Question

3O₂ (g) ⇌ 2O₃(g) for the above reaction at 298 K, K_c is found to be 3.0 x 10^-59. If the conc. of O₂ at equilibrium is 0.040 M then concentration of O₃ in M is

• A. 4.38 x 10^-32
• B. 1.9 x 10^-63
• C. 2.4 x 10³¹
• D. 1.2 x 10²¹

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The equilibrium constant $K_{c}$ relates the concentrations of products and reactants at equilibrium.
Key Formula or Approach:
\[ K_{c} = \frac{[O_{3}]^{2}}{[O_{2}]^{3}} \]
Step 2: Detailed Explanation:
Given:
$K_{c} = 3.0 \times 10^{-59}$
$[O_{2}] = 0.040$ M = $4 \times 10^{-2}$ M
Calculation:
\[ 3.0 \times 10^{-59} = \frac{[O_{3}]^{2}}{(4 \times 10^{-2})^{3}} \]
\[ 3.0 \times 10^{-59} = \frac{[O_{3}]^{2}}{64 \times 10^{-6}} \]
\[ [O_{3}]^{2} = 3.0 \times 10^{-59} \times 64 \times 10^{-6} \]
\[ [O_{3}]^{2} = 192 \times 10^{-65} \]
\[ [O_{3}]^{2} = 19.2 \times 10^{-64} \]
Taking square root:
\[ [O_{3}] = \sqrt{19.2 \times 10^{-64}} \]
\[ [O_{3}] \approx 4.38 \times 10^{-32} \text{ M} \]
Step 3: Final Answer:
The concentration of $O_{3}$ is $4.38 \times 10^{-32}$ M. Option (1) is correct.