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Neutralisation Reaction

Set up a ther...

Question

Set up a thermodynamic cycle for determining the enthalpy of hydration of ${C a}^{2 +}$ ions using the following data:
Enthalpy of sublimation of $C a (s), + 178.2 k J$ ${m o l}^{- 1}$ , first and second ionization enthalpies of $C a (g), 589.7 k J$ ${m o l}^{- 1}$ and $1145 k J$ ${m o l}^{- 1}$ ; enthalpy of vaporization of bromine, $+ 30.91 k J$ ${m o l}^{- 1}$ ; dissociation enthalpy of ${B r}_{2} (g), + 192.9 k J$ ${m o l}^{- 1}$ ; electron gain enthalpy of $B r (g), - 331.0 k J$ ${m o l}^{- 1}$ ; enthalpy of solution of $C a {B r}_{2} (s), - 103.1 k J$ ${m o l}^{- 1}$ ; enthalpy of hydration of ${B r}^{- -} (g), - 337 k J$ ${m o l}^{- 1}$ (only magnitude in nearest integer in kj/mol).

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Solution

Hydration of $C a^{+ 2}$
$R : C a^{+ 2} (g) \to C a^{+ 2} (a q .); Δ H$
Given
sublimation of Ca(s)
$R_{1} : C a (s) \to C a (g); Δ H_{1} = + 178.2 k J m o l^{- 1}$
First ionization
$R_{2} : C a (g) \to C a^{+} (g); Δ H_{2} = + 589.7 k J m o l^{- 1}$
Second Ionization
$R_{3} : C a^{+} (g) \to C a^{2 +} (g); Δ H_{3} = + 1145 k J m o l^{- 1}$
Vaporization of bromine (Bromine exits as diatomic liquid)
$R_{4} : B r_{2} (l) \to B r_{2} (g); Δ H_{4} = + 30.91 k J m o l^{- 1}$
Dissociation of $B r_{2}$
$R_{5} : B r_{2} (g) \to 2 B r (g); Δ H_{5} = + 192.9 k J m o l^{- 1}$
Electron gain by Br
$R_{6} : B r (g) \to B r^{-} (g); Δ H_{6} = - 331.0 k J m o l^{- 1}$
Hydration of $B r^{-}$
$R_{7} : B r^{-} (g) \to B r^{-} (a q .); Δ H_{7} = - 337.0 k J m o l^{- 1}$
Dissolution of $C a B r_{2} (s)$
$R_{8} : C a B r_{2} (s) \to C a^{+ 2} (a q .) + 2 B r^{-} (a q .); Δ H_{8} = - 103.1 k J m o l^{- 1}$
Formation of $C a B r_{2}$
$R_{9} : C a (s) + B r_{2} (l) \to C a B r_{2} (s); Δ H_{9} = - 683 k J m o l^{- 1}$
Now
$- R = R_{1} + R_{2} + R_{3} + R_{4} + R_{5} + 2 R_{6} + 2 R_{7} - R_{8} - R_{9}$
$\Rightarrow R = - R_{1} - R_{2} - R_{3} - R_{4} - R_{5} - 2 R_{6} - 2 R_{7} + R_{8} + R_{9}$
$\Rightarrow Δ H = - Δ H_{1} - Δ H_{2} - Δ H_{3} - Δ H_{4} - Δ H_{5} - 2 Δ H_{6} - 2 Δ H_{7} + Δ H_{8} + Δ H_{9}$
$\Rightarrow Δ H = - 178.2 - 589.7 - 1145 - 30.91 - 192.9 + 2 \times 331.0 + 2 \times 337.0 - 103.1 - 683 = 1587.19 k J m o l^{- 1}$
Answer = 1587

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C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

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Q. Calculate the lattice formation enthalpy in

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