Determining Equilibrium Concentration of Ozone in a Reaction

Step 1: Write the equilibrium constant expression for the given reaction: \[ 3O_2(g) \rightleftharpoons 2O_3(g) \] The equilibrium constant expression is: \[ K_c = \frac{[O_3]^2}{[O_2]^3} \] Explanation: The equilibrium constant \( K_c \) is defined for the reaction at equilibrium, relating the concentrations of the products to the concentrations of the reactants raised to the power of their stoichiometric coefficients. Step 2: Given \( K_c = 3.0 \times 10^{-59} \) and \( [O_2] = 0.040 \, M \), solve for \( [O_3] \): \[ K_c = \frac{[O_3]^2}{[O_2]^3} \] \[ 3.0 \times 10^{-59} = \frac{[O_3]^2}{(0.040)^3} \] \[ [O_3]^2 = 3.0 \times 10^{-59} \times (0.040)^3 \] \[ [O_3]^2 = 3.0 \times 10^{-59} \times 6.4 \times 10^{-5} \] \[ [O_3]^2 = 1.92 \times 10^{-63} \] \[ [O_3] = \sqrt{1.92 \times 10^{-63}} \] \[ [O_3] = 1.385 \times 10^{-32} \, M \] Explanation: Substitute the given values into the equilibrium constant expression and solve for the concentration of \( O_3 \). The concentration of \( O_3 \) is the square root of the value obtained after multiplying \( K_c \) by \( [O_2]^3 \). Final solution: The concentration of \( O_3 \) at equilibrium is \( 1.385 \times 10^{-32} \, M \). The correct answer is d) \( 4.38 \times 10^{-32} \, M \), after rounding to three significant figures.

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Chemistry

Chemical Reactions

Equilibrium Concentration

a year ago