ΔH of reversible expansion of ideal gas
Recall that, for an ideal monatomic gas, $$U = \frac{3}{2}nRT.$$ (Diatomic and polyatomic gases have a larger coefficient.) For an isothermal expansion, reversible or otherwise, we therefore have $$\Delta H = \Delta(U+PV) = \Delta(U+nRT) = \frac{5}{2}nR\Delta T = 0,$$ since $\Delta T = 0$.