Question

Suppose 4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50-L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O2(g) Picture 2NO2(g) is reached when [NO] = 1.6 M. Calculate Kc for the reaction at this temperature.

Answer 1

The first thing you do is you rewrite the balanced equation in the form of an ICE table. To get the initial concentrations of NO and oxygen, you divide the moles by the volume and then record them in the initial row of the table. To get x, you set 8.4-2x=1.6. You write your Kc expression and plug the values in. The value of Kc is 3.9.

Answer 2

Answer : The value of $K_c$ for the reaction is, 3.9

Solution :  Given,

Moles of $O_2$ = 4.2 mol

Moles of $NO$ = 4.0 mol

First we have to calculate the concentration of $O_2\text{ and }NO$.

$\text{Concentration of }O_2=\frac{Moles}{Volume}=\frac{4.2mol}{0.50L}=8.4M$

$\text{Concentration of }NO=\frac{Moles}{Volume}=\frac{4.0mol}{0.50L}=8.0M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

                          $2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$

Initially conc.       8.4       8.0          0

At eqm.             (8.4-2x)  (8.0-x)     2x

The expression of $K_c$ will be,

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

$K_c=\frac{(2x)^2}{(8.4-2x)^2\times (8.0-x)}$       .......(1)

The concentration of NO at equilibrium = 1.6 M = (8.4-2x)

So,

8.4 - 2x = 1.6

x = 3.4

Now put the value of 'x' in the above equation 1, we get:

$K_c=\frac{(2\times 3.4)^2}{(8.4-2\times 3.4)^2\times (8.0-3.4)}$

$K_c=3.9$

Therefore, the value of $K_c$ for the reaction is, 3.9