Question

For following reaction, relation between $\Delta H$ and $\Delta U$ is

$SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$

Answer 1

ΔH = ΔU

Answer 2

For a reaction at constant pressure, the relation is given by

$$\Delta H = \Delta U + \Delta (PV) = \Delta U + \Delta n\,RT.$$

For the reaction

$$\text{SO}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\,\text{SO}_{3(g)},$$

the initial moles of gas are $1 + 1 = 2$ and the final moles are $2$. Thus,

$$\Delta n = 2 - 2 = 0.$$

Therefore,

$$\Delta H = \Delta U + 0 \cdot RT = \Delta U.$$

ΔH = ΔU + ΔnRT; here Δn = 0, so ΔH = ΔU.