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Enthalpy of Combustion

The heat of s...

Question

The heat of solution of anhydrous $C u S O_{4}$ and $C u S O_{4} .5 H_{2} O$ are - 15.89 and 2.80 $K c a l {m o l}_{- 1}$ respective. What will be heat of hydration of anhydrous $C u S O_{4}$ ?

A

-18.69 KCal

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B

18.69 KCal

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C

-28.96 KCal

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D

28.96 KCal

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Solution

The correct option is A -18.69 KCal
Solution:- (A) $- 18.69 k c a l$
${C u S O_{4}}_{(s)} + a q . ⟶ {C u S O_{4}}_{a q .} Δ H = - 15.89 k c a l / m o l . . . . . (1)$
$C u S O_{4} .5 H_{2} O + a q . ⟶ {C u S O_{4}}_{(a q .)} + 5 H_{2} O Δ H = 2.8 k c a l / m o l$
${C u S O_{4}}_{(a q .)} + 5 H_{2} O ⟶ C u S O_{4} .5 H_{2} O + a q . Δ H = - 2.8 k c a l . . . . . (2)$
Adding ${e q}^{n} (1) & (2)$ , we have
${C u S O_{4}}_{(s)} + a q . + {C u S O_{4}}_{(a q .)} + 5 {H_{2} O}_{(l)} ⟶ {C u S O_{4}}_{a q .} + C u S O_{4} .5 H_{2} O + a q . Δ H = (- 15.89) + (- 2.8)$
${C u S O_{4}}_{(s)} + 5 {H_{2} O}_{(l)} ⟶ C u S O_{4} .5 H_{2} O Δ H = - 18.69 k c a l$
Hence the heat of hydration of anhydrous $C a S O_{4}$ is $- 18.69 k c a l$ .

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