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JEE Main 2017 · 2 Apr · Shift 1, Chemistry question 31

Given

CX(graphite)+OX2(g)→COX2(g)\ce{C_{(graphite)} + O2(g) -> CO2(g)} ; ΔrH∘=−393.5\Delta_rH^\circ=-393.5 kJ mol−1^{-1}

HX2(g)+12 OX2(g)→HX2O(l)\ce{H2(g) + 1/2O2(g) -> H2O(l)} ; ΔrH∘=−285.8\Delta_rH^\circ=-285.8 kJ mol−1^{-1}

COX2(g)+2 HX2O(l)→CHX4(g)+2 OX2(g)\ce{CO2(g) + 2H2O(l) -> CH4(g) + 2O2(g)} ; ΔrH∘=+890.3\Delta_rH^\circ=+890.3 kJ mol−1^{-1}

Based on the above thermochemical equations, the value of ΔrH∘\Delta_rH^\circ at 298 K for the reaction CX(graphite)+2 HX2(g)→CHX4(g)\ce{C_{(graphite)} + 2H2(g) -> CH4(g)} will be :

Choose one option
Answer and worked solution

Official answer: A

Add the first equation, twice the second and the third:
ΔH=−393.5+2(−285.8)+890.3=−74.8 kJ mol−1.\Delta H=-393.5+2(-285.8)+890.3=-74.8\text{ kJ mol}^{-1}.

Topics:Hess's law and enthalpies of reaction
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