Answer :
Answer:
The equilibrium constant of the reaction at this temperature is [tex]2.147 \times 10^{-50}[/tex]
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
[tex]3O_2 <--->2O_3[/tex]
[tex]K_c[/tex] 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 [tex]K_c[/tex] expressed as
[tex]K_c = \frac {[Products concentration]}{[Reactants concentration]}[/tex]
Plugging in the values given in the question, we have
[tex]K_c =\frac {[O_3 ]^2}{[O_2 ]^3}[/tex]
[tex]=\frac {((2.86\times10^{-28})^2)}{(1.6\times10^{-2})^3}[/tex]
[tex]=\frac {(8.1796\times10^{-56})}{(4.096\times10^{-6})}[/tex]
[tex]=2.147 \times 10^{-50}[/tex] (Answer)
