Question

Consider the following reaction:

Answer 1

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

                        $SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Initial conc.    $2.4\times 10^{-2}$          0             0

At eqm.          $(2.4\times 10^{-2}-x)$   x              x

As we are given:

Concentration of $Cl_2$ at equilibrium = $1.3\times 10^{-2}M$

That means,

$x=1.3\times 10^{-2}M$

The expression for equilibrium constant is:

$K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$

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

$K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}$

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}$

$K_c=0.0154$

Thus, the value of equilibrium constant (Kc) is, 0.0154