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For the reaction 3O2(g)⇌2O3(g)3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g) at 298 K298\text{ K}, KcK_c is found to be 3.0×10−593.0 \times 10^{-59}. If the concentration of O2\text{O}_2 at equilibrium is 0.040 M0.040\text{ M}, then the concentration of O3\text{O}_3 in M\text{M} is:

A

1.2×10211.2 \times 10^{21}

B

4.38×10−324.38 \times 10^{-32}

C

1.9×10−631.9 \times 10^{-63}

D

2.4×10312.4 \times 10^{31}

Step-by-Step Solution

The given balanced chemical equation is: 3O2(g)⇌2O3(g)3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g)

For this reaction, the equilibrium constant expression (KcK_c) is written as the ratio of the concentration of the product raised to its stoichiometric coefficient to the concentration of the reactant raised to its stoichiometric coefficient: Kc=[O3]2[O2]3K_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}

Given the values at equilibrium: Kc=3.0×10−59K_c = 3.0 \times 10^{-59} [O2]=0.040 M=4×10−2 M[\text{O}_2] = 0.040\text{ M} = 4 \times 10^{-2}\text{ M}

Substituting these values into the equilibrium constant expression: 3.0×10−59=[O3]2(4×10−2)33.0 \times 10^{-59} = \frac{[\text{O}_3]^2}{(4 \times 10^{-2})^3} [O3]2=3.0×10−59×(4×10−2)3[\text{O}_3]^2 = 3.0 \times 10^{-59} \times (4 \times 10^{-2})^3 [O3]2=3.0×10−59×64×10−6[\text{O}_3]^2 = 3.0 \times 10^{-59} \times 64 \times 10^{-6} [O3]2=192×10−65[\text{O}_3]^2 = 192 \times 10^{-65} [O3]2=19.2×10−64[\text{O}_3]^2 = 19.2 \times 10^{-64}

Taking the square root on both sides to find the concentration of ozone: [O3]=19.2×10−64=19.2×10−32[\text{O}_3] = \sqrt{19.2 \times 10^{-64}} = \sqrt{19.2} \times 10^{-32} Since 16=4\sqrt{16} = 4 and 25=5\sqrt{25} = 5, 19.2\sqrt{19.2} is approximately 4.384.38. [O3]≈4.38×10−32 M[\text{O}_3] \approx 4.38 \times 10^{-32}\text{ M}

Thus, the concentration of O3\text{O}_3 is 4.38×10−32 M4.38 \times 10^{-32}\text{ M}.

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