$$2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$$
In an experiment, $$2.0$$ moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be $$0.4$$ mol/L. The equilibrium constant at $$30°$$C is ______ $$\times 10^{-4}$$.

Given reaction: $$2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$$

Initial moles of NOCl = 2.0 mol in a 1 L flask, so initial concentration = 2.0 M

At equilibrium, concentration of NO = 0.4 mol/L

Setting up the ICE table:

$$2NOCl \rightleftharpoons 2NO + Cl_2$$

Initial: 2.0, 0, 0

Change: $$-2x$$, $$+2x$$, $$+x$$

Equilibrium: $$2.0 - 2x$$, $$2x$$, $$x$$

Since $$[NO] = 2x = 0.4$$ mol/L:

$$x = 0.2$$ mol/L

At equilibrium:

$$[NOCl] = 2.0 - 2(0.2) = 1.6$$ mol/L

$$[NO] = 0.4$$ mol/L

$$[Cl_2] = 0.2$$ mol/L

The equilibrium constant expression is:

$$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$$

Substituting the values:

$$K_c = \frac{(0.4)^2 \times (0.2)}{(1.6)^2}$$

$$K_c = \frac{0.16 \times 0.2}{2.56}$$

$$K_c = \frac{0.032}{2.56}$$

$$K_c = 0.0125 = 125 \times 10^{-4}$$

Hence, the answer is 125.