Oxides of nitrogen are formed in air at the high temperatures generated in lightning flashes according to the equation $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ $K_{f} = 5 \times 10^{-3}$ at 3000°C At 3000°C, the equilibrium constant $K_{2}$ for the reaction $4NO(g) \rightleftharpoons 2N_2(g) + 2O_2(g)$ would be A - VCE - SSCE Chemistry - Question 15 - 2003 - Paper 1

For the first reaction:

$K_f = \frac{[NO]^2}{[N_2][O_2]}$

The reaction we are interested in is the reverse reaction:

$4NO(g) \rightleftharpoons 2N_2(g) + 2O_2(g)$

The equilibrium constant $K_2$ will be given by:

$K_2 = \frac{[N_2]^2[O_2]^2}{[NO]^4}$

Since the second reaction is the reverse of the first, we can relate $K_2$ to $K_f$ using the following formula:

$K_2 = \frac{1}{K_f^2}$

Substituting the value of $K_f$ into this equation gives:

$K_2 = \frac{1}{(5 \times 10^{-3})^2}$

Now, we calculate:

$K_2 = \frac{1}{25 \times 10^{-6}} = 4 \times 10^{4}$

Thus, the equilibrium constant $K_2$ for the reaction at 3000°C is:

A. $4 \times 10^{4}$