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Given C(graphite) + O2(g) → CO2(g); ΔrH° = −393.5 kJ mol−1 H2(g) + 1/2 O2(g) → H2O(l); ΔrH° = −285.8 kJ mol−1 CO2(g) + 2H2O(l) → CH4(g) + 2O2(g); ΔrH° = +890.3 kJ mol−1 Based on the above thermochemical equations, the value of ΔfH° at 298 K for the reaction C(graphite) + 2H2(g) → CH4(g) will be:

1. +74.8 kJ mol−1
2. +144.0 kJ mol−1
3. −74.8 kJ mol−1
4. −144.0 kJ mol−1

Correct answer: −74.8 kJ mol−1

Solution

C + 2H2 -> CH4 = (C+O2->CO2) + 2(H2+1/2O2->H2O) + (CO2+2H2O->CH4+2O2) = (-393.5) + 2(-285.8) + (+890.3) = -74.8 kJ/mol. So dfH = -74.8 kJ/mol.

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