Question

The value of $\Delta {{H}_{sol}}$ of $BaC{{l}_{2}}$(s) and $BaC{{l}_{2}}.2{{H}_{2}}O$ (s) are a kJ and b kJ respectively. The value of $\Delta {{H}_{hydration}}$ of $BaC{{l}_{2}}$(s) is:
(a)- b-a
(b)- a+b
(c)- -a-b
(d)- a-b

Answer 1

Hint: Write the equation of solution for enthalpy of solution of barium chloride and hydrated barium chloride. Now, add both equations in such a way that this equation is equal to the equation of hydration of barium chloride.

Complete answer:
The enthalpy of solution of a substance in a particular solvent is defined as the enthalpy change (i.e., amount of heat evolved or absorbed) when 1 mole of the substance is dissolved in a specific amount of solvent. However, if such a large volume of the solvent is taken that further addition of the solvent does not produce any more heat change, it is called enthalpy of solution at infinite dilution.
The equation of the solution of barium chloride will be:
$BaC{{l}_{2}}(s)+aq\to BaC{{l}_{2}}(aq)$
The enthalpy of the solution of barium chloride given is a kJ. This is equation 1.
The equation of the solution of hydrated barium chloride will be:
$BaC{{l}_{2}}.2{{H}_{2}}O(s)+aq\to BaC{{l}_{2}}(aq)+2{{H}_{2}}O$
The enthalpy of the solution of hydrated barium chloride given is b kJ. This is equation 2.
The amount of enthalpy change (i.e., the heat evolved or absorbed) when one mole of the anhydrous salt combines with the required number of moles of water to change into the hydrated salt, is called the enthalpy of hydration or heat of hydration.
The equation of hydration of barium chloride is:
$BaC{{l}_{2}}(s)+2{{H}_{2}}O\to BaC{{l}_{2}}.2{{H}_{2}}O$
For calculating the enthalpy of hydration of barium chloride we have to add equations 1 and 2.
But before that, equation 2 is reversed.
$BaC{{l}_{2}}(aq)+2{{H}_{2}}O\to BaC{{l}_{2}}.2{{H}_{2}}O(s)+aq$
Then the enthalpy of the solution of hydrated barium chloride will be –b kJ. This is equation 3.
Now add equation 1 and 3, we get
$BaC{{l}_{2}}(s)+aq+BaC{{l}_{2}}(aq)+2{{H}_{2}}O\to BaC{{l}_{2}}.2{{H}_{2}}O(s)+aq+BaC{{l}_{2}}(aq)$
$BaC{{l}_{2}}(s)+2{{H}_{2}}O\to BaC{{l}_{2}}.2{{H}_{2}}O$
So, the enthalpy will be: $a+(-b)$ $=\ a-b$
Hence, the enthalpy of hydration of barium chloride is $a-b$

Hence, the correct answer is an option (d)- a-b

Note:
If the enthalpy of the reaction is negative then the reaction is exothermic i.e., heat is evolved during the reaction. If the enthalpy of the reaction is positive then the reaction is endothermic i.e., heat is absorbed during the reaction. Don’t forget to change the sign of enthalpy of the reaction when it is reversed.