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Enthalpies of solution of \small BaCl_2 and \small BaCl_2.2H_2O are -19 KJ/mol and 7 KJ/mol, respectively. Calculate enthalpy of hydration (in KJ/mol) of \small BaCl_2 to \small BaCl_2.2H_2O.

(Response should be an integer value)

Option: 1

-12


Option: 2

12


Option: 3

-26


Option: 4

26


Answers (1)

best_answer

Enthalpy of hydration of  BaCl2  is given as,

\mathrm{BaCl_{2(s)}+2H_2O_{(l)}\rightarrow BaCl_2.2H_2O_{(s)}\: \: \: \: \Delta _{hyd}H=?}

We have given enthalpy of solution of BaCl2 and BaCl2.2H2O.

\mathrm{(i).BaCl_{2(s)}+aq+2H_2O_{(l)}\rightarrow BaCl_2.2H_2O_{(aq)}\: \: \: \: \Delta H=-19KJ/mol}\mathrm{(ii).BaCl_{2}.2H_2O_{(s)}+aq\rightarrow BaCl_2.2H_2O_{(aq)}\: \: \: \: \Delta H=+7KJ/mol}

Subtracting equation (ii) from (i) we get the resultant equation as,

\mathrm{BaCl_{2(s)}+2H_2O_{(l)}\rightarrow BaCl_2.2H_2O_{(s)}\: \: \: \: \Delta H=(-19-7)KJ/mol}

The above reaction is Enthalpy of hydration of  BaCl2  .

So,

\mathrm{\Rightarrow \Delta H=-26KJ/mol}

Posted by

himanshu.meshram

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