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  • What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) I_2 (g) + Cl_2 (g); K_c = 0.14

7.16     What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?

               2ICl _{(g)}\rightleftharpoons I_{2}_{(g)}+Cl_{2}_{(g)}; K_{c}=0.14

Answers (1)

best_answer

The given reaction is:
                                2ICl (g) \rightleftharpoons I_2 (g) + Cl_2 (g)
Initial conc.                0.78 M           0           0
At equilibrium         (0.78 - 2x) M     x M        x M

The value of K_c = 0.14

Now we can write,
K_c = \frac{[I_2][Cl_2]}{[ICl]^2}
\\0.14 = \frac{x^2}{(.78-x)^2}\\ 0.374 = \frac{x}{(.78-x)}

By solving this we can get the value of x = 0.167

Posted by

manish

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