`alpha`- maltose `(C_(12)H_(22)O_(11))` can be hydrolysed to glucose `(C_(6)H_(12)O_(6))` according to the following reaction.
`C_(12)H_(22)(O_(11)(aq)+H_(2)O(l)to2C_(6)H_(12)O_(6)(aq)`
Given:
Standard enthalpy of formation of `C_(12)H_(22)O_(11)(aq)=-2238(kJ)/(mol)`
Standard enthalpy of formation of `H_(2)O(l)=-285(kJ)/(mol)`
Standard enthalpy of formation of `C_(6)H_(12)O_(6)(aq)=-1263(kJ)/(mol)`
Which of the following statements (s) is/are true?

Options

A. The hydrolysis reaction is exothermic
B. Heat liberated in combustion of 1.0 mol of `alpha`- maltose is greater than the heat liberated in combustion of 2.0 mole of glucose
C. Increasing temperature will increase the degree of hydrolysis of `alpha`- maltose
D. Enthalpy of reaction will remain same even if solid `alpha`-maltose is taken in the reaction

Answer

A, B

Explanation

(a) `Delta_(r )H^(@) = 2(-1263) -(-285) -(-2238) = -3 implies "Reaction is slightly exothermic"`.
(B) `C_(12)H_(22)O_(11)(aq) + 12O_(2)(g) rarr 12 CO_(2)(g) + 11 H_(2)O (l) " "Delta_(c) H_(1)^(@)`
`2C_(6)H_(12)O_(6)(aq.) + 12O_(2)(g) rarr 12CO_(2)(g) + 12 H_(2)O(l) " "Delta_(c)H_(2)^(@)`
`Delta H_(1)^(@) = 2(x) + 11(-285)-(-280)-(-2238) = 12x - 897` and
`Delta_(c )H_(2)^(@) = 2(x) + 12(-285) - 2(-1263) = 12x - 894`
Clearly: `Delta_(c )H_(1)^(@) > Delta_(c )H_(12)^(@)` [as x will be negative].
( C) Increasing T will shift the reaction in backward d irection. So, ‘h’ of `alpha`- maltose will decrease with temperature.
(D) `Delta_(r )H^(@)` will change if solid `alpha`- maltose is taken instead of aq.