For the reaction at , is found to be . If the concentration of at equilibrium is , then concentratio — Chemical Equilibrium Chemistry Question
Question
For the reaction $3O_2(g) \rightleftharpoons 2O_3(g)$ at $298\text{ K}$, $K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040\text{ M}$, then concentration of $O_3$ in M is approximately:
Answer: A
💡 Solution & Explanation
$K_c = [O_3]^2 / [O_2]^3 \Rightarrow 3.0 \times 10^{-59} = [O_3]^2 / (0.040)^3$. $[O_3]^2 = 3.0 \times 10^{-59} \times 6.4 \times 10^{-5} = 19.2 \times 10^{-64} = 1.92 \times 10^{-63}$. taking square root gives $[O_3] \approx 4.38 \times 10^{-32}\text{ M}$.
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