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) $\ce{Br2(l) &lt;=&gt; Br2(g)}$
  
  (B) $\ce{2C(s) + O2(g) r 2CO(g)}$
  
  (C) $\ce{H2(g) + F2(g) -&gt; 2HF(g)}$
  
  (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?
You are right. B is the correct choice.

$$\Delta H = \Delta E + \Delta PV$$

Since $\Delta E $ is more negative than $\Delta H$, $$\Delta H &gt; \Delta E$$.
Hence $$\Delta PV&gt;0$$

$\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 $\Delta PV$.