    Question

# The standard enthalpies of formation of CO2(g),H2O(l) and glucose(s) at 25∘C are – 400 kJ/mol, - 300 kJ/mol and – 1300 kJ/mol, respectively. The standard enthalpy of combusion per gram of glucose at 25∘C is

A
+ 2900kJ
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B
– 2900 kJ
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C
– 16.11 kJ
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D
+ 16.11 kJ
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Solution

## The correct option is C – 16.11 kJΔc H∘ (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised. Also standard heat of formation (ΔfH∘) can be taken as the standard of that substance. H∘CO2=ΔFH∘(CO2)=−400 kJ mol−1 H∘H2O=ΔfH∘(H2O)=300 kJ mol−1 H∘glucose=ΔfH∘(glucose)=−1300 kJ mol−1 H∘O2=ΔfH∘(O2)=0.00 C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l) ΔCH∘(glucose)=6[ΔfH∘(CO2)+ΔfH∘(H2O)] −[ΔfH∘(C6H12O6)+6ΔfH∘(O2)] =−2900 kJ mol−1 Molar mass of C6H12O6=180g mol−1 Thus,standard heat of combustion of glucose per gram =−2900180=−16.11 kJ g−1 To solve such heat of combustion of glucose per gram =−2900180=−16.11 kJ g−1 To solve such problem , students are advised to keep much importance in unit conversion. As here, value of R (8.314 JK−1mol−1) in JK−1 mol−1 must be converted into kJ by dividing the unit by 1000.  Suggest Corrections  0      Similar questions
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