Assuming ideal gas behavior, how can I tell when the change in enthalpy is equal to the change in the inner energy?

Question

If ideal gas behavior is assumed, for which of the following reactions does $\Delta H = \Delta U$?

\begin{align} \tag{a} \ce{N2O4 &-> 2 NO2(g)}\\ \tag{b} \ce{CH4(g) + 2 O2(g) &-> CO2(g) + 2 H2O (l)}\\ \tag{c} \ce{SO2(g) + 1/2 O2(g) &-> SO3(g)}\\ \tag{d} \ce{Br2(l) + 3 Cl2(g) &-> 2 BrCl3(g)}\\ \tag{e} \ce{Cl2(g) + F2(g) &-> 2 ClF (g)}\\ \end{align}

The correct answer is (e), and I just want to make sure I understand why. For $\Delta U = \Delta H$, no work can be done as $\Delta U = H - W$ (where $W$ is work done by the gas).

This means that the total number of molecules can't change, otherwise a change in volume will occur, causing work to be done upon molecules or the molecules to do work. Likewise, phase changes will also require work to be done on the molecules. Thus, we rule out (a), (b), (c), and (d)?

I'm not sure if this is too simplistic or the correct way of thinking about this problem.

Answer 1

You got it right, and there is really nothing to add. Same number of molecules means no change in volume, hence no work is done, hence $\Delta H=\Delta U$.

On a side note, I find it hardly believable that burning of methane (b) will result in liquid water, but if the problem statement was given to you exactly like this, well, so be it.

Answer 2

The heat of reaction, by convention, is defined in terms of the amount of heat that needs to be added in bringing about the transition at constant temperature and pressure from a thermodynamic equilibrium state of pure products to a thermodynamic equilibrium state of pure reactants. If the initial and final states happen to be at room temperature, then specifying liquid water for the products is perfectly reasonable.