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

Explanation:

Since, it is given that concentration of nitrogen is 0.033 M and concentration of oxygen is 0.00180 M. Value of is

Also, the given reaction is as follows.

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.

As, <<<< 1. So, x <<<< 1. Therefore, (0.033 - x) = 0.033 and (0.00180 - x) = 0.00180.

Therefore,          

                                x = M

Since, concentration of NO equals 2x. So, this is equal to M =

Thus, we can conclude that the equilibrium concentration of NO at 25 °C is .

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