The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.323 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K.

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Alleei Alleei · Answer 1 · 2019-07-09T09:06:23+02:00

Answer : The equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.

Explanation : Given,

Moles of [tex]COCl_2[/tex] = 0.323 mole

Volume of solution = 1 L

Initial concentration of [tex]COCl_2[/tex] = 0.323 M

Let the moles of [tex]CO\text{ and }Cl_2[/tex] be, 'x'. So,

Concentration of [tex]CO[/tex] = x M

Concentration of [tex]Cl_2[/tex] = x M

The given balanced equilibrium reaction is,

[tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]

Initial conc. 0.323 M 0 0

At eqm. conc. (0.323-x) M (x M) (x M)

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]

Now put all the given values in this expression, we get :

[tex]1.29\times 10^{-2}=\frac{(x)\times (x)}{(0.323-x)}[/tex]

By solving the term 'x', we get :

x = 0.0584

Thus, the concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] at equilibrium are :

Concentration of [tex]COCl_2[/tex] = (0.323-x) M = (0.323-0.0584) M = 0.2646 M

Concentration of [tex]CO[/tex] = x M = 0.0584 M

Concentration of [tex]Cl_2[/tex] = x M = 0.0584 M

Therefore, the equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.