<h3>
Answer:</h3>
#a. 1.95 kJ
#b. 39 kJ/mol
<h3>
Explanation:</h3>
We are given;
Volume of HCl =100 mL
Molarity of HCl = 0.5 M
Volume of KOH = 100 mL
Molarity of KOH = 0.5 M
Change in temperature, Δt, = 2.5 °C
Specific heat capacity of the solution = 3.97 J/g°C
We are required to calculate the heat of the reaction and the molar enthalpy of the reaction in kJ/mol
Step-by-step solution;
<h3>
Step 1 : Mass of the solution </h3>
Total volume of the solution = 200 mL
Taking the density as 1 g/mL
Mass of the solution = 200 mL× 1 g/mL
= 200 g
<h3>
Step 2: Heat of the reaction in kJ</h3>
Heat change = m×c×Δt
Therefore;
Heat change, Q = 200 g × 3.9 J/g°C × 2.5 °C
= 1950 Joules
But, 1 kJ = 1000 J
Therefore, heat change = 1.95 kJ
<h3>Step 3: Molar enthalpy of the reaction (kJ/mol)</h3>
Moles = Molarity × Volume
Moles of HCl = 0.5 × 0.1 L
= 0.05 moles
But since moles of the acid are equal to the moles of water produced.
Moles of solution = 0.05 moles
But;
ΔH = heat change ÷ Number of moles
= 1.950 kJ÷ 0.05 moles
= 39 kJ/mol
Therefore, the enthalpy of the reaction is 39 kJ/mol