Question

One mole of anhydrous salt AB dissolves in water and liberates 21.0 J mol $^{-1}$ of heat. The value of $\Delta$ H $_{hydration}$ AB is -29.4 J mol $^{-1}$ . The heat of dissolution of hydrated salt AB.2H $_2$ O $_{(s)}$ is:

A. 50.4 J mol $^{-1}$

B. 8.4 J mol $^{-1}$

C. -50.4 J mol $^{-1}$

D. 8.4 J mol $^{-1}$

Answer 1

The heat of hydration is defined in the terms of heat enthalpy of solution. The heat enthalpy of solution (H $_{solution}$ ) is the sum of the lattice (H $_{lattice\,energy}$ ), and the hydration energies (H $_{hydration}$ ) of the solution.

Complete step by step answer:

First, let us define the heat of hydration. It is defined as the amount of energy liberated when 1 mole of ions undergo hydration. It is considered to be a special type of dissolution energy, where the solvent is water.

Now, we will write the reaction of dissolution of anhydrous salt AB, i.e.

AB + 2H $_2$ O $\rightarrow$ AB.2H $_2$ O (aq), $\Delta$ H $_1$ = -21.0 J/ mol

Now, let us write the reaction of hydration of AB salt, i.e.

AB + 2H $_2$ O $\rightarrow$ AB.2H $_2$ O (s), $\Delta$ H $_2$ = -29.4 J/ mol

We have to find the heat of dissolution of hydrated salt; the chemical reaction can be written as:

AB.2H $_2$ O (s) + (aq) $\rightarrow$ AB.2H $_2$ O (aq),

Now, as mentioned

$\Delta$ H $_{req}$ = $\Delta$ H $_1$ - $\Delta$ H $_2$

$\Delta$ H $_{req}$ = (-21.0 J/mol) - (-29.4 J/mol)

$\Delta$ H $_{req}$ = 8.4 J/mol; $\Delta$ H $_{req}$ represents the heat of dissolution.

Therefore, the heat of dissolution of hydrated salt is AB.2H $_2$ O (s) is 8.4 J/mol, or 8.4 J mol $^{-1}$ .

The correct option is B.

Note: Now, the confusion can arise that why negative sign is placed before the value of $\Delta$ H $_1$ , because we are given in the question that it is the amount of heat liberated by the one mole of anhydrous salt when dissolved in water, or we can say that the enthalpy of hydration is exothermic reaction for ions dissolving in water.