Solveeit Logo
Login

Question

Question: he enthalpy changes of some process are given below: A.\[\alpha -D-glucos{{e}_{(s)}}+{{H}_{2}}O\to...

he enthalpy changes of some process are given below:
A.α−D−glucose(s)+H2O→α−D−glucose(aq);ΔHdissolution=10.84kJ\alpha -D-glucos{{e}_{(s)}}+{{H}_{2}}O\to \alpha -D-glucos{{e}_{(aq)}};\Delta {{H}_{dissolution}}=10.84kJ
B.β−D−glucose(s)+H2O→β−D−glucose(aq);ΔHdissolution=4.68kJ.\beta -D-glucos{{e}_{(s)}}+{{H}_{2}}O\to \beta -D-glucos{{e}_{(aq)}};\Delta {{H}_{dissolution}}=4.68kJ.
C.α−D−glucose(aq)→β−D−glucose(aq);ΔHmutarotation=−1.16kJ\alpha -D-glucos{{e}_{(aq)}}\to \beta -D-glucos{{e}_{(aq)}};\Delta {{H}_{mutarotation}}=-1.16kJ
D.The ΔH∘ for α−D−glucose→β−D−glucose is:The~\Delta {{H}^{\circ }}~for~\alpha -D-glucose\to \beta -D-glucose~is{ }:

Explanation

Solution

Enthalpy of dissolution is also called enthalpy of solution or heat of solution.
Heat of solution can be defined as the change in enthalpy which is associated with the dissolution of a solute in a solvent at constant pressure resulting in infinite dissolution.
The change in optical activity as a monosaccharide reaches equilibrium between its α and β-anomers is called mutarotation.Enthalpy of mutarotation is the energy used in mutarotation.

Complete step by step answer:
To find  ΔH∘ for α−D−glucose→β−D−glucose ~\Delta {{H}^{\circ }}~for~\alpha -D-glucose\to \beta -D-glucose~we need to solve the equations given in the question using algebra
α−D−glucose(s)+H2O→α−D−glucose(aq);ΔHdissolution=10.84kJ\alpha -D-glucos{{e}_{(s)}}+{{H}_{2}}O\to \alpha -D-glucos{{e}_{(aq)}};\Delta {{H}_{dissolution}}=10.84kJ ……..(i)
β−D−glucose(s)+H2O→β−D−glucose(aq);ΔHdissolution=4.68kJ.\beta -D-glucos{{e}_{(s)}}+{{H}_{2}}O\to \beta -D-glucos{{e}_{(aq)}};\Delta {{H}_{dissolution}}=4.68kJ. ……….(ii)
α−D−glucose(aq)→β−D−glucose(aq);ΔHmutarotation=−1.16kJ\alpha -D-glucos{{e}_{(aq)}}\to \beta -D-glucos{{e}_{(aq)}};\Delta {{H}_{mutarotation}}=-1.16kJ ……………(iii)
 α−D−glucose→β−D−glucose ~\alpha -D-glucose\to \beta -D-glucose~ ………..(iv)
Subtracting equation (ii) from equating (i) and then after adding equation (iii) to the product of (ii)-(i) we will get equation (iv)

Hence, ΔHo=10.84kJ−4.68kJ−1.15kJ=5kJ\Delta {{H}^{o}}=10.84kJ-4.68kJ-1.15kJ=5kJ.

Additional information:
Some common type of enthalpy are:
Enthalpy of formation: It is the heat involved in the formation of one mole of substance (product) from its constituent particles.
Enthalpy of combustion: It is the heat evolved when one mole of substance is completely burned out in presence of oxygen.
Enthalpy of bond: It is the heat absorbed in breaking of one mole of bond.
Enthalpy of atomisation: It is the amount of heat required to break one mole of substance into its constituent particles.
Enthalpy of vaporization: The amount of heat required by a substance to change its liquid state to gaseous state.

Note:
Mutarotation is the change in the optical rotation because of the change in the equilibrium between two anomers when the corresponding stereocenters interconvert.
Reducing sugars can undergo mutarotation. Non-reducing sugars cannot undergo mutarotation.