Using the following data, calculate equilibrium constant K for the reaction
$AgC l_{(s)} ⇌ {Ag}_{(a q)}^{+} + {Cl}_{(a q)}^{-}$
$Δ G_{f (AgCl)}^{\circ} = - 109.7 kJ, Δ G_{f (A g^{+})}^{\circ} = 77.1 kJ,$
$Δ G_{f (C l^{-})}^{\circ} = - 131.2 kJ,$

55.6 \times 10^{3}

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2.95 \times 10^{- 41}

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1.78 \times 10^{- 10}

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9.75

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Solution

The correct option is D $1.78 \times 10^{- 10}$
$Δ G_{r e a c t i o n}^{\circ} = \sum Δ G_{f}^{\circ} (p r o d u c t s) - \sum Δ G_{f}^{\circ} (r e a c t a n t s) = Δ G_{f}^{\circ} ({A g}^{+}) + Δ G_{f}^{\circ} ({C l}^{-}) - Δ G_{f}^{\circ} (A g C l) = 77.1 + (- 131.2) - (- 109.7) = 55.6 k J$
Now, $Δ G = Δ G^{\circ} + R T ㏑ K$
For equilibrium, $Δ G = 0$
$∴ Δ G^{\circ} = - R T ㏑ K_{e q} ∴ \frac{- 55.6 \times 10^{3}}{8.314 \times 298} = ㏑ K_{e q} ∴ K_{e q} = e^{- \frac{55600}{8.314 \times 298}} = 1.78 \times 10^{- 10}$

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Similar questions

A g C l_{s} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}

x \times 10^{- 10}

x

: Δ G_{f (A g C l)}^{0} = - 109.7 k J;

Δ G_{f (A g^{+})}^{0} = 77.1 k J;

Δ G_{f (C l^{-})}^{0} = - 131.2 k J

A g C l_{(s)} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}

: Δ G_{f A g C l}^{0} = - 109.7 k J

Δ G_{f A g^{+}}^{0} = 77.1 k J

Δ G_{f C l^{-}}^{0} = - 131.2 k J

If ∆G° for a reaction is 1.93×10^6. Calculate the equilibrium constant if coditions are standard.

Δ G^{\circ}

δ G

Δ G^{\circ}

(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)

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