# Relating change in internal energy and to that in enthalpy

The question is

For which exothermic reaction is $\Delta\,E$ more negative than $\Delta\,H$?

(A) $\ce{Br2(l) <=> Br2(g)}$

(B) $\ce{2C(s) + O2(g) r 2CO(g)}$

(C) $\ce{H2(g) + F2(g) -> 2HF(g)}$

(D) $\ce{2SO2(g) + O2(g) r 2SO2(g)}$

r is supposed to be reaction? Either a one headed or equilibrium arrows. The "correct" answer is supposed to be D.

However, using the formula $H = U + PV$ to $\Delta H = \Delta U + \Delta PV$, I found that $\Delta U$ would be less if $\Delta V$ was positive.

Thus, more gaseous products should be produced. As a result, I chose B.

Note, A is not exothermic, so would not be considered.

Could anyone confirm my answer or offer an explanation for my answer being wrong?

• The last reaction should have $\ce {SO3}$ as the product – Binary Geek Apr 3 '15 at 6:43

$$\Delta H = \Delta E + \Delta PV$$
Since $\Delta E$ is more negative than $\Delta H$, $$\Delta H > \Delta E$$. Hence $$\Delta PV>0$$
$\Delta E \space$has to be negative. So option A is ruled out.
Among the other choices, B is the only one where number of gaseous moles are increasing from left to right, leading to a positive value of $\Delta V$ and hence $\Delta PV$.