Question

For the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} M\) What is the concentration of \(\mathrm{O}_{2}(g)\) under these conditions?

Short answer

The concentration of \(\text{O}_{2}(g)\) under the given equilibrium conditions is approximately \(8.04 \times 10^{-2} M\).

Step-by-step solution

Write the equilibrium constant expression

For the given chemical equilibrium reaction, we can write the equilibrium constant expression as: \[K = \frac{[\text{H}_{2}]^2[\text{O}_{2}]}{[\text{H}_{2}\text{O}]^2}\]

Plug in the given values and solve for [O₂(g)]

We are given \(K = 2.4 \times 10^{-3}\), \([\text{H}_{2}\text{O}(g)] = 1.1 \times 10^{-1} M\), and \([\text{H}_{2}(g)] = 1.9 \times 10^{-2} M\). We can plug these values into the equilibrium constant expression and solve for \([\text{O}_{2}(g)]\): \[ 2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2 [\text{O}_{2}]}{(1.1 \times 10^{-1})^2} \] Now, we will solve for \([\text{O}_{2}(g)]\): \[ [\text{O}_{2}] = \frac{2.4 \times 10^{-3} (1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2} \]

Calculate the concentration of O₂(g)

Perform the calculation: \[ [\text{O}_{2}] = \frac{2.4 \times 10^{-3} (1.21 \times 10^{-2})}{(3.61 \times 10^{-4})} \] \[ [\text{O}_{2}] = \frac{2.904 \times 10^{-5}}{3.61 \times 10^{-4}} \] \[ [\text{O}_{2}] \approx 8.04 \times 10^{-2} M \] The concentration of \(\text{O}_{2}(g)\) under the given equilibrium conditions is approximately \(8.04 \times 10^{-2} M\).

Key concepts

Chemical Equilibrium

In a chemical equilibrium, the forward and reverse reactions occur at the same rate, which means that the concentrations of reactants and products remain constant over time. This state of balance can be represented by an equilibrium constant (\( K \)), which is a numerical value that indicates the ratio of product concentrations to reactant concentrations for a given chemical reaction at equilibrium.

Consider the reaction:\[2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \]At equilibrium, while the reaction continues to move forward and backward simultaneously, there is no net change in the concentrations of the substances involved. The equilibrium constant expression for this reaction is given by:\[K = \frac{[\text{H}_{2}]^2[\text{O}_{2}]}{[\text{H}_{2}\text{O}]^2}\]The value of \( K \) tells us about the position of equilibrium:

• A large \( K \) value indicates that the forward reaction is favored, and products are predominant at equilibrium.
• A small \( K \) value suggests that the reverse reaction is favored, and reactants are predominant.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of the reactants and products when a reaction has reached equilibrium. Knowing these concentrations allows us to understand the dynamics of the reaction and how the reactants and products coexist.

In the given exercise, we're tasked with finding the equilibrium concentration of \( \text{O}_{2} \) when \( [\text{H}_{2}\text{O}(g)] = 1.1 \times 10^{-1} M \) and \( [\text{H}_{2}(g)] = 1.9 \times 10^{-2} M \).With the equilibrium constant \( K = 2.4 \times 10^{-3} \), we can set up the equation:\[K = \frac{(1.9 \times 10^{-2})^2 [\text{O}_{2}]}{(1.1 \times 10^{-1})^2}\]By plugging in the known values, we solve for \([\text{O}_{2}]\) to determine its equilibrium concentration. Following the calculation steps carefully ensures the right concentration, which is vital for understanding the proportions of reactants and products at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It helps in understanding how the amount of each substance is interrelated according to the balanced equation.

For the equilibrium reaction:\[2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g)\]The stoichiometry tells us that for every two moles of \( \text{H}_{2}\text{O} \) that decompose, two moles of \( \text{H}_{2} \) and one mole of \( \text{O}_{2} \) are formed. Understanding this relationship helps in setting up the equilibrium expression using the concentrations. When we solve for \([\text{O}_{2}]\), we apply the stoichiometric coefficients:

• Square the concentration of \( \text{H}_{2} \) because there are two moles of it.
• Do the same for \( \text{H}_{2}\text{O} \) since there are two moles decomposing.

Stoichiometry ensures the balance and accuracy of our calculations, reflecting the true nature of the reaction at equilibrium.