Question

Part A:

A hypothetical solution forms between a solid and a liquid. The values of the thermodynamic quantities involved in the process are shown in the following table.

Action	Enthalpy
Separation of Solute	17.5 kJ/mol
Separation of Solvent	28.8 kJ/mol
Formation of Solute-Solvent Interactions	-90.7 kJ/mol solute

Calculate the enthalpy of solution in kilojoules per mole of solute.

Enter your answer numerically in kilojoules per mole of solute.

Part B:

A 4.00 m solution of KCl was prepared using 1.00 kg of water at 25.0 ∘C. Once the solid had all dissolved, the temperature of the solution was 12.3 ∘C. Calculate the heat of solution, ΔHsoln, of KCl. Assume that the specific heat of the solution is identical to that of water, 4.18 J/(g⋅∘C).

Express your answer numerically in kilojoules per mole.

Answer 1

We know the enthalpy of the solution is given by following equation.

Delta Hsol = Delta H product – Delta Hreactant.

Here product is solute and solvent system,

Reactants : We have to use separation value of solute and solvent together.

Lets plug given value in above equation.

Delta H of solution = ( - 90.7 ) – ( 17.5 + 28.8 )

= -137 kJ / mol

Question 2 :

Given : Temperature change = (12.3 – 25.0 ) deg C

Molarity = 4.0 m

Mass of water = 1.0 kg

Lets first find heat of solution.

q (solution) = m C delta T

here C is specific heat of solution, m is mass of solution in g, Delta T is change in T

q = (mass of water in g + mass of KCl in g ) x 4.184 J / g deg C x (12.3 – 25.0 ) deg C

Lets calculate mass of KCl

Number of moles of KCl = 4.0 mol

Mass of KCl = Moles x molar mass

= 4.0 mol x 74.5513 g / mol

= 298.205 g

q (sol) = (1000 + 298.205 ) x 4.184 x (12.3-25.0)

= -68982.50 J

Here sign is negative since temperature of the solution is decreased.

Lets calculated delta H of solution

Delta H of solution =- q solution / n

n is number of moles of solute.

Delta H solution = - ( -68982.50 J ) / 4.0 mol

= 17245.6 J / mol

Ans in kJ /mol

= 17.24 kJ /mol