Enthalpy of solution (ΔH) for BaCl2 . 2H2O & BaCl2 are 8.8 & -20.6kJ mol-1 reapectively. Calculate the heat of hydration of BaCl2 to BaCl2 . 2H2O................

Hi Siddhartha,

 

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 .

 

Hopes this will help u sid...!!@@!!

 

Cheerrzzzzzzz.........!!@@!! :-)

  • 42

Thanx olivea :):)

  • -9

no need of thanx dear...its my pleasure to help u out :))

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