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 combustion per gram of glucose at 25°C is (a) +2900 kJ (b) –2900 kJ(c) –16.11 kJ (d) +16.11 kJ

Answer 1

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Answer 2

Answer:

C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)

ΔrH∘=ΔcH∘=[6Δf(CO2)+6ΔfH∘(H2O)]−[ΔfH∘(C6H12O6)+6ΔfH∘(O2)]

=6×(−400)+6(−300)−(−1300)

=−2900kJ

=−2900/180kJ/g=−16.11kJ/g