ThermochemistryHard
Question
The value of ΔHsol of anhydrous copper (II) sulphate is −66.11 kJ. Dissolution of 1 mole of blue vitriol, [Copper (II) sulphate pentahydrate] is followed by the absorption of 11.5 kJ of heat. The enthalpy of dehydration of blue vitriol is
Options
A.−77.61 kJ
B.+77.61 kJ
C.−54.61 kJ
D.+54.61 kJ
Solution
Given T.C.E. are
(a) $CuSO_{4}(s) + aq \rightarrow CuSO_{4}(aq);\Delta H = - 66.11\text{ kJ}$
(b) $CuSO_{4}.5H_{2}O(s) + aq \rightarrow CuSO_{4}(aq);\Delta H = + 11.5\text{ kJ}$
Required T.C.E. is
$CuSO_{4}.5H_{2}O(s) \rightarrow CuSO_{4}(s);\Delta H = ? $$$\text{From }(b) - (a);\Delta H = 11.5 - ( - 66.11) = 77.61\text{ kJ}$$
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