Certainly! Let's find the equilibrium constant ([tex]\( K_c \)[/tex]) for the given reaction. The reaction is:
[tex]\[ 3 O_2(g) \longleftrightarrow 2 O_3(g) \][/tex]
Given the equilibrium concentrations at [tex]\( 298 \, K \)[/tex]:
- The equilibrium concentration of [tex]\( O_2 \)[/tex] is [tex]\( [O_2] = 1.6 \times 10^{-2} \, \text{M} \)[/tex].
- The equilibrium concentration of [tex]\( O_3 \)[/tex] is [tex]\( [O_3] = 2.86 \times 10^{-20} \, \text{M} \)[/tex].
The equilibrium constant expression for the reaction is:
[tex]\[ K_c = \frac{[O_3]^2}{[O_2]^3} \][/tex]
We will substitute the given concentrations into this expression.
1. Substitute the equilibrium concentration of [tex]\( O_3 \)[/tex]:
[tex]\[ [O_3] = 2.86 \times 10^{-20} \, \text{M} \][/tex]
[tex]\[ [O_3]^2 = (2.86 \times 10^{-20} \, \text{M})^2 = 8.1796 \times 10^{-40} \, \text{M}^2 \][/tex]
2. Substitute the equilibrium concentration of [tex]\( O_2 \)[/tex]:
[tex]\[ [O_2] = 1.6 \times 10^{-2} \, \text{M} \][/tex]
[tex]\[ [O_2]^3 = (1.6 \times 10^{-2} \, \text{M})^3 = 4.096 \times 10^{-6} \, \text{M}^3 \][/tex]
3. Now, substitute these values into the equilibrium constant expression:
[tex]\[ K_c = \frac{[O_3]^2}{[O_2]^3} = \frac{8.1796 \times 10^{-40} \, \text{M}^2}{4.096 \times 10^{-6} \, \text{M}^3} \][/tex]
4. Perform the division:
[tex]\[ K_c = \frac{8.1796 \times 10^{-40}}{4.096 \times 10^{-6}} \approx 1.9969726562499997 \times 10^{-34} \][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction at [tex]\( 298 \, K \)[/tex] is:
[tex]\[ K_c \approx 1.9969726562499997 \times 10^{-34} \][/tex]
That is your final answer.
