Anhydrous magnesium chloride, MgCl₂, can absorb water to form the hydrated salt, MgCl₂·4H₂O - AQA - A-Level Chemistry - Question 1 - 2017 - Paper 3

To calculate the enthalpy change for the reaction, we use the enthalpies of solution:

$\Delta H = \Delta H_{solution} (MgCl_2 \cdot 4H_2O) - \Delta H_{solution} (MgCl_2)$

Substituting the values:

$\Delta H = -39 - (-155) = -39 + 155 = 116 \text{ kJ mol}^{-1}$

Thus, the enthalpy change is +116 kJ mol⁻¹.

To determine the enthalpy of solution of anhydrous magnesium chloride, follow these steps:

Stage 1: Method

1. Measure a specific volume of distilled water (about 100 mL) in a calorimeter.
2. Record the initial temperature of the water.
3. Weigh out approximately 0.8 g of anhydrous magnesium chloride.
4. Gradually add the magnesium chloride to the water, stirring constantly.
5. Record the maximum temperature attained once the salt has completely dissolved.

Stage 2: Measurements

1. Determine the change in temperature (ΔT) by subtracting the initial temperature from the maximum temperature.
2. Use the mass of the water and the specific heat capacity (4.18 J g⁻¹ K⁻¹) to calculate the heat change (q) using the formula:

$q = m \cdot c \cdot \Delta T$

where m is the mass of the water and c is the specific heat capacity.

Stage 3: Results

1. The enthalpy of solution is then calculated by dividing the heat change by the number of moles of MgCl₂ dissolved.

This experiment allows us to calculate the enthalpy change associated with the dissolution of magnesium chloride in water, which can then be used in conjunction with the data from the previous questions.

To calculate the gradient, we will plot the free-energy change (ΔG) against temperature (T). From the table, we will use the points (240, -602.2) and (298, -592.5).

Using the formula for the gradient:

$\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{(-592.5) - (-602.2)}{298 - 240} = \frac{9.7}{58} = 0.166 \text{ kJ K}^{-1} \text{ mol}^{-1}$

Thus, the gradient is 0.166 kJ K⁻¹ mol⁻¹.

Entropy Change

To find the entropy change ΔS:

$\Delta S = \text{gradient} \times 1000 = 0.166 \times 1000 = 166 \text{ J K}^{-1} \text{ mol}^{-1}$

Therefore, ΔS for the formation of anhydrous magnesium chloride is 166 J K⁻¹ mol⁻¹.