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Heat of Reaction

Enthalpies of...

Question

Enthalpies of solution of $B a C l_{2} .2 H_{2} O$ and $B a C l_{2}$ are $8.8$ and $- 20.6 k J m o l^{- 1}$ respectively. Calculate the heat of hydration of $B a C l_{2}$ to $B a {C l}_{2} .2 H_{2} O$ .

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Solution

Solution:-
${B a {C l}_{2}}_{(s)} + a q . ⟶ {B a {C l}_{2}}_{a q .} Δ H = - 20.6 k J / m o l . . . . . (1)$
${B a {C l}_{2} .2 H_{2} O}_{(s)} + a q . ⟶ {B a {C l}_{2}}_{(a q .)} + 2 H_{2} O Δ H = 8 k J / m o l$
${B a {C l}_{2}}_{(a q .)} + 2 H_{2} O ⟶ {B a {C l}_{2} .2 H_{2} O}_{(s)} + a q . Δ H = - 8 k J / m o l . . . . . (2)$
Adding ${e q}^{n} (1) & (2)$ , we have
${B a {C l}_{2}}_{(s)} + a q . + {B a {C l}_{2}}_{(a q .)} + 2 {H_{2} O}_{(l)} ⟶ {B a {C l}_{2}}_{a q .} + {B a {C l}_{2} .2 H_{2} O}_{(s)} + a q . Δ H = (- 20.6) + (- 8)$
${B a {C l}_{2}}_{(s)} + 2 {H_{2} O}_{(l)} ⟶ B a {C l}_{2} .2 H_{2} O Δ H = - 28.6 k J / m o l$
Hence the heat of hydration of anhydrous $B a {C l}_{2}$ is $- 28.6 k J / m o l$ .

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