Chemistry - Thermochemistry Question with Solution | TestHub
Cellobiose can be hydrolysed to glucose according to the following reaction:
C12H22O11(aq) + H2O(l) → 2C6H12O6(aq)
The standard enthalpy of formation of H2O(l), C6H12O6(aq) and C12H22O11(aq) (cellobiose) are -285, -1263 and -2250 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:
ΔH = 2(-1263) - (-285) - (-2250) = -2526 + 285 + 2250 = +9 kJ, endothermic (a true). Let X be the combustion enthalpy of 2 mol glucose. Cellobiose + H2O → 2 glucose (+9), then burning (X) gives Δc H(cellobiose) = X + 9, which is 9 kJ less exothermic than X, so cellobiose liberates less heat (b true). For an endothermic reaction, a higher temperature shifts the equilibrium forward and increases hydrolysis (c true). Solid cellobiose has a different enthalpy of formation (heat of solution is involved), so ΔH changes (d false). Ans (a),(b),(c).
