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ChemistrySingle correct+2Question 27 of 60

JEE Advanced 2013 · Paper 1, Chemistry question 27

The standard enthalpies of formation of COX2(g)\ce{CO2}(g), HX2O(l)\ce{H2O}(l) and glucose(s) at 25 °C are −400 kJ/mol, −300 kJ/mol and −1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25 °C is

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Answer and worked solution

Official answer: C

Combustion. CX6HX12OX6+6 OX2→6 COX2+6 HX2O\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}:
ΔcH=6(−400)+6(−300)−(−1300)=−2900 kJ mol−1.\Delta_cH=6(-400)+6(-300)-(-1300)=-2900\ \text{kJ mol}^{-1}.

Per gram. −2900180=−16.11\dfrac{-2900}{180}=-16.11 kJ.

(C)

Topics:Hess's law and enthalpies of reaction
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