The disachharide $α$ - maltose can be hydrolysed to glucose according to the equation (only magnitude in nearest integer in kj/mol)
$C_{12} H_{22} O_{11} (a q) + H_{2} O (l) ⟶ 2 C_{6} H_{12} O_{6} (a q)$
Using the following values, calculate the standard enthalpy change in this reaction:
$Δ_{f} H^{o} (H_{2} O, l) = - 285.85 k J$ ${m o l}^{- 1}$
$Δ_{f} H^{o} (C_{6} H_{12} O_{6}, a q) = - 1263.1 k J$ ${m o l}^{- 1}$
$Δ_{f} H^{o} (C_{12} H_{22} O_{11}, a q) = - 2238.3.1 k J$ ${m o l}^{- 1}$

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Solution

$Δ H_{r e a c t i o n} = Δ_{f} H_{P r o d u c t s}^{o} - Δ_{f} H_{R e a c t a n t s}^{o}$
$\Rightarrow Δ H_{r e a c t i o n} = 2 Δ_{f} H^{o} (C_{6} H_{12} O_{6}, l) - [Δ_{f} H^{o} (C_{12} H_{22} O_{11}, a q .) + Δ_{f} H^{o} (H_{2} O, a q .)]$
$\Rightarrow Δ H_{r e a c t i o n} = 2 (- 1263.1) - (- 2238.3 - 285.85) = 2.05 k J m o l e^{- 1}$
Answer = 2

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