# 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.........!!@@!! :-)

• 46

Thanx olivea :):)

• -7

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

• 0
Dont know
• -17
what
• 2
What are you looking for?