Question

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Short answer

The equilibrium concentration of \(O_3\) is approximately \(3.26\times10^{-4}\mathrm{M}\).

Step-by-step solution

Write the expression for the equilibrium constant (K)

The expression for K can be obtained from the balanced equation. For a general equilibrium reaction, it can be written as:$$K=\frac{([C]^{c}[D]^{d})}{([A]^{a}[B]^{b})}$$In our case, the reaction is: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. We can write the expression for K as:$$K=\frac{[\mathrm{O}_{3}]^{2}}{[\mathrm{O}_{2}]^{3}}$$

Substitute the given values

We are given that $$ K=1.8 \times 10^{-7}$$ and the equilibrium concentration of $$O_2$$ is $$0.062 \mathrm{M}$$. We can substitute these values into the expression for K we found in Step 1.$$1.8\times10^{-7}=\frac{[\mathrm{O}_{3}]^{2}}{(0.062)^{3}}$$

Solve for the concentration of $$O_3$$

Now, we need to solve for the equilibrium concentration of $$O_3$$. We will first simplify the given equation and then find the value for $$[\mathrm{O}_{3}]$$. $$[\mathrm{O}_{3}]^{2}=(1.8\times10^{-7})(0.062^{3})$$Now, take the square root of both sides to solve for $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})}$$

Calculate the equilibrium concentration of $$O_3$$

Now, use a calculator to find the equilibrium concentration of $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})} \approx 3.26\times10^{-4} \mathrm{M}$$ So, the equilibrium concentration of $$O_3$$ is $$3.26\times10^{-4}\mathrm{M}$$.

Key concepts

Ozone Concentration

In chemical reactions involving gases, it is essential to monitor the concentration of the species involved. For this exercise, we focus on ozone, i.e., \( \mathrm{O}_3 \), which is formed from dioxygen, \( \mathrm{O}_2 \), through a reversible reaction:\[3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)\]The concentration of ozone at equilibrium helps us understand the extent to which this reaction moves forward or backward under a given set of conditions. The equilibrium concentration tells us about the dynamic balance between the formation and decomposition of ozone under specific conditions, such as pressure and temperature.

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of all reactants and products remain constant over time.

• In our example, the ozone equation reaches equilibrium when the amount of \( \mathrm{O}_2 \) converting to \( \mathrm{O}_3 \) is equal to the amount of \( \mathrm{O}_3 \) decomposing back into \( \mathrm{O}_2 \).
• This does not mean the amounts of reactants and products are equal, only that their rates of change are balanced.

It is essential to recognize that reaching equilibrium does not stop the reaction but keeps it in a stable dynamic state where concentrations do not fluctuate anymore.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Understanding this concept helps us set up our calculations correctly when working with equilibrium problems.For the reaction: \[3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g)\]

• The coefficients in the balanced equation (3 for \( \mathrm{O}_2 \) and 2 for \( \mathrm{O}_3 \)) are crucial. They indicate the ratio in which reactants combine and products are formed.
• These coefficients also play a significant role while writing the equilibrium expression and eventually solving for unknown concentrations.

Using stoichiometry, we identify that for every 3 moles of \( \mathrm{O}_2 \) consumed, 2 moles of \( \mathrm{O}_3 \) are produced, which is essential in deriving consistent equilibrium calculations.

Equilibrium Expression

An equilibrium expression shows the relationship between the concentrations of reactants and products at equilibrium. It utilizes the coefficients from the balanced equation, as powers, in a specific mathematical formula.For our reaction:\[3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g)\]The equilibrium constant \( K \) is expressed as:\[K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\]

• This formula helps calculate the concentration of any species in the reaction if the \( K \) value and other species' concentrations are known.
• The exponents (2 for \( \mathrm{O}_3 \) and 3 for \( \mathrm{O}_2 \)) come from the stoichiometric coefficients of the balanced equation.

By substituting known values into this expression, we can solve for unknown concentrations, like that of \( \mathrm{O}_3 \), and understand the dynamics of the equilibrium state.