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Hess' Law

KCls⟶ K + + C...

Question

$K C l_{(s)} ⟶ K^{+} + C l^{-}$

The heat evolved in kcal for the process at $250^{\circ} C$ of dissolving 1.0 mole of KCl in excess of water is __________.

[Given: $Δ H_{f}^{\circ}$ of $K_{(a q)}^{+}, C l_{(a q)}^{-}$ and $K C l_{(s)}$ are -60, -40 and -104 kcal/mole]

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Solution

The heat evolved for the process is
$Δ H = Δ H_{f} (K^{+}) + Δ H_{f} (C l^{-}) - Δ H_{f} (K C l) = - 60 k c a l / m o l - 40 k c a l / m o l - (- 104 k c a l / m o l) = 4 k c a l / m o l$

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