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Equilibrium Constant and Standard Free Energy Change

Using the dat...

Question

Using the data given below, predict $Δ H$ for the reaction in $k J {m o l}^{- 1}$ .
$M g (g) + 2 F (g) ⟶ {M g}^{2 +} (g) + 2 F^{-} (g)$
${I E}_{2}$ and ${I E}_{2}$ of $M g (g)$ are $737.7$ and $1451 k J$ ${m o l}^{- 1}$ . ${E A}_{1}$ for $F (g)$ is $- 328 k J$ ${m o l}^{- 1}$ .
[Take the nearest integer]

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Solution

$Δ H = I E_{1} + {I E}_{2} + 2 \times {E A}_{1}$
$= 737.7 + 1451 + 2 \times (- 328)$
$= + 1532.7 k J$ ${m o l}^{- 1} \equiv 1533 k J m o l^{- 1}$

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Q. Use (IE) and (EA) listed below to determine whether the following process is endothermic or exothermic:

M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g)

(I E)_{1}

M g (g) = 737.7 k J m o l^{- 1}

(I E)_{2}

M g (g) = 1451 k J m o l^{- 1}

(E A)

F (g) = - 328 k J m o l^{- 1}

Q. Calculate the enthalpy of formation of

M g F_{2}

from the following data-
Enthalpy of sublimation for

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

Enthalpy of dissociation of

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

Ionisation energy of

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

Electron gain enthalpy of

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

Lattice enthalpy of

M g F_{2} = - 2922.5 k J m o l^{- 1}

I E_{1}

I E_{2}

of magnesium are

178

348

The energy for the reaction

M g_{(g)} \to M g_{(g)}^{2 +} + 2 e^{-}

△_{i} H_{1}

△_{i} H_{2}

of Mg(g) are 740 and 1450 kJ

△_{i} H_{2}

{m o l}^{- 1}

. Calculate the percentage of

{M g}^{+} (g)

{M g}^{2 +} (g)

if 1 g of Mg(g) absorbs 50 kJ of energy.

I E

A l (g) ⟶ {A l}^{+} (g) + e

577.5 k J

{m o l}^{- 1}

Δ H

A l (g) ⟶ {A l}^{3 +} (g) + 3 e

5140 k J

{m o l}^{- 1}

. If second and third

I E

values are in the ratio

2 : 3

{I E}_{2}

{I E}_{3} - 0.5

k J

{m o l}^{- 1}

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