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

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alexiaalfaro alexiaalfaro · Answer 1 · 2018-10-31T21:24:32+01:00

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

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BarrettArcher BarrettArcher · Answer 2 · 2019-03-08T06:34:44+01:00

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

Solution : Given,

Concentration of [tex]O_2[/tex] = [tex]1.6\times 10^{-2}M[/tex]

Concentration of [tex]O_3[/tex] = [tex]2.86\times 10^{-28}M[/tex]

The given balanced equilibrium reaction is,

[tex]3O_2(g)\rightleftharpoons 2O_3(g)[/tex]

The expression for equilibrium constant will be,

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

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

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

[tex]K_c=1.99\times 10^{-50}=2.0\times 10^{-50}[/tex]

Therefore, the value of equilibrium constant is, [tex]2.0\times 10^{-50}[/tex]