1. Given reaction

$2\,\mathrm{NO}(g)+\mathrm{O_2}(g)\rightleftharpoons 2\,\mathrm{NO_2}(g)$

We need to find the equilibrium constant at $600\,\mathrm{K}$.

2. Initial moles

$n_{\mathrm{NO}}=2,\quad n_{\mathrm{O_2}}=1,\quad n_{\mathrm{NO_2}}=0$

At equilibrium, oxygen present is $0.6$ mol.

So oxygen consumed is

$1-0.6=0.4\text{ mol}$

3. Use stoichiometry

From the reaction,

$1\text{ mol }\mathrm{O_2} \text{ consumed} \Rightarrow 2\text{ mol }\mathrm{NO} \text{ consumed and }2\text{ mol }\mathrm{NO_2} \text{ formed}$

Therefore, if $0.4$ mol $\mathrm{O_2}$ is consumed:

$\mathrm{NO}\text{ consumed}=2\times 0.4=0.8$ $\mathrm{NO_2}\text{ formed}=2\times 0.4=0.8$

Thus equilibrium moles are:

$n_{\mathrm{NO}}=2-0.8=1.2$ $n_{\mathrm{O_2}}=0.6$ $n_{\mathrm{NO_2}}=0.8$

4. Total moles at equilibrium

$n_{\text{total}}=1.2+0.6+0.8=2.6$

Total pressure is $1$ atm, so partial pressures are mole fractions times total pressure.

$P_{\mathrm{NO}}=\frac{1.2}{2.6}\times 1=\frac{1.2}{2.6}$ $P_{\mathrm{O_2}}=\frac{0.6}{2.6}\times 1=\frac{0.6}{2.6}$ $P_{\mathrm{NO_2}}=\frac{0.8}{2.6}\times 1=\frac{0.8}{2.6}$

5. Expression for equilibrium constant

For the reaction,

$K_p=\frac{\left(P_{\mathrm{NO_2}}\right)^2}{\left(P_{\mathrm{NO}}\right)^2\left(P_{\mathrm{O_2}}\right)}$

Substitute values:

$K_p=\frac{\left(\frac{0.8}{2.6}\right)^2}{\left(\frac{1.2}{2.6}\right)^2\left(\frac{0.6}{2.6}\right)}$

Simplify:

$K_p=\frac{0.8^2}{1.2^2\cdot 0.6}\times 2.6$

Since

$0.8^2=0.64,\quad 1.2^2=1.44,\quad 1.44\times 0.6=0.864$

So

$K_p=\frac{0.64}{0.864}\times 2.6$

$K_p\approx 0.7407\times 2.6\approx 1.93$

6. Nearest integer

$K_p\approx 2$

7. Comparison with stored answer

Stored correct answer = $2$.

Our derived answer matches it.