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Question

The standard enthalpies of formation of CO2(g),H2O(l) and glucose(s) at 25oC are 400kJ/mol,300kJ/mol and 1300kJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 25oC is:

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

The correct option is B 16.11kJ
C6H12O6(s)+6O2(g)6CO2+6H2O(l)
ΔH=[6ΔfHCO2+ΔfHH2O][ΔfHC6H12O6+6ΔfHO2]
=[6(400)+6(300)][1300+0]
=4200+1300=2900kJ/mol
Enthalpy oc combustion per gram =2900180=16.11kJ/g

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