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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?

User Gnanam
by
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1 Answer

1 vote

Answer:

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 &nbsp; <--->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)

User L L
by
8.7k points
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