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.

Step-by-step 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=([SO_3][NO])/([SO_2][NO_2])`

We are given:

`K_c=2.50`

Putting values in above equation, we get:

`2.50=((0.005+x)* (0.005+x))/((0.500-x)* (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.