
Given that the enthalpy of solution of the hydrated copper (II) sulphate is +11.3 kJ/mol, what is the enthalpy of hydration of the anhydrous solid? (Enthalpy of solution of anhydrous copper (II) sulphate is -66.5 kJ/mol).
(A) 97.321 kJ/mol
(B) 87.321 kJ/mol
(C) 77.971 kJ/mol
(D) None of these
Answer
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Hint: Enthalpy of solution $(\Delta {H_{sol}})$ is the enthalpy change occurs when sample dissociates in the solvent. Enthalpy of hydration $(\Delta {H_{hyd}})$ is the heat released when an anhydrous molecule gets hydrated. By relating the given enthalpy of solution data we can obtain the enthalpy of hydration.
Complete step by step answer:
Step-1.
First of all write chemical equations for every possible reaction according to the question.
As given in the question, enthalpy of solution of the hydrated copper (II) sulphate is +11.3 kJ/mol. We know enthalpy of solution$(\Delta {H_{sol}})$ is the enthalpy change occurs when a sample is dissolved in the solvent. So we can write the equation as follows:
$CuS{O_4}.5{H_2}{O_{(s)}} + aq \to CuS{O_4} + 5{H_2}O$ $(\Delta {H_1} = + 11.3kJ/mol)$…. (i)
In similar way enthalpy of solution of anhydrous copper (II) sulphate can be represented as:
$CuS{O_4}_{(s)} + aq \to CuS{O_4}_{(aq)}$ $(\Delta {H_2} = - 66.5kJ/mol)$…. (ii)
Now the question is to find out enthalpy of hydration of the anhydrous copper (II) sulphate i.e.
$CuS{O_4}_{(s)} + 5{H_2}O \to CuS{O_4}.5{H_2}O$ $(\Delta {H_3} = ?)$…. (iii)
Step-2.
Relate equation (i) and (ii) to get equation (iii). On subtracting equation (i) from equation (ii) we can get equation (iii).
So the enthalpy relation will be:
$\Delta {H_3} = \Delta {H_2} - \Delta {H_1}$…. (iv)
On putting values of the enthalpies from equation (i) and (ii) to equation (iv)
$\Delta {H_3} = - 66.5kJ/mol - ( + 11.3kJ/mol)$
$\Delta {H_3} = - 77.8kJ/mol$
So, the correct answer is “Option D”.
Note: Don’t get confused with enthalpy of hydration with enthalpy of solution as the sound is similar in case when solvent is water. Another point to keep in mind is the relation of reactions to obtain the required equation, as if they are related incorrectly the answer will be wrong.
Complete step by step answer:
Step-1.
First of all write chemical equations for every possible reaction according to the question.
As given in the question, enthalpy of solution of the hydrated copper (II) sulphate is +11.3 kJ/mol. We know enthalpy of solution$(\Delta {H_{sol}})$ is the enthalpy change occurs when a sample is dissolved in the solvent. So we can write the equation as follows:
$CuS{O_4}.5{H_2}{O_{(s)}} + aq \to CuS{O_4} + 5{H_2}O$ $(\Delta {H_1} = + 11.3kJ/mol)$…. (i)
In similar way enthalpy of solution of anhydrous copper (II) sulphate can be represented as:
$CuS{O_4}_{(s)} + aq \to CuS{O_4}_{(aq)}$ $(\Delta {H_2} = - 66.5kJ/mol)$…. (ii)
Now the question is to find out enthalpy of hydration of the anhydrous copper (II) sulphate i.e.
$CuS{O_4}_{(s)} + 5{H_2}O \to CuS{O_4}.5{H_2}O$ $(\Delta {H_3} = ?)$…. (iii)
Step-2.
Relate equation (i) and (ii) to get equation (iii). On subtracting equation (i) from equation (ii) we can get equation (iii).
So the enthalpy relation will be:
$\Delta {H_3} = \Delta {H_2} - \Delta {H_1}$…. (iv)
On putting values of the enthalpies from equation (i) and (ii) to equation (iv)
$\Delta {H_3} = - 66.5kJ/mol - ( + 11.3kJ/mol)$
$\Delta {H_3} = - 77.8kJ/mol$
So, the correct answer is “Option D”.
Note: Don’t get confused with enthalpy of hydration with enthalpy of solution as the sound is similar in case when solvent is water. Another point to keep in mind is the relation of reactions to obtain the required equation, as if they are related incorrectly the answer will be wrong.
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