Heat of Solution Formula

The heat of solution is defined as the difference in enthalpy between the dissolving material and a solvent under constant pressure, resulting in infinite dilution. When the solute is dissolved in the solvent, the enthalpy change is detected, which is referred to as heat dissolution or heat of solution. It is equal to the product of mass, temperature change and specific heat of the solvent. It is denoted by the symbol ΔH. Its standard unit of measurement is KJ/mol.

Formula

ΔH = m × ΔT × S

where,

ΔH is the heat of solution,

m is the mass of solvent,

ΔT is the change in temperature,

S is the specific heat of solvent.

Sample Problems

Problem 1. Calculate the heat of solution when a hydrated salt is dissolved in water at 300 K and rate of 43 KJ/mol. The specific heat of water is 0.004184 kJ/g^∘C.

Solution:

We have,

m = 43

T = 300

S = 0.004184

Using the formula we have,

ΔH = m × ΔT × S

= 43 × 300 × 0.004184

= 53.97 KJ/mol

Problem 2. Calculate the heat of solution when a hydrated salt is dissolved in water at 250 K and rate of 65 KJ/mol. The specific heat of water is 0.004184 kJ/g^∘C.

Solution:

We have,

m = 65

T = 250

S = 0.004184

Using the formula we have,

ΔH = m × ΔT × S

= 65 × 250 × 0.004184

= 68 KJ/mol

Problem 3. Calculate the heat of solution when a hydrated salt is dissolved in a solvent at 150 K and rate of 20 KJ/mol. The specific heat of solvent is 0.062 kJ/g^∘C.

Solution:

We have,

m = 20

T = 150 

S = 0.062

Using the formula we have,

ΔH = m × ΔT × S

= 20 × 150 × 0.062

= 186 KJ/mol

Problem 4. Calculate the mass of solvent when a hydrated salt is dissolved in it at a heat of solution 150 KJ/mol and 100 K. The specific heat of solvent is 0.165 kJ/g^∘C.

Solution:

We have,

ΔH = 150

T = 100

S = 0.165

Using the formula we have,

ΔH = m × ΔT × S

=> 150 = m × 100 × 0.165

=> m = 150/16.5

=> m = 9.09 KJ

Problem 5. Calculate the mass of solvent when a hydrated salt is dissolved in it at a heat of solution 450 KJ/mol and 500 K. The specific heat of solvent is 0.542 kJ/g^∘C.

Solution:

We have,

ΔH = 450

T = 500

S = 0.542

Using the formula we have,

ΔH = m × ΔT × S

=> 450 = m × 500 × 0.542

=> m = 450/271

=> m = 1.66 KJ

Problem 6. Calculate the temperature change for solvent when a hydrated salt is dissolved in it at a heat of solution 170 KJ/mol and rate of 70 KJ per mole. The specific heat of solvent is 0.004 kJ/g^∘C.

Solution:

We have,

ΔH = 170

m = 70

S = 0.004

Using the formula we have,

ΔH = m × ΔT × S

=> 170 = 70 × ΔT × 0.004

=> ΔT = 170/0.28

=> ΔT = 607 K

Problem 7. Calculate the temperature change for solvent when a hydrated salt is dissolved in it at a heat of solution 600 KJ/mol and rate of 120 KJ per mole. The specific heat of solvent is 0.076 kJ/g^∘C.

Solution:

We have,

ΔH = 600

m = 120

S = 0.076

Using the formula we have,

ΔH = m × ΔT × S

=> 600 = 120 × ΔT × 0.076

=> ΔT = 600/9.12

=> ΔT = 65.78 K