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 equilibrium concentration of \(\mathrm{O}_{2} (g)\) under these conditions is \(8.05 \times 10^{-2} \text{ M}\).

Step-by-step solution

Write the expression for the equilibrium constant

The equilibrium constant \(K\) is equal to the ratio of the product concentrations raised to their stoichiometric coefficients, divided by the ratio of the reactant concentrations raised to their stoichiometric coefficients. For the given reaction, the equilibrium constant expression is: \[K = \frac{[\mathrm{H}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{H}_{2} \mathrm{O}]^2}\]

Insert the known values

We know the values of \(K\), \([\mathrm{H}_{2} \mathrm{O}]\), and \([\mathrm{H}_{2}]\). Now, we can insert these values into the equation: \(2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2[\mathrm{O}_{2}]}{(1.1 \times 10^{-1})^2}\)

Solve for \([\mathrm{O}_{2}]\)

The only unknown variable in the equation is \([\mathrm{O}_{2}]\). We can rearrange the equation to isolate this variable, then solve for it: \[\frac{(1.9 \times 10^{-2})^2[\mathrm{O}_{2}]}{(1.1 \times 10^{-1})^2} = 2.4 \times 10^{-3}\] \[[\mathrm{O}_{2}] = \frac{2.4 \times 10^{-3}(1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2}\] Now, solve for \([\mathrm{O}_{2}]\): \[[\mathrm{O}_{2}] = \frac{2.4 \times 10^{-3}(1.21 \times 10^{-2})}{(3.61 \times 10^{-4})}\] \[[\mathrm{O}_{2}] = \frac{2.904 \times 10^{-5}}{3.61 \times 10^{-4}}\] \[[\mathrm{O}_{2}] = 8.05 \times 10^{-2} \text{ M}\] The equilibrium concentration of \(\mathrm{O}_{2} (g)\) under these conditions is \(8.05 \times 10^{-2} \text{ M}\).

Key concepts

Reaction Stoichiometry

In the world of chemistry, reaction stoichiometry is a key concept that helps us understand the quantitative relationships in a chemical reaction. Simply put, it tells us how the reactants and products relate to each other in terms of amount. For the given reaction: \[2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\] Each molecule of water (\(\mathrm{H}_2\mathrm{O}\)) breaks down into hydrogen (\(\mathrm{H}_2\)) and oxygen (\(\mathrm{O}_2\)). According to the stoichiometric coefficients, 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen. This tells us that:

• For every 2 molecules of water, we produce 2 molecules of hydrogen.
• Simultaneously, we produce 1 molecule of oxygen from the breakdown of 2 molecules of water.
• The coefficients (2 for water and hydrogen, 1 for oxygen) indicate how the moles of reactants and products interconvert.

Understanding these ratios helps us predict the amounts of products formed or reactants consumed when the reaction reaches equilibrium.

Concentration Calculations

Calculating concentrations in a chemical reaction is an essential part of understanding equilibrium dynamics. Concentration refers to the amount of a substance in a given volume. When given partial concentrations in a reaction, like \[\left[\mathrm{H}_{2} \mathrm{O} \right] = 1.1 \times 10^{-1} \text{ M}\] \[\left[\mathrm{H}_{2} \right] = 1.9 \times 10^{-2} \text{ M}\]we can use these values to calculate the unknown concentrations, such as \([\mathrm{O}_2]\) using equilibrium expressions. The unit "M" (Molarity) is used to denote concentration in moles per liter. To calculate the concentration of \([\mathrm{O}_{2} ]\), we follow these steps:

• Use the equilibrium constant expression and the known values of concentration.
• Plug in the known values and rearrange to solve for the unknown concentration.
• Apply mathematical operations such as squaring, multiplication, division to isolate the desired concentration term.

These calculations reveal the interdependency of concentrations within the equilibrium framework.

Equilibrium Expressions

Equilibrium expressions are indispensable in tying all elements of a chemical reaction at equilibrium together. An equilibrium expression allows us to systematically express and calculate the concentration relationships between products and reactants. For our reaction, the equilibrium expression is: \[K = \frac{[\mathrm{H}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{H}_{2} \mathrm{O}]^2}\] This formula provides a snapshot of the chemical system at equilibrium, defined by the equilibrium constant (\(K\)). Here's what you need to know about equilibrium expressions:

• The numerator involves the concentrations of products raised to their stoichiometric coefficients, which reflects their impact on the reaction balance.
• The denominator relates to the reactants, again raised to their respective stoichiometric coefficients.
• \(K\) tells us the extent of the reaction at equilibrium—smaller values indicate that the reactants are favored, while larger values suggest product dominance.

Understanding how to correctly set up and manipulate equilibrium expressions is crucial for solving equilibrium concentration questions and predicting how a reaction can shift under different conditions.