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Q.The enthalpies of combustion of methane, graphite and hydrogen at 298 K are -890.3, -393.5 and -285.8 in kJ/mol respectively. Enthalpy of formation of CH4(g) will be

(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
Jharkhand JacJAC Intermediate Board (1st Year) 2025MCQ· 1mImportance★★★★★
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Combine the three given combustion enthalpies via Hess's law to build the target formation reaction C(graphite) + 2H2(g) -> CH4(g) — the sum works out to -74.8 kJ/mol.

We want the enthalpy of formation of methane:

Target: C(graphite) + 2H2(g) -> CH4(g), Delta_f H = ?

Given combustion reactions:

(1) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), Delta H1 = -890.3 kJ/mol

(2) C(graphite) + O2(g) -> CO2(g), Delta H2 = -393.5 kJ/mol

(3) H2(g) + 1/2 O2(g) -> H2O(l), Delta H3 = -285.8 kJ/mol

By Hess's law, construct the target equation from these three: Target = (2) + 2x(3) - (1)

Check the atoms cancel correctly:

(2): C + O2 -> CO2

2x(3): 2H2 + O2 -> 2H2O

Sum: C + 2H2 + 2O2 -> CO2 + 2H2O …

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