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Question: Relation between $\Delta H$ and $\Delta U$ for reaction, $2SO_{3(g)} \rightarrow 2SO_{2(g)} + O_{2(g...

Relation between ΔH\Delta H and ΔU\Delta U for reaction, 2SO3(g)→2SO2(g)+O2(g)2SO_{3(g)} \rightarrow 2SO_{2(g)} + O_{2(g)} is

A

ΔH+ΔU=RT\Delta H + \Delta U = RT

B

ΔH+ΔU=−RT\Delta H + \Delta U = -RT

C

ΔH−ΔU=2RT\Delta H - \Delta U = 2RT

D

ΔH−ΔU=−RT\Delta H - \Delta U = -RT

Answer

None of the options is correct. The correct relation is ΔH−ΔU=RT\Delta H - \Delta U = RT

Explanation

Solution

For a reaction at constant pressure, the relation between enthalpy and internal energy is given by

ΔH=ΔU+Δn RT\Delta H = \Delta U + \Delta n\,RT

where Δn\Delta n is the change in the number of moles of gas.

For the reaction

2 SO3(g)→2 SO2(g)+O2(g)2\,SO_{3(g)} \rightarrow 2\,SO_{2(g)} + O_{2(g)}

  • Moles of gas in reactants = 2 (from 2 SO32\,SO_3)
  • Moles of gas in products = 2 (from 2 SO22\,SO_2) + 1 (from O2O_2) = 3

Thus,

Δn=3−2=1\Delta n = 3 - 2 = 1

Substitute into the relation:

ΔH=ΔU+RT⟹ΔH−ΔU=RT\Delta H = \Delta U + RT \quad \Longrightarrow \quad \Delta H - \Delta U = RT