Question

In air, at 25 ºC and 1.00 atm, the concentrations of N2 and O2 are 0.033 M and 0.00180, respectively. The reaction N2(g) + O2 (g) 2 NO (g) has Kc= 4.8 x 10-11 at 25 °C Taking the given concentrations as the initial concentrations, calculate the equilibrium concentration of NO at 25 °C

Answer 1

Explanation:

Since, it is given that concentration of nitrogen is 0.033 M and concentration of oxygen is 0.00180 M. Value of $K_{c}$ is $4.8 \times 10^{-11}$

Also, the given reaction is as follows.

          $N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)$

At initial :      0.033       0.00180                       x

At equilibrium : 0.033 - x     0.00180 - x         2x

Expression for equilibrium constant for the given reaction will be as follows.

               $K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$

                 $4.8 \times 10^{-11} = \frac{(2x)^{2}}{(0.033 - x)(0.00180 - x)}$

As, $K_{c}$ <<<< 1. So, x <<<< 1. Therefore, (0.033 - x) = 0.033 and (0.00180 - x) = 0.00180.

Therefore,          $K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$

                          $4.8 \times 10^{-11} = \frac{4x^{2}}{(0.033)(0.00180)}$

                                x = $2.7 \times 10^{-8}$ M

Since, concentration of NO equals 2x. So, this is equal to $2 \times 2.7 \times 10^{-8}$ M = $5.4 \times 10^{-8}$

Thus, we can conclude that the equilibrium concentration of NO at 25 °C is $5.4 \times 10^{-8}$.