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

Step-by-step 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 * 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) = ([NO]^(2))/([N_(2)][O_(2)])`

`4.8 * 10^(-11) = ((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) = ([NO]^(2))/([N_(2)][O_(2)])`

`4.8 * 10^(-11) = (4x^(2))/((0.033)(0.00180))`

x =
`2.7 * 10^(-8)` M

Since, concentration of NO equals 2x. So, this is equal to
`2 * 2.7 * 10^(-8)` M =
`5.4 * 10^(-8)`

Thus, we can conclude that the equilibrium concentration of NO at 25 °C is
`5.4 * 10^(-8)`.