Consider the exothermic reaction COCl42– (aq) + 6H2O (ℓ) ⇌ [CO(H2O)6]2+ (aq) + 4Cl– (aq) [Consider initially only COCl42

Engineering

Chemistry

Large or Small Equilibrium Constant Problems

Question

Consider the exothermic reaction

COCl₄^2– (aq) + 6H₂O (ℓ) ⇌ [CO(H₂O)₆]²⁺ (aq) + 4Cl^– (aq)

[Consider initially only COCl₄^2– is added to water]

On increasing temperature the concentration of CoCl₄^2– will decrease

If HCl is added to above equilibrium then concentration of CoCl₄^2– (aq) will increase

On doubling the volume of solution by addition water the final concentration of [Co(H₂O)₆]²⁺ will be more than its initial concentration

On dilution with water the concentration of CoCl₄^2– (aq) will decrease

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On adding Cl^– ion the equilibrium will shift in the backward direction so. [COCl₄]^2– will increase on dilution the volume will increase.

Now as $K_{C} = \frac{n {[CO {(H_{2} O)}_{6}]}^{2 +} . n_{{Cl}^{-}}^{4}}{{n_{COCl}}_{4}^{2 -}} . \frac{1}{V^{4}}$

so. Q will get decreased and equilibrium will shift in the forward direction. Though the amount of [CO(H₂O)]²⁺ will get increased but its concentration can not get doubled as volume has increased.

As reaction is exothermic in nature so increasing temp will shift equilibrium is backward direction.

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