$3 O _2(g) \longleftrightarrow 2 O _3(g)$ 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. $2.0 \times 10^{50}$
C. $1.8 \times 10^{-26}$
D. $1.8 \times 10^{26}$
Answer (2)
The equilibrium constant K e q for the reaction 3 O 2 ( g ) ⟷ 2 O 3 ( g ) at 298 K is approximately 2.0 × 1 0 − 50 , calculated using the equilibrium concentrations of [ O 2 ] and [ O 3 ] . Therefore, the correct answer is option A: 2.0 × 1 0 − 50 .
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Write the equilibrium constant expression: K e q = [ O 2 ] 3 [ O 3 ] 2 .
Substitute the given concentrations: K e q = ( 1.6 × 1 0 − 2 ) 3 ( 2.86 × 1 0 − 28 ) 2 .
Calculate the equilibrium constant: K e q ≈ 2.0 × 1 0 − 50 .
The equilibrium constant of the reaction at 298 K is 2.0 × 1 0 − 50 .
Explanation
Problem Setup and Goal We are given the equilibrium concentrations of O 2 and O 3 for the reaction 3 O 2 ( g ) ⟷ 2 O 3 ( g ) at 298 K. Our goal is to calculate the equilibrium constant, K e q , for this reaction.
Equilibrium Constant Expression The equilibrium constant expression for the given reaction is: K e q = [ O 2 ] 3 [ O 3 ] 2
Substitution of Given Values We are given that [ O 2 ] = 1.6 × 1 0 − 2 M and [ O 3 ] = 2.86 × 1 0 − 28 M . We substitute these values into the expression for K e q : K e q = ( 1.6 × 1 0 − 2 ) 3 ( 2.86 × 1 0 − 28 ) 2
Calculation of Equilibrium Constant Now, we calculate the value of K e q : K e q = ( 1.6 × 1 0 − 2 ) 3 ( 2.86 × 1 0 − 28 ) 2 = 4.096 × 1 0 − 6 8.1796 × 1 0 − 56 = 1.99697265625 × 1 0 − 50 ≈ 2.0 × 1 0 − 50
Final Answer Therefore, the equilibrium constant K e q for the reaction at 298 K is approximately 2.0 × 1 0 − 50 .
Examples
Understanding equilibrium constants is crucial in many real-world applications. For instance, in environmental science, it helps predict the levels of ozone in the atmosphere, which is vital for blocking harmful UV radiation. In industrial chemistry, it aids in optimizing reaction conditions to maximize product yield, making processes more efficient and cost-effective. Also, in biochemistry, it helps understand enzyme-substrate interactions, which is essential for drug development and understanding metabolic pathways. The equilibrium constant provides valuable insights into the extent to which a reaction will proceed under specific conditions.
