Chemistry: 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. 2.0*10∧-50b. 2.0*10∧50c. 1.8*10∧-26d. 1.8*10∧26

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At 298 k, the equilibrium concentration of o2 is - id298212582370, ketricduggerp2ciuc

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03.01.2022

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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