Question

At 298 K, the equilibrium concentration of O2 is 1.6 x 10-2 M, and the equilibrium concentration of O3 is 2.86 x 10-28 M. What is the equilibrium constant of the reaction at this temperature?

A. 2.0*10∧-50
B. 2.0*10∧50
C. 1.8*10∧-26
D. 1.8*10∧26

Answer 1

On edg the correct answer is A. I just took the unit test review and got it wrong.

Answer 2

Answer : The correct option is, (A)
`2.0* 10^(-50)`

Solution : Given,

Concentration of
`O_2` =
`1.6* 10^(-2)M`

Concentration of
`O_3` =
`2.86* 10^(-28)M`

The given balanced equilibrium reaction is,

`3O_2(g)\rightleftharpoons 2O_3(g)`

The expression for equilibrium constant will be,

`K_c=([O_3]^2)/([O_2]^3)`

Now put all the given values in this formula, we get

`K_c=((2.86* 10^(-28))^2)/((1.6* 10^(-2))^3)`

`K_c=1.99* 10^(-50)=2.0* 10^(-50)`

Therefore, the value of equilibrium constant is,
`2.0* 10^(-50)`