Potassium hydroxide and hydrochloric acid react in aqueous solution according to the following equation - VCE - SSCE Chemistry - Question 10 - 2009 - Paper 1

Assuming the density of the solution is similar to that of water, the mass is given by:

$\text{mass} = \text{volume} \times \text{density} = 100 \text{ mL} \times 1 \text{ g/mL} = 100 \, \text{g}$.

Using the formula for heat transfer:

$q = m imes c imes \Delta T$ where:

• $m$ = mass (100 g)
• $c$ = specific heat capacity (4.18 J/g°C)
• $\Delta T$ = change in temperature (3.5°C)

Thus,

$q = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times 3.5 \, \text{°C} = 1463 \, \text{J}$.

To convert to kilojoules:

$1463 \, \text{J} = 1.463 \, \text{kJ}.$

Both KOH and HCl are present in 0.025 mol, so the reaction has complete consumption of both reactants for the limiting case. Therefore, moles of the limiting reactant = 0.025 mol.

Using the enthalpy change formula:

$\Delta H = \frac{q}{\text{moles}} = \frac{-1.463 \, \text{kJ}}{0.025 \, \text{mol}} = -58.52 \, \text{kJ/mol}$. Rounding this gives approximately -59 kJ/mol.

The value closest to -59 kJ/mol is: B. -59