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Question

Consider the following data

(i) $2Al(s) + 6HCl(aq) \rightarrow {Al}_{2}{Cl}_{6}(aq) + 3H_{2}(\text{ }g) + 1200\text{ }kJ/mol$.

(ii) $H_{2}(\text{ }g) + {Cl}_{2}(\text{ }g) \rightarrow 2HCl(g) + 164\text{ }kJ/mol$.

(iii) $HCl(g) + aq \rightarrow HCl(aq) + 83\text{ }kJ/mol$.

(iv) ${Al}_{2}{Cl}_{6}(\text{ }s) + aq \rightarrow {Al}_{2}{Cl}_{6}(aq) + 663\text{ }kJ/mol$.

The enthalpy of formation of anhydrous solid ${Al}_{2}{Cl}_{6}$ is :

Options

A.$- 648\text{ }kJ{\text{ }mol}^{- 1}$
B.$- 1350\text{ }kJ{\text{ }mol}^{- 1}$
C.$- 2002\text{ }kJ{\text{ }mol}^{- 1}$
D.$- 1527\text{ }kJ{\text{ }mol}^{- 1}$

Solution

Sol. [Key Concept: Apply Hess's law — combine the given reactions to build $2Al + 3{Cl}_{2} \rightarrow \ {Al}_{2}{Cl}_{6}$.]

$$2Al(s) + 6HCl(aq) \rightarrow {Al}_{2}{Cl}_{6}(aq) + 3H_{2}(\text{ }g)$$

$$\begin{matrix} & \ \Delta_{r}H_{1} = - 1200\text{ }kJ/mol \\ & \left\lbrack H_{2}(\text{ }g) + {Cl}_{2}(\text{ }g) \rightarrow 2HCl(\text{ }g) \right\rbrack \times 3 \\ & \ \Delta_{r}H_{2} = - 163 \times 3\text{ }kJ/mol \\ & \lbrack HCl(\text{ }g) + aq \rightarrow HCl(aq)\rbrack \times 6 \\ & \ \Delta_{r}H_{3} = - 83 \times 6\text{ }kJ/mol \\ & {Al}_{2}{Cl}_{6}(aq) \rightarrow {Al}_{2}{Cl}_{6}(\text{ }s) + aq.\Delta_{r}H_{4} = + 663\text{ }kJ/mole \\ & 2Al(\text{ }s) + 3{Cl}_{2} \rightarrow {Al}_{2}{Cl}_{6}(\text{ }s)\ \Delta_{r}H \\ & \begin{matrix} \Delta_{r}H = \Delta_{r}H_{1} + \Delta_{r}H_{2} + \Delta_{r}H_{3} + \Delta_{r}H_{4} \\ \ = - 1200 - 163 \times 3 - 83 \times 6 + 663 \\ \ = - 1527\text{ }kJ/mole \end{matrix} \end{matrix}$$

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