Question

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

Short answer

The equilibrium concentrations of the species are approximately: \([\text{CO}_{2}] \approx 0.397 \, \text{mol/L}\), \([\text{CO}] \approx 1.06 \times 10^{-3} \, \text{mol/L}\), and \([\text{O}_{2}] \approx 5.3 \times 10^{-4} \, \text{mol/L}\).

Step-by-step solution

Set up ICE table

First, we need to create an ICE (Initial, Change, Equilibrium) table with the given information. In this table, the initial concentrations of CO and O\(_2\) are not given, so we'll assume they are 0. \[ \begin{array}{|c|c|c|c|} \hline \textbf{Species} & \boldsymbol{[\textbf{CO}_{2}]} &\boldsymbol{[\textbf{CO}]} &\boldsymbol{[\textbf{O}_{2}]} \\ \hline \textbf{Initial} & \frac{2.0 \text{mol}}{5.0 \text{L}} & 0 & 0 \\ \hline \textbf{Change} & -2x & +2x & +x \\ \hline \textbf{Equilibrium} & \frac{2.0 \text{mol}}{5.0 \text{L}}-2x & 2x & x \\ \hline \end{array} \] Here, x represents the change in concentration of each species during the process of reaching equilibrium.

Write the equilibrium expression

Next, we need to write the equilibrium constant expression using the concentrations from the ICE table, and the given value of K: \(K = \frac{[\text{CO}]^{2}[\text{O}_{2}]}{[\text{CO}_{2}]^{2}}\)

Substitute equilibrium concentrations into the equilibrium expression

Now, replace the species' concentrations from the ICE table with the equilibrium concentrations in terms of x: \(2.0 \times 10^{-6} = \frac{(2x)^{2}(x)}{(\frac{2.0 \text{mol}}{5.0 \text{L}}-2x)^{2}}\)

Solve for x

At this point, we need to solve the equation for x. This might be a little bit complicated to solve directly; however, we can make some assumptions to simplify the calculation: As K is very small (2.0 x 10^{-6}), the reaction will not move significantly towards products. Thus, the values of 2x and x will be much smaller compared to 2.0 mol/5.0 L, which lets us ignore the 2x in the denominator: \(2.0 \times 10^{-6} \approx \frac{(2x)^{2}(x)}{(\frac{2.0 \text{mol}}{5.0 \text{L}})^{2}}\) Now, simplify and solve for x: \(x = \sqrt[3]{2.0 \times 10^{-6} \cdot (\frac{2.0 \text{mol}}{5.0 \text{L}})^{2}}\) \(x \approx 5.3 \times 10^{-4} \, \text{mol/L}\)

Calculate equilibrium concentrations

Finally, calculate the equilibrium concentrations of all species using the value of x: -\([CO_{2}] = \frac{2.0 \text{mol}}{5.0 \text{L}}-2x = \frac{2.0 \text{mol}}{5.0 \text{L}}-2(5.3 \times 10^{-4} \, \text{mol/L}) \approx 0.397 \, \text{mol/L}\) -\([CO] = 2x = 2(5.3 \times 10^{-4} \, \text{mol/L}) \approx 1.06 \times 10^{-3} \, \text{mol/L}\) -\([O_{2}] = x = 5.3 \times 10^{-4} \, \text{mol/L}\) So, the equilibrium concentrations are approximately: \( \begin{aligned} [\text{CO}_{2}] &= 0.397\, \text{mol/L}\\ [\text{CO}] &= 1.06 \times 10^{-3}\, \text{mol/L}\\ [\text{O}_{2}] &= 5.3 \times 10^{-4}\, \text{mol/L} \end{aligned} \)

Key concepts

ICE table

An ICE table is a tool chemists use to track the changes in concentration of reactants and products during a chemical reaction as it reaches equilibrium. "ICE" stands for Initial, Change, and Equilibrium:

• Initial: This row lists the initial concentrations of the substances before the reaction begins. In our exercise, we started with 0.4 mol/L of CO₂, and both CO and O₂ were initially at 0, as they had not formed yet.
• Change: This row shows how much the concentrations increase or decrease as the reaction proceeds. Typically, the changes are expressed as multiples of a variable, often labeled 'x.'
• Equilibrium: This final row displays the concentrations when the reaction has reached equilibrium. You calculate these by adding the changes to the initial concentrations.

By filling in these rows carefully, we can easily clarify how each substance's concentration changes over time. Using the variable 'x' helps us express these changes neatly, providing all the information needed to solve for unknown concentrations.

equilibrium expression

The equilibrium expression is a mathematical statement that relates the concentrations of reactants and products in a reaction at equilibrium to the equilibrium constant, K. This expression derives from the Law of Mass Action, which states that for a reversible chemical reaction at equilibrium, the ratio of the products to the reactants raised to the power of their stoichiometric coefficients is constant.

In our example, the chemical equation is:
\[2 ext{CO}_{2}(g) ightleftharpoons 2 ext{CO}(g) + ext{O}_{2}(g)\]

Using the coefficients from this equation, we write the equilibrium expression as:
\[K = \frac{[ ext{CO}]^{2}[ ext{O}_{2}]}{[ ext{CO}_{2}]^{2}}\]

Here, the square brackets denote concentration, and the exponents close connection to their coefficients in the balanced reaction equation. This expression guides finding the equilibrium concentrations, as known values allow solving for unknown concentrations in terms of x.

equilibrium constant (K)

The equilibrium constant, denoted as K, is a measure of the extent to which a chemical reaction proceeds to completion. It varies with temperature, reflecting how much of the reactants are typically turned into products at equilibrium.

In our problem, the value of K is given as a tiny number, \(2.0 \times 10^{-6}\). This small K indicates that, at equilibrium, only a minuscule proportion of CO₂ converts into CO and O₂. Most of the CO₂ remains unreacted.

When K is small, we can make simplifications in our calculations, assuming the changes in initial concentrations are negligible (as demonstrated in the step-by-step solution). Through such estimations, the math involved in solving for the changes becomes more manageable while remaining accurate.

chemical reaction

Understanding the basics of the chemical reaction in question is pivotal. A chemical reaction involves the transformation of reactants into products. Here, we examine the reversible reaction:

\[2 ext{CO}_{2}(g) \rightleftharpoons 2 ext{CO}(g) + ext{O}_{2}(g)\]

This means that carbon dioxide molecules are breaking apart to form carbon monoxide and oxygen molecules. However, it's crucial to remember this process is reversible, indicating that the products can also revert to the reactants.

Studying such reactions helps us understand various concepts, such as how reaction dynamics and equilibrium are influenced by factors like pressure, temperature, and concentration. Recognizing the behavior of the gases involved in the reaction gives us insights into how equilibrium concentrations are established.