Chemistry - Thermochemistry Question with Solution | TestHub
α-maltose can be hydrolysed to glucose according to the following reaction:
α-C12H22O11(aq) + H2O(l) → 2C6H12O6(aq)
The standard enthalpy of formation of H2O(l), C6H12O6(aq) and α-C12H22O11(aq) are -285, -1263 and -2238 kJ/mol, respectively. Which of the following statement(s) is/are true? (one or more than one correct)
Options:(select one or more)
Answer:
Solution:
The enthalpy change (Δ H) for the reaction is calculated as follows:
Δ H = 2(-1263) - (-285) - (-2238) = -2526 + 285 + 2238 = -3 kJ
This indicates that the hydrolysis is exothermic (statement 'a' is true).
Let X be the enthalpy of combustion for 2 moles of glucose. The overall process can be described as:
Maltose + H₂O → 2 Glucose (Δ H = -3 kJ)
followed by the combustion of glucose:
2 Glucose + 12O₂ → 12CO₂ + 11H₂O (with an enthalpy change of X)
Therefore, the enthalpy of combustion for maltose (Δc H(maltose)) is X - 3. This means maltose liberates 3 kJ more energy than 2 moles of glucose, not less (statement 'b' is false).
Since the reaction is exothermic, a higher temperature will shift the equilibrium backward, thus lowering the extent of hydrolysis (statement 'c' is false).
Solid maltose has a different enthalpy of formation compared to aqueous maltose due to the involvement of the heat of solution. Consequently, the Δ H for the reaction would change if solid maltose were used instead of aqueous maltose (statement 'd' is false).
The correct answer is (a).
