A group of students carried out an investigation into the energy changes that take place when metal hydroxides dissolve in water - Scottish Highers Chemistry - Question 9 - 2022

The experiment was carried out in a polystyrene cup with a lid to minimize heat loss to the surroundings. Polystyrene is a poor conductor of heat, which helps to insulate the contents and maintain the temperature changes accurately. The lid further prevents heat from escaping and ensures that the temperature readings reflect only the reaction taking place within the cup.

To find the energy released per mole of potassium hydroxide, we first note that 5.61 g of KOH releases 5.25 kJ. The molar mass of KOH is approximately 56.11 g/mol. The number of moles in 5.61 g is calculated as:

$n = \frac{5.61}{56.11} \approx 0.1000 \text{ mol}$

Now, we can find the energy released per mole:

$\text{Energy per mole} = \frac{5.25 \text{ kJ}}{0.1000 \text{ mol}} \approx 52.5 \text{ kJ mol}^{-1}$

Therefore, approximately 52.5 kJ of energy is released when one mole of potassium hydroxide dissolves in water.

To calculate the enthalpy change for the reaction:

$\text{Ca}(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(s) + \text{H}_2(g)$

We can use the enthalpies of formation given:

1. For the formation of water: $\Delta H_{H2O} = -286 \text{ kJ mol}^{-1}$
2. For the formation of calcium hydroxide: $\Delta H_{Ca(OH)2} = -986 \text{ kJ mol}^{-1}$

We apply Hess's law:

$\Delta H = \Delta H_{Ca(OH)2} - [\Delta H_{H2O}] \times 2$

Substituting the values:

$\Delta H = -986 - 2(-286)$ $\Delta H = -986 + 572 = -414 \text{ kJ mol}^{-1}$

Thus, the enthalpy change for the reaction is -414 kJ mol⁻¹.