1
Question
The enthalpy changes of some process are given below :
α−D−glucose(s)+H2O⟶α−D−glucose(aq);ΔHdissolution=10.84kJ.
β−D−glucose(s)+H2O⟶β−D−glucose(aq);ΔHdissolution=4.68kJ.
α−D−glucose(aq)⟶β−D−glucose(aq);ΔHmutarotation=−1.16kJ
The ΔH∘ for α−D−glucose⟶β−D−glucose is :
α−D−glucose(s)+H2O⟶α−D−glucose(aq);ΔHdissolution=10.84kJ.
β−D−glucose(s)+H2O⟶β−D−glucose(aq);ΔHdissolution=4.68kJ.
α−D−glucose(aq)⟶β−D−glucose(aq);ΔHmutarotation=−1.16kJ
The ΔH∘ for α−D−glucose⟶β−D−glucose is :
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Solution
α−D−glucose(s)+H2O⟶α−D−glucose(aq);Heatofdissolution=10.84kJ ......(1)
β−D−glucose(s)+H2O⟶β−D−glucose(aq);Heatofdissolution=4.68kJ......(2)
α−D−glucose(aq)⟶β−D−glucose(aq);Heatofmutarotation=−1.16kJ .....(3)
α−D−glucose(s)⟶β−D−glucose(s)......(4)
Equation (1) - equation (2) + equation (3) gives equation (4).
Hence, ΔHo=10.84kJ−4.68kJ−1.15kJ=5kJ
β−D−glucose(s)+H2O⟶β−D−glucose(aq);Heatofdissolution=4.68kJ......(2)
α−D−glucose(aq)⟶β−D−glucose(aq);Heatofmutarotation=−1.16kJ .....(3)
α−D−glucose(s)⟶β−D−glucose(s)......(4)
Equation (1) - equation (2) + equation (3) gives equation (4).
Hence, ΔHo=10.84kJ−4.68kJ−1.15kJ=5kJ
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