Question

One mole of I2 vapours are placed in one litre evacuated flask at a certain temperature. When it gets 10% dissociated as I2(g) =2I(g) The value of Kc is a) 4.44 x 10-4 b) 2.22 x 10-2 c) 4.44 x 10-2 d) 8.88 x 10-2​

Answer 1

The answer is option c)$= 4.44 \times 10^{-2}$

Explanation:

The dissociation of iodine vapors is taking place as follows:

$I_2(g) \rightleftharpoons 2I(g)$

To find the equilibrium constant$K_c$, we need to prepare the ICE table.

                                                $I_2(g)$             $\rightleftharpoons$          $2I(g)$

Initial Concentration (I)           1mol/L                          0

Change (C) taking place        10% or -0.1               +0.2

At equilibrium (E)                    1-0.1 = 0.9                  0.2

Substituting these values in the equilibrium constant expression for the reaction, we obtain

$K_c =\frac{ [I]^2}{[I_2]}$

$K_c =\frac{ [0.2]^2}{[0.9]}$

$K_c =\frac{ [0.2\times 0.2]}{[0.9]}$

$K_c =0.0444$$= 4.44 \times 10^{-2}$

Hence option c) is correct.