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enthalpy of solution of delta H for bacl 2.2 H2O and bacl2 are 8.8 and 20.6 kg joule per mole calculate heat of hydration of bacl2 bacl 2.2 H2O
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 atBaCl2(s) + 2H2O ----------> BaCl2 . 2H2O(s), Δhyd Ho = ? ......(3) Equation (2) may be writen in two steps asBaCl2(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 .
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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