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 .

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be