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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?

A. 2.010∧-50
B. 2.010∧50
C. 1.810∧-26
D. 1.810∧26

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 th class=

Verified Answer :

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

Answer : The correct option is, (A) \(2.0\times 10^{-50}\)

Solution : Given,

Concentration of \(O_2\) = \(1.6\times 10^{-2}M\)

Concentration of \(O_3\) = \(2.86\times 10^{-28}M\)

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant will be,

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

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

\(K_c=\frac{(2.86\times 10^{-28})^2}{(1.6\times 10^{-2})^3}\)

\(K_c=1.99\times 10^{-50}=2.0\times 10^{-50}\)

Therefore, the value of equilibrium constant is, \(2.0\times 10^{-50}\)

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