Question

The ΔH°soln of HNO3 is –33.3 kJ/mol. 10.0 mL of 12.0 M HNO3 is dissolved in 100.0 mL of distilled water initially at 25°C.

How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to 25°C after dissolution of the acid and equilibrium with the ice is reached? The molar heat capacity is 80.8 J/(mol·°C) for the solution, and the molar heat capacity is 75.3 J/(mol·°C) for pure water..

Answer 1

Moles of HNO₃ = (10 mL/1000 mL) L x 12 M = 0.12 Moles

Heat imparted = 0.12 moles x (33.3 kJ/mol HNO₃) = 3.996 kJ = 3996 J

Assuming conservation of energy, this heat goes into heating 100.00 mL of water

Using density of water as 0.9970749 g/mL at 25 ^oC and by assuming constant density through the temperature change

q = n_water C_p x T_water

3996 J = 100 mL x 0.9970749 g/mL x 75.3 J/mol ^oC/18.015g x (T_f-25 ^oC)

On solving, T_f = 34.58 ^oC

Now, this solution should be returned to 25 ^oC by adding ice.

Since the solution had a higher heat capacity(80.8 J/mol), new volume (110 mL) and a new density (0.9938447 g/mL), more heat than 3996 J would be required.

According to the law of conservation of energy,

qsoln + qice = 0

n_soln C_p x T_soln + q1 + q2 = 0

Now, the heat that must be transferred out of the solution can be calculated as,

q_soln = 110 mL x (0.9938447 g/mL) x (1 mol/18.015 g) x 80.8 J ^oC (25^oC – 34.58 ^oC) = 4697 J

Now, total heat coming in to the ice can be calculated as,

q_ice = q₁ + q₂ = (m_ice/18.015)[H_fus + C_p water x T_water]

Since,

-q_soln = q_ice

4697 J = (m_ice/18.015)[H_fus + C_p water x T_water]

Now, [H_fus + C_p water x T_water] = 6010 J/mol + 75.3 J/mol(25^oC -0) = 7892.5 J/mol

The mass of ice, mice = (4697/7892.5) x 18.015g/mol = 10.72 g of ice

Mass of ice required = 10.72 g