Questions: For the following reaction 3 O2(g) rightleftharpoons 2 O3(g) Kc=2.10 times 10^-7 at a certain temperature. If [O2]=0.0435 M when at equilibrium, what is the equilibrium O3 concentration?

For the following reaction
3 O2(g) rightleftharpoons 2 O3(g)
Kc=2.10 times 10^-7 at a certain temperature. If [O2]=0.0435 M when at equilibrium, what is the equilibrium O3 concentration?

Transcript text: For the following reaction \[ 3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{~g}) \] $\mathrm{Kc}=2.10 \times 10^{-7}$ at a certain temperature . If $\left[\mathrm{O}_{2}\right]=0.0435 \mathrm{M}$ when at equilibrium, what is the equilibrium $\mathrm{O}_{3}$ concentration?

Solution

Solution Steps

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the given reaction is: \[ \mathrm{Kc} = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3} \]

Step 2: Substitute Known Values into the Equilibrium Expression

We are given: \[ \mathrm{Kc} = 2.10 \times 10^{-7} \] \[ [\mathrm{O}_2] = 0.0435 \, \mathrm{M} \]

Substitute these values into the equilibrium expression: \[ 2.10 \times 10^{-7} = \frac{[\mathrm{O}_3]^2}{(0.0435)^3} \]

Step 3: Solve for the Equilibrium Concentration of $\mathrm{O}_3$

First, calculate $(0.0435)^3$: \[ (0.0435)^3 = 0.0000823 \]

Next, solve for $[\mathrm{O}_3]^2$: \[ 2.10 \times 10^{-7} = \frac{[\mathrm{O}_3]^2}{0.0000823} \] \[ [\mathrm{O}_3]^2 = 2.10 \times 10^{-7} \times 0.0000823 \] \[ [\mathrm{O}_3]^2 = 1.7283 \times 10^{-11} \]

Finally, take the square root to find $[\mathrm{O}_3]$: \[ [\mathrm{O}_3] = \sqrt{1.7283 \times 10^{-11}} \] \[ [\mathrm{O}_3] = 1.3146 \times 10^{-6} \, \mathrm{M} \]