Question

Consider the following reaction at a certain temperature:$$4\mathrm{Fe}(s)+3{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{Fe}}_2{\mathrm{O}}_3(s)$$.An equilibrium mixture contains 1.0 mole of $\mathrm{Fe},1.0\times {10}^{-3}$ mole of ${\mathrm{O}}_2,$ and 2.0 moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ all in a $2.0-\mathrm{L}$ container. Calculate the value of $K$ for this reaction.

Short answer

The value of the equilibrium constant (K) for the given reaction is $8.0\times {10}^9$.

Step-by-step solution

Write the balanced chemical equation

For this problem, we are given the balanced chemical equation: $$4\mathrm{Fe}(s)+3{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{Fe}}_2{\mathrm{O}}_3(s)$$

Calculate the concentrations of Fe, O₂, and Fe₂O₃

We know the amount of each substance and the volume of the container (2.0 L). The concentrations can be calculated as follows: - Fe(s): $1.0\text{mole}$; $\frac{1.0\text{mole}}{2.0\text{L}}=0.5\text{M}$ - O₂(g): $1.0\times {10}^{-3}\text{mole}$; $\frac{1.0\times {10}^{-3}\text{mole}}{2.0\text{L}}=5.0\times {10}^{-4}\text{M}$ - Fe₂O₃(s): $2.0\text{moles}$; $\frac{2.0\text{moles}}{2.0\text{L}}=1.0\text{M}$

Apply the equilibrium constant formula

For the reaction equilibrium constant $K_c$, the formula is: $$K_c=\frac{[{\mathrm{Fe}}_2{\mathrm{O}}_3{]}^2}{[\mathrm{Fe}{]}^4[{\mathrm{O}}_2{]}^3}$$ However, we don't include solids (Fe and Fe₂O₃) in the equilibrium expression. Hence, the formula becomes: $$K_c=\frac{1}{[{\mathrm{O}}_2{]}^3}$$

Substitute the values and calculate K

Now we substitute the concentration of O₂ into the formula and calculate K: $$K_c=\frac{1}{(5.0\times {10}^{-4}{)}^3}=\frac{1}{1.25\times {10}^{-10}}=8.0\times {10}^9$$ Thus, the value of the equilibrium constant (K) for this reaction is $8.0\times {10}^9$.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as $K$, is an important concept in chemical equilibrium. It reflects the ratio of the concentration of products to reactants when a reaction has reached equilibrium. In an equilibrium state, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time.

To calculate the equilibrium constant for a reaction, use the formula:

• $K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

Here, $[A],[B],[C],[D]$ are the molar concentrations, and $a,b,c,d$ are the coefficients from the balanced chemical equation. For reactions involving gases, the equilibrium constant can also be expressed in terms of partial pressures, known as $K_p$. In the example reaction, only equilibrium concentrations of gaseous components should be used in the expression for $K_c$. Solids and pure liquids are not included.

Balanced Chemical Equation

A balanced chemical equation is a fundamental aspect of understanding chemical reactions. It represents both the reactants and products in a reaction with their respective quantities indicated. Balancing ensures mass conservation by having the same number of each type of atom on both sides of the equation.

For the example reaction:

• $4\mathrm{Fe}(s)+3{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{Fe}}_2{\mathrm{O}}_3(s)$

The equation is balanced because both sides have equal numbers of each atom type.

• 4 iron (Fe) atoms on the left equal 4 iron atoms on the right.
• 3 oxygen (O) molecules, each having 2 oxygen atoms (total of 6 oxygen atoms), equal the 6 oxygen atoms in the products.

Balancing is crucial for determining the correct stoichiometry for the calculation of the equilibrium constant.

Reaction Concentration

Reaction concentration refers to the amount of a substance in a given volume of a solution or mixture, often expressed in molarity (M), which is moles per liter. To calculate the concentration, divide the number of moles of the substance by the volume in liters.

In the given example:

• Iron: $\frac{1.0\text{mole}}{2.0\text{L}}=0.5\text{M}$
• Oxygen: $\frac{1.0\times {10}^{-3}\text{mole}}{2.0\text{L}}=5.0\times {10}^{-4}\text{M}$
• Ferric oxide: $\frac{2.0\text{moles}}{2.0\text{L}}=1.0\text{M}$

Understanding reaction concentration is essential for calculating the equilibrium constant and understanding how concentrations change as a reaction approaches equilibrium.

Equilibrium Mixture

An equilibrium mixture contains both reactants and products that exist in concentrations where the rates of the forward and reverse reactions are equal. At equilibrium, there is no net change in the concentration of reactants and products over time.

In the context of the example:

• The equilibrium mixture includes 0.5 M of Fe, $5.0\times {10}^{-4}\text{M}$ of ${\mathrm{O}}_2$, and 1.0 M of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in a 2.0 L container.

The concentrations remain consistent within this mixture as long as conditions such as temperature and pressure remain constant.

Understanding the nature of equilibrium mixtures helps predict how changes in conditions can shift the equilibrium position, altering the concentrations of substances in the mixture.