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?

Answer 1

The equilibrium constant of the reaction at this temperature is
`2.147 * 10^(-50)`

Step-by-step explanation:

Equilibrium is the condition at which the reactants and products concentration is constant.

At equilibrium rate of the forward reaction = rate of the backward or reverse reaction

The Chemical equilibrium is

`3O_2   <--->2O_3`

`K_c` is the equilibrium constant and is defined as products concentration over reactant concentration and the coefficient is raised to its power. Thus we have the
`K_c` expressed as

`K_c = \frac {[Products concentration]}{[Reactants concentration]}`

Plugging in the values given in the question, we have

`K_c =\frac {[O_3 ]^2}{[O_2 ]^3}`

`=\frac {((2.86*10^(-28))^2)}{(1.6*10^(-2))^3}`

`=\frac {(8.1796*10^(-56))}{(4.096*10^(-6))}`

`=2.147 * 10^(-50)` (Answer)