The standard enthalpy of formation of octane (C8H18) is −250 kJ/mol. Calculate the enthalpy of combustion of C8H18. Given that enthalpy of formation of CO2(g) and H2O(l) are −394 kJ/mol and −286 kJ/mol respectively.
A
−5200 kJ/mol
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B
−5726 kJ/mol
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C
−5476 kJ/mol
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D
−5310 kJ/mol
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Solution
The correct option is C−5476 kJ/mol According to the question, 8C(graphite)+9H2(g)→C8H18(g);ΔH∘f=−250 kJ/mol......(i)C(graphite)+O2(g)→CO2(g);ΔH∘f=−394 kJ/mol......(ii)H2(g)+12O2(g)→H2O(l);ΔH∘f=−286 kJ/mol......(iii)
The reaction of combustion of octane is C8H18(g)+252O2→8CO2(g)+9H2O(l).....(iv)
The equation(iv) can be obtained by 8×equation(ii)+9×equation(iii)−equation(i) ∴ΔH∘combustion=8×(−394)+9×(−286)−(−250)=−5476 kJ/mol