Question

When a 5.00 g 5.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 2.26 kJ ⋅ K − 1 , 2.26 kJ⋅K−1, the temperature decreases by 0.370 K . 0.370 K. Calculate the molar heat of solution of KBr .

Answer 1

Answer : The molar heat of solution of KBr is, 19.9 kJ/mol

Explanation :

First we have to calculate the heat absorbed by calorimeter.

$q=c\times (\Delta T)$

where,

q = heat absorbed = ?

c = specific heat capacity of calorimeter = $2.26kJ/K$

$\Delta T$ = change in temperature = 0.370 K

Now put all the given values in the above formula, we get:

$q=2.26kJ/K\times 0.370K$

$q=0.8362kJ$

Now we have to calculate the molar heat of solution of KBr.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 0.8362kJ

m = mass of $KBr$ = 5.00 g

Molar mass of $KBr$ = 119 g/mol

$\text{Moles of }KBr=\frac{\text{Mass of }KBr}{\text{Molar mass of }KBr}=\frac{5.00g}{119g/mole}=0.0420mole$

Now put all the given values in the above formula, we get:

$\Delta H=\frac{0.8362kJ}{0.0420mole}$

$\Delta H=19.9kJ/mol$

Thus, the molar heat of solution of KBr is, 19.9 kJ/mol