When a 14.50 g of potassium chlorate is dissolved in 200.0 mL of water in a vacuum flask, the temperature falls from 25.0°C to 19.0°C. The heat capacity of the solution is 4.184 J.g^-1.°C^-1.

What is the enthalpy change ΔH_diss (in kJ.mol^-1) for the dissolution of KClO₃ in water?

Question

When a 14.50 g of potassium chlorate is dissolved in 200.0 mL of water in a vacuum flask, the temperature falls from 25.0°C to 19.0°C. The heat capacity of the solution is 4.184 J.g^-1.°C^-1.

What is the enthalpy change ΔH_diss (in kJ.mol^-1) for the dissolution of KClO₃ in water?

| General Chemistry 2

Accepted Answer

c_S = $\frac{c_{P}}{m}$ = $\frac{q_{P}}{∆ T \times m}$

c_S = specific heat capacity of the solution (in J.g^-1.°C^-1)
q_P = energy added as heat (in J) = n_KClO3 ΔH_diss
n = number of moles (in mol)
ΔT = Tf – Ti (in °C)
m = mass (in g)

c_S = $\frac{n_{KClO 3} \times ∆ H}{∆ T \times m_{solution}}$ ⇒ ΔH = $\frac{c_{S} \times ∆ T \times m_{solution}}{n_{KClO 3}}$

n_KClO3 = $\frac{m_{KClO 3}}{M_{KClO 3}}$ = $\frac{14.50}{122.6}$ = 0.118 mol

ΔH = $\frac{4.184 \times 6.0 \times 214.50}{0.118}$ = 45.6 x 10⁴ J.mol^-1 = 45.6 kJ.mol^-1

Midterm 2 - Question 1 | General Chemistry 2

General Chemistry 2 - Midterm 2