consider the reaction. 3o₂(g)⇌2o₃(g) at 298 k, the equilibrium concentration of o₂ is 1.6 x 10⁻² m, and the…

consider the reaction. 3o₂(g)⇌2o₃(g) 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? 2.0 x 10⁻⁵⁰ 2.0 x 10⁵⁰ 1.8 x 10⁻²⁶ 1.8 x 10²⁶

consider the reaction. 3o₂(g)⇌2o₃(g) 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? 2.0 x 10⁻⁵⁰ 2.0 x 10⁵⁰ 1.8 x 10⁻²⁶ 1.8 x 10²⁶

Answer

Explanation:

Step1: Write equilibrium - constant expression

For the reaction $3O_2(g)\rightleftharpoons2O_3(g)$, the equilibrium - constant expression $K_c$ is given by $K_c=\frac{[O_3]^2}{[O_2]^3}$.

Step2: Substitute the given equilibrium concentrations

We are given that $[O_2]=1.6\times 10^{-2}\ M$ and $[O_3]=2.86\times 10^{-28}\ M$. Substitute these values into the equilibrium - constant expression: [ \begin{align*} K_c&=\frac{(2.86\times 10^{-28})^2}{(1.6\times 10^{-2})^3}\ &=\frac{8.1796\times 10^{-56}}{4.096\times 10^{-6}}\ & = 2.0\times10^{-50} \end{align*} ]

Answer:

$2.0\times 10^{-50}$