Ncert Exercises & Solutions

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) I2 (g) + Cl2 (g); Kc = 0.14

The given reaction is:

2 {ICI}_{(g)} \leftrightarrow I_{2 (g)} {Cl}_{2 (g)}

Initial conc . 0.78 M 0 0

At equilibrium (0.78 - 2 x) M xM xM

Now, we can write,

\frac{[I_{2}] [{Cl}_{2}]}{{[ICl]}^{2}} = K_{c}

\begin{array}{l} \Rightarrow \frac{x \times x}{(0.78 - 2 x)^{2}} = 0.14 \\ \Rightarrow \frac{x^{2}}{(0.78 - 2 x)^{2}} = 0.14 \\ \Rightarrow \frac{x}{0.78 - 2 x} = 0.374 \\ \Rightarrow x = 0.292 - 0.748 x \\ \Rightarrow 1.748 x = 0.292 \\ \Rightarrow x = 0.167 \end{array}

Hence, at equilibrium,

\begin{array}{l} [H_{2}] = [I_{2}] = 0.167 M \\ [HI] = (0.78 - 2 \times 0.167) M \\ = 0.446 M \end{array}