Question

Calculate ΔH for the reaction 2 al (s) + 3 Cl₂ (g) → 2 alCl₃ (s) from the following data.

Answer 1

The enthalpy change (ΔH) for the reaction
`\(2 \text{Al} (\text{s}) + 3 \text{Cl}_2 (\text{g}) \rightarrow 2 \text{AlCl}_3 (\text{s})\)`is
`\(-1592 \text{ kJ}\).`

Step-by-step explanation:

In order to calculate the enthalpy change
`(\(ΔH\))`for the given reaction, we can use Hess's law. Hess's law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken to achieve the final reaction.

The given reaction can be broken down into two steps:

1.
`\(2 \text{Al} (\text{s}) + 3 \text{Cl}_2 (\text{g}) \rightarrow 2 \text{AlCl}_3 (\text{g})\)`

2.
`\(2 \text{AlCl}_3 (\text{g}) \rightarrow 2 \text{AlCl}_3 (\text{s})\)`

The enthalpy change for each step is given as follows:

`1. \(\Delta H_1 = -1331 \text{ kJ}\)`(Given data)

`2. \(\Delta H_2 = -261 \text{ kJ}\)` (Given data)

To find the enthalpy change for the overall reaction, we add the enthalpies of the individual steps:

`\[ΔH_{\text{overall}} = ΔH_1 + ΔH_2 = -1331 \text{ kJ} - 261 \text{ kJ} = -1592 \text{ kJ}.\]`

Therefore, the enthalpy change for the reaction
`\(2 \text{Al} (\text{s}) + 3 \text{Cl}_2 (\text{g}) \rightarrow 2 \text{AlCl}_3 (\text{s})\)` is
`\(-1592 \text{ kJ}\).` This negative value indicates that the reaction is exothermic, releasing heat to the surroundings. The calculated value demonstrates the net heat change associated with the formation of
`\(2 \text{AlCl}_3 (\text{s})\)`from its elemental components.