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We are given,
1. BaCl2 . 2H2O(s) + aq ------------> BaCl2 (aq),
Δsol Ho = 8.8kJ mol-1
2. BaCl2(s) + aq ---------> BaCl2(aq),
Δsol Ho = -20.6kJ mol-1
We aim at
BaCl2(s) + 2H2O ----------> BaCl2 . 2H2O(s),
Δhyd Ho = ? ......(3)
Equation (2) may be writen in two steps as
BaCl2(s) + 2H2O -------> BaCl2 . 2H2O(s),
ΔH = Δr H1o (say).......(4)
BaCl2 . 2H2O(s) + aq ----------> BaCl2(aq),
ΔH = Δr H2o (say).......(5)
Then according to Hess's Law,
Δr H1o + Δr H2o = -20.6kJ
But Δr H2o = 8.8kJ mol-1
[therefore, Equation(1) = Equation(5)]
therefore, Δr H1o = -20.6 - 8.8 = -29.4kJ mol-1
But Equation(3) = Equation(4)
Hence, the heat of hydration of BaCl2
Δhyd Ho = -29.4kJ mol-1 .
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