Question

A 1.00 liter flask initially contained 0.24 mol NO2 at 700 K which decomposed according to the following equation. When equilibrium was achieved, 0.14 mol NO was present. Calculate Kc. 2NO2(g) ↔ 2NO(g) + O2(g)

Answer 1

Answer:

$Kc=0.14$

Explanation:

Hello,

In this case, for the given reaction, the equilibrium expression is:

$Kc=\frac{[NO]^2[O_2]}{[NO_2]^2}$

That in terms of the reaction extent $x$ is written as (initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L):

$Kc=\frac{(2*x)^2(x)}{(0.24M-2*x)^2}$

Moreover, since at the equilibrium 0.14 moles of NO are present (a 0.14-M concentration), we can compute the reaction extent as shown below:

$[NO]=2*x=0.14M$

$x=0.14M/2=0.07M$

In such a way, knowing $x$, we compute Kc as shown below:

$Kc=\frac{(2*0.07)^2(0.07)}{(0.24M-2*0.07)^2}\\\\Kc=0.14$

Regards.