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 in a \(2.0-\) L container it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\) Calculate the moles of \(\mathrm{O}_{2}(g)\) present under these conditions.

Short answer

Under these conditions, there are approximately \(1.6 \times 10^{-2}\) moles of \(\mathrm{O}_2(g)\) present at equilibrium in the 2.0 L container.

Step-by-step solution

Write the expression for the equilibrium constant, K

Using the given reaction, the expression for K can be written as: \[K = \frac{[\mathrm{H}_2]^2[\mathrm{O}_2]}{[\mathrm{H_2O}]^2}\]

Plug in the given values

We are given \(K = 2.4 \times 10^{-3}\), $\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} \mathrm{M}$. Plug these values into the equilibrium constant expression: \[ 2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2[\mathrm{O}_2]}{(1.1 \times 10^{-1})^2} \]

Solve for the concentration of O₂

Now we need to find the value of \([\mathrm{O}_2]\) by isolating it in the equation. \[ [\mathrm{O}_2]= \frac{2.4 \times 10^{-3}(1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2} \] Calculate the value: \[ [\mathrm{O}_2]\approx 7.8 \times 10^{-3} \mathrm{M} \]

Convert concentration to moles

Since the container has a volume of 2 L, we can find the moles of \(\mathrm{O}_2(g)\) using the calculated concentration: Moles of $\mathrm{O}_2 = [\mathrm{O}_2] \times V\] \[ = (7.8 \times 10^{-3}\mathrm{M}) \times (2\:\mathrm{L}) \] Calculate the moles of \(\mathrm{O}_2(g)\): \[ \text{Moles of}\:\mathrm{O}_2\approx 1.6 \times 10^{-2}\:\text{moles} \] Under these conditions, there are approximately \(1.6 \times 10^{-2}\) moles of \(\mathrm{O}_2(g)\) present at equilibrium in the 2.0 L container.

Key concepts

Equilibrium Constant (K)

The equilibrium constant, denoted as \(K\), provides a snapshot of a chemical reaction at equilibrium. It is a numerical value that connects the concentrations of products and reactants in a balanced reaction. For the reaction in question, the equilibrium constant is expressed as:

• \[K = \frac{[\mathrm{H}_2]^2[\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}]^2}\]

The values in the numerator and denominator represent the concentrations of products and reactants raised to the power corresponding to their coefficients in the balanced equation. Here, we see the equilibrium constant incorporates the concentrations of hydrogen and oxygen gas over the concentration of water vapor squared.
This ratio remains constant at equilibrium at a given temperature, encapsulating the idea that reactions move toward equilibrium regardless of starting conditions.

Reaction Stoichiometry

Stoichiometry is the calculation method in chemistry adjusting amounts based on balanced equations. For our reaction:

• \(2\, \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons 2\, \mathrm{H}_2(g) + \mathrm{O}_2(g)\)

The stoichiometry tells us the relative amounts of reactants and products. Given that two moles of water decompose to form two moles of hydrogen and one mole of oxygen, the stoichiometric coefficients (2, 2, 1) are essential in forming the equilibrium expression for \(K\) and understanding the transformation of matter.
Through stoichiometry, we can predict the relationship among amounts of different substances and use this relationship to calculate various properties, such as how changes in one substance might affect another.

Gas Concentrations

Concentration is a measure of how much of a substance is present in a given volume of solution or gas. For gases, concentrations are typically expressed in molarity (M), defined as moles of gas per liter of volume (mol/L). In chemical equilibrium, it is crucial to calculate the concentration of reactants and products accurately:

• For water vapour, \([\mathrm{H}_2\mathrm{O}] = 1.1 \times 10^{-1} \: \mathrm{M}\)
• For hydrogen gas, \([\mathrm{H}_2] = 1.9 \times 10^{-2} \: \mathrm{M}\)

These concentrations are used to plug into the equilibrium expression to help us find unknown concentrations, like that of oxygen. Understanding gas concentrations is critical in predicting how the reaction components interact and establish equilibrium state.

Mole Calculation

To find how many moles of a gas are present in a container at equilibrium, we use the concentration (in M) and the volume (in L) of the container. The formula is straightforward:

• Moles = Concentration \( \times \) Volume

For example, after finding the concentration of \(\mathrm{O}_2\) to be approximately \(7.8 \times 10^{-3} \: \mathrm{M}\), we multiply by the container volume, 2 L, to calculate moles:

• \(1.6 \times 10^{-2}\) moles of \(\mathrm{O}_2\)

This simple multiplication shows the power of stoichiometry and equilibrium calculations to inform us about not just ratios, but actual physical entities in a closed system. Performing mole calculations is foundational in understanding material balances in chemical reactions.