Answered

Consider the reaction.

$3 O_2(g) \longleftrightarrow 2 O_3(g)$



At 298 K, the equilibrium concentration of $O _2$ is $1.6 \times 10^{-2} M$, and the equilibrium concentration of $O _3$ is $2.86 \times 10^{-28} M$. What is the equilibrium constant of the reaction at this temperature?


A$2.0 \times 10^{-50}$
B$1.8 \times 10^{26}$
C$1.8 \times 10^{-26}$
D$2.0 \times 10^{50}$

Verified Answer :

  • The formula for the equilibrium constant is \(K_c = \frac{[O_3]^2}{[O_2]^3}\).
  • Substituting the given concentrations \([O_3] = 2.86 \times 10^{-28}\) M and \([O_2] = 1.6 \times 10^{-2}\) M, we get \(K_c = \frac{(2.86 \times 10^{-28})^2}{(1.6 \times 10^{-2})^3}\).
  • After performing the calculation, the result is approximately \(1.997 \times 10^{-50}\).
  • The correct choice is \(\boxed{2.0 \times 10^{-50}}\) (Option a).

Explanation

  1. Identify the equilibrium constant expression

To determine the equilibrium constant (\(K_c\)) for the reaction \(3 O_2(g) \longleftrightarrow 2 O_3(g)\), we first identify the formula for the equilibrium constant based on the law of mass action. For a general reaction \(aA \longleftrightarrow bB\), the expression is \(K_c = \frac{[B]^b}{[A]^a}\). Therefore, for the given reaction, the equilibrium constant is expressed as:
\[K_c = \frac{[O_3]^2}{[O_2]^3}\]
We are provided with the equilibrium concentrations at \(298\) K: \([O_2] = 1.6 \times 10^{-2}\) M and \([O_3] = 2.86 \times 10^{-28}\) M.

  1. Calculate the equilibrium constant

Now, substitute the provided equilibrium concentration values into the formula:
\[K_c = \frac{(2.86 \times 10^{-28})^2}{(1.6 \times 10^{-2})^3}\]
First, compute the numerator:
\[(2.86 \times 10^{-28})^2 = 8.1796 \times 10^{-56}\]
Next, compute the denominator:
\[(1.6 \times 10^{-2})^3 = 4.096 \times 10^{-6}\]
Dividing the numerator by the denominator:
\[K_c = \frac{8.1796 \times 10^{-56}}{4.096 \times 10^{-6}} \approx 1.997 \times 10^{-50}\]
Rounding this result gives us approximately \(2.0 \times 10^{-50}\). Comparing this to the given choices, it matches option a.

Examples

The concept of the equilibrium constant is vital in environmental chemistry, especially when studying the ozone layer. The reaction between oxygen (\(O_2\)) and ozone (\(O_3\)) is constantly occurring in the stratosphere. By calculating the equilibrium constant at different temperatures, scientists can determine how concentrations of ozone might shift under various environmental conditions, helping us understand the stability of the protective ozone layer and the impact of pollutants.

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