Understanding the Enthalpy Change Reaction

The problem asks for the enthalpy change ($\Delta H$) when $\alpha$-D-glucose ($C_6H_{12}O_6$) is converted to maltose ($C_{12}H_{22}O_{11}$) and water ($H_2O$). This is a dehydration synthesis reaction where two glucose molecules combine, eliminating one water molecule.

The balanced chemical equation for this conversion is:

$2 C_6H_{12}O_6(s) \rightarrow C_{12}H_{22}O_{11}(s) + H_2O(l)$

We are given the enthalpy changes for the complete combustion of glucose and maltose.

Applying Hess's Law for Enthalpy Calculation

We can use Hess's Law and the provided combustion data to find the enthalpy change for the target reaction. The combustion reactions are:

1. Combustion of $\alpha$-D-glucose:

   $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l); \quad \Delta H_{comb, glucose} = -2809.1\text{ kJ mol}^{-1}$

2. Combustion of Maltose:

   $C_{12}H_{22}O_{11}(s) + 12O_2(g) \rightarrow 12CO_2(g) + 11H_2O(l); \quad \Delta H_{comb, maltose} = -5645.5\text{ kJ mol}^{-1}$

To find the enthalpy change for $2 C_6H_{12}O_6 \rightarrow C_{12}H_{22}O_{11} + H_2O$, we manipulate the combustion equations:

Multiply the glucose combustion equation by 2:

$2C_6H_{12}O_6(s) + 12O_2(g) \rightarrow 12CO_2(g) + 12H_2O(l); \quad 2 \times \Delta H_{comb, glucose} = 2 \times (-2809.1) = -5618.2\text{ kJ mol}^{-1}$

The target reaction requires maltose as a product. We can express the target reaction enthalpy ($\Delta H_{target}$) using the combustion enthalpies:

$\Delta H_{target} = [ \Delta H_{comb, maltose} + \Delta H_{comb, H_2O} ] - [ 2 \times \Delta H_{comb, glucose} ]$

However, a simpler way using Hess's Law based on the relation derived from formation enthalpies is:

$\Delta H_{target} = 2 \times \Delta H_{comb, glucose} - \Delta H_{comb, maltose}$

Substitute the given values:

$\Delta H_{target} = 2 \times (-2809.1\text{ kJ mol}^{-1}) - (-5645.5\text{ kJ mol}^{-1})$

$\Delta H_{target} = -5618.2\text{ kJ mol}^{-1} + 5645.5\text{ kJ mol}^{-1}$

$\Delta H_{target} = +27.3\text{ kJ mol}^{-1}$

This enthalpy change ($+27.3\text{ kJ mol}^{-1}$) is for the conversion of 2 moles of glucose to 1 mole of maltose and 1 mole of water.

Determining Enthalpy Change Per Mole of Glucose

The question asks for the enthalpy change accompanying the conversion of *1 mol* of $\alpha$-D-glucose.

Since the calculated $\Delta H_{target}$ ($+27.3\text{ kJ mol}^{-1}$) corresponds to the reaction involving 2 moles of glucose, we need to find the enthalpy change per mole of glucose.

Enthalpy change per mole of glucose = $\frac{\Delta H_{target}}{2}$

Enthalpy change per mole of glucose = $\frac{+27.3\text{ kJ mol}^{-1}}{2 \text{ mol glucose}}$

Enthalpy change per mole of glucose = $+13.65\text{ kJ mol}^{-1}$

Rounding to one decimal place, the value is $+13.7\text{ kJ mol}^{-1}$.