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Question

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, 890.3kJ mol1393.5kJ mol1, and 285.8 kJ mol1 respectively. Enthalpy of formation of CH4(g) will be
(i) 74.8kJ mol1
(ii) 52.27kJ mol1
(iii) +74.8kJ mol1
(iv) +52.26kJmol1

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Solution

According to the question,

CH4(g)+2O2(g)CO2(g)+2H2O(l) ,ΔH=893kJmol1.....(i)
C(s)+O2(g)CO2(g), ΔH=393.5kJmol1.........(ii)
H2+12O2(g)H2O(l),ΔH=285.8kJmol1.......(iii)
C(s)+2H2(g)CH4(g), ΔH=?......(iv)
Equation (ii)+2× Equation (iii) -Equation (i) gives the required equation (iv) with ΔH = −393.5+2(−285.8)−(−890.3)kJmol1 = -74.8kJmol1


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