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NCERT Exemplar · Q6

Q.ΔfU° of formation of CH4(g) at certain temperature is -393 kJ mol^-1. The value of ΔfH° is

(i) zero
(ii) < ΔfU°
(iii) > ΔfU°
(iv) equal to ΔfU°
Mizoram MbseMCQ· 1mImportance★★★★★est
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For a reaction involving gases, ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT. Forming CH4(g)\text{CH}_4(g) from its elements has Δng=−1\Delta n_g = -1, so ΔfH∘=ΔfU∘−RT<ΔfU∘\Delta_f H^\circ = \Delta_f U^\circ - RT < \Delta_f U^\circ. The correct option is (ii).

Enthalpy and internal energy are linked, for ideal gases, by:

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

where Δng\Delta n_g is the change in the number of moles of gas (products − reactants). The numerical value ΔfU∘=−393 kJ mol−1\Delta_f U^\circ = -393\ \text{kJ mol}^{-1} is not needed; only the sign of Δng\Delta n_g matters.

The standard formation reaction of methane, using elements in their standard states, is:

C(s)+2 H2(g)⟶CH4(g)\text{C(s)} + 2\,\text{H}_2(g) \longrightarrow \text{CH}_4(g)

Carbon is a solid, so only the gaseous species count:

Δng=1 (CH4)−2 (H2)=−1\Delta n_g = 1\ (\text{CH}_4) - 2\ (\text{H}_2) = -1 …

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