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 = ? $$$\text{From }(b) - (a);\Delta H = 11.5 - ( - 66.11) = 77.61\text{ kJ}$$
Create a free account to view solution
View Solution FreeMore Thermochemistry Questions
Enthalpy change for the reaction,4H(g) → 2H2 (g) is - 869.6 kJ The dissociation energy of H - H bond is...Which of the following property is correct for the compound having maximum covalent character among the following series...Formation of ozone from oxygen is an endothermic process. In the upper atmosphere, ultraviolet is the source of energy t...The species which by definition has ZERO standard molar enthalpy of formation at 298 K is...Calculate the enthalpy of the following homogeneous gaseous reactionCH3COCH3 + 2O2 → CH3COOH + CO2 + H2Ofrom the followi...