The heat of solution of anhydrous $C u S O_{4}$ is $- 15.9 k c a l$ and that of $C u S O_{4} .5 H_{2} O$ is $2.8 k c a l$ . Calculate the heat of hydration of $C u S O_{4}$ .

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Solution

1) $C u S O_{4} (s) \to C u S O_{4} (a q) △ H_{f} = - 15.9 K c a l$ (given)
2) $C u S O_{4} .5 H_{2} O (s) \to C u S O_{4} (a q) + 5 H_{2} O △ H_{2} = + 2.8 k c a l$ (given)
To find hydration enthalpy of $C u S O_{4}$
3) $C u S O_{4} (s) + 5 H_{2} O △ H a t m - ---- \to C u S O_{4} .5 H_{2} O (s)$
Reversing equation (2)
4) $C u S O_{4} (a q) + 5 H_{2} O △ H_{h y d r a t i o n} - ------ \to C u S O_{4} .5 H_{2} O (s) △ H_{4} = - 2.8 k c a l$
Adding (4) + (1)
$C u S O_{4} (s) + 5 H_{2} O △ H_{h y d r a t i o n} - ------ \to C u S O_{4} .5 H_{2} O (s)$
$△ H_{h y d r a t i o n} = △ H_{1} + △ H_{4} = - 18.7 k c a l$

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