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Thermodynamics
Question
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Given
C(grahite) + O2(g) → CO2(g)
ΔrH°  = - 393.5 kJ mol-1
H2(g) + 1/2O2(g) → H2O (l)
ΔrH°  = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction
C(grahite) + 2H2(g) →CH4 will be

  •  +74.8 kJ mol–1 

  • +144.0 kJ mol–1

  • –74.8 kJ mol–1

  • –144.0 kJ mol–1

Answer

C.

–74.8 kJ mol–1

CO2(g) + 2H2O (l)→ CH4 (g) + 2O2(g);
ΔrH°= 890.3
ΔfH° –393.5 –285.8 ? 0
ΔrH°= Σ(ΔfH°)products -Σ(ΔfH°)reactant
890.3 = [ 1 x(ΔfH°)CH4 + 2x0]-[1x(-393.5)+2(-285.8)]
(ΔfH°)CH4 = 890.3-965.1 = -74.8 kJ/mol

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