${H_2}\left( g \right) + B{r_2}\left( g \right) \to 2HBr\left( g \right)$$
B.
$$C\left( s \right) + 2{H_2}O\left( g \right) \to 2{H_2}\left( g \right) + C{O_2}\left( g \right)$$
C.
$$PC{l_5}\left( g \right) \to PC{l_3}\left( g \right) + C{l_2}\left( g \right)$$
D.
$$2CO\left( g \right) + {O_2}\left( g \right) \to 2\,C{O_2}\left( g \right)$$
Answer :
$${H_2}\left( g \right) + B{r_2}\left( g \right) \to 2HBr\left( g \right)$$
Solution :
As we know that,
$$\Delta H = \Delta E + p\Delta V\,\,{\text{or}}\,\,\Delta H = \Delta E + \Delta {n_g}RT$$
where, $$\Delta {n_g} \to $$ number of gaseous moles of product - number of gaseous moles of reactant
If $$\Delta ng = 0$$ ( for reactions in which the total number of moles of gaseous products are equal to total number of moles of gaseous reactants ), therefore $$\Delta H = \Delta E$$
So, for reaction $$\left( a \right)\Delta n = 2 - 2 = 0$$
Hence, for reaction $$\left( a \right),\Delta H = \Delta E$$