Question

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500 M SO2, 0.500 M NO2, 0.00500 M SO3, and 0.00500 M NO?

Answer 1

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

                         $SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial:             0.500  0.500      0.005   0.005

At eqllm:      0.500-x  0.500-x   0.005+x  0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.