Question

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Short answer

The equilibrium constant for the reaction \(\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)\) is approximately \(3.944 \times 10^{-82}\).

Step-by-step solution

Adding the first two reactions together

Add reaction 2 and reaction 3: $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g)$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ To add the reactions, we need to sum them: $$\mathrm{NO}_{2}(g) - \mathrm{NO}_{2}(g) + \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}(g) + \mathrm{NO}(g) - \mathrm{NO}(g) + \mathrm{O}_{2}(g)$$

Simplify the reaction

Simplify the reaction by canceling out the common terms on both sides: $$\mathrm{O}_{2}(g) + \mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ This is our desired reaction.

Calculate the equilibrium constant for the combined reaction

To find the equilibrium constant for the combined reaction, multiply the equilibrium constants of reactions 2 and 3: $$K = K_1 \times K_2 = (6.8 \times 10^{-49}) \times (5.8 \times 10^{-34})$$

Calculate the final value for the equilibrium constant

Calculate the equilibrium constant for the combined reaction: $$K = (6.8 \times 5.8) \times 10^{-49-34} = 39.44 \times 10^{-83}$$ The equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ is approximately $$3.944 \times 10^{-82}$$.

Key concepts

Reaction Equilibrium

Understanding reaction equilibrium is crucial when studying chemical reactions. In a reaction, when the rate of the forward reaction equals the rate of the backward reaction, the system is said to be at equilibrium. At this point, the concentrations of the reactants and products remain constant over time.
It’s important to note that while the concentrations remain constant, they aren’t necessarily equal.

• Equilibrium can only be achieved in a closed system where no substances enter or leave.
• The state of equilibrium is dynamic, meaning that the reactions are continuously occurring but no net change is observed.

Recognizing when a reaction has reached equilibrium is key in calculating equilibrium constants and understanding how a reaction proceeds under given conditions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and they occur when chemical bonds are broken and formed. In understanding these processes, it's crucial to consider each aspect contributing to the reaction.

• Chemical reactions can be classified based on various criteria, such as their energy profiles (exothermic or endothermic) or the phases of the reactants and products (gas, liquid, solid).
• The balanced chemical equation is vital in representing the process accurately, showing the relationship between reactants and products.

In our example, we look at a reaction where oxygen molecules and atoms react to form ozone. This transformation is achieved by adding two known reactions and then simplifying them to get the desired chemical equation. Understanding the nature of reactions helps in predicting the behavior of reactants under different conditions.

Equilibrium Expression

An equilibrium expression is essential for calculating an equilibrium constant, which provides insights into the reaction dynamics. For a given chemical equilibrium, the equilibrium expression is derived from the balanced chemical equation.

• The equilibrium expression for a reaction of type \(aA + bB \rightleftharpoons cC + dD\) is given by \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where \([A], [B], [C], [D]\) are the molar concentrations of the species at equilibrium.
• The equilibrium constant \(K\) indicates the extent to which a reaction proceeds to form products.

In the problem example, to find the equilibrium constant for the reaction, we multiplied the constants of the individual reactions involved. This multiplicative nature reflects how the contributions of each reaction to the overall balance affect the final equilibrium state. Knowing how to efficiently use the equilibrium expression is a potent tool in predicting and manipulating chemical reactions in practical applications.