Question

The ΔH°soln of HNO3 is –33.3 kJ/mol. 13.0 mL of 14.0 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

Answer:

15,4g of ice

Explanation:

The dissolution of HNO₃ increases the temperature of the solution thus:

Q = -C×m×ΔT (1)

Where Q is heat produced:

Q = -33 kJ/mol×(0,0130L×14,0M) = -6,00kJ

C is molar heat capacity of solution: 80,8 J/mol°C

m are moles of solution ≈ moles of water = 100mL≡100g×(1mol/18g) = 5,56 moles

And ΔT is final temperature - Initial temperature (X-25°C)

Replacing in (1):

-6000J = -80,8J/mol°C×5,56mol×(X-25°C)

13,4 = X-25°C

X = 38,4°C

Knowing that you want to return the temperature of the system to 25°C, the ice must to absorb 6000J of energy (In the fussion process and increasing each temperature until equilibrium) produced in the dissolution of HNO₃:

6000J = ΔH°fus×X + C×X×ΔT

-Where X are moles of ice-

Replacing:

6000J = 6010J/mol×X + 75,3J/mol°C×X×(38,4°C-25°C)

6000J = 6010J×X + 1006J×X

0,855 = moles of X

In grams:

0,855 moles×(18g/1mol)= 15,4g of ice

I hope it helps!