Problem 59 - Chemistry An Atoms First Approach

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 - \(\mathrm{L}\) vessel, calculate the equilibrium concentrations of all species.

Step-by-Step Solution

Answer

At the given temperature, the equilibrium concentrations of CO2, CO, and O2 are approximately \( CO2 = 0.40 M - 2x, CO = 2x, O2 = x\). Solving the equilibrium constant equation, \( 2.0 * 10^{-6} = \frac{(2x)^2 * x}{(0.40 - 2x)^2}\), we find \(x \approx 3.97 * 10^{-5}\). Therefore, the equilibrium concentrations of CO2, CO, and O2 are approximately 0.3992 M, 7.94 * 10^{-5} M, and 3.97 * 10^{-5} M respectively.

Step 1: Find the initial concentrations of the species

In this step, we calculate the initial concentration of CO2. We are given the amount of CO2 in moles (2.0 moles) and the volume of the vessel (5.0 L). To find the initial concentration, we use the formula: Concentration = (moles of substance) / (volume of solution in liters) We find the initial concentration of CO2: CO2_initial = (2.0 moles) / (5.0 L) = 0.40 M Since no CO and O2 are mentioned, their initial concentrations will be 0.

Step 2: Set up the ICE table

An initial-change-equilibrium (ICE) table will help us analyze the changes in concentrations of the reactants and products as the reaction proceeds. We will use x as the change in concentration of CO, CO2, and O2 from their initial values. | | CO2 | CO | O2 | |---------|--------------|------------|--------- | | Initial | 0.40 M | 0 M | 0 M | | Change | -2x | +2x | +x | | Equilibrium | 0.40 M - 2x | 0 M + 2x | 0 M + x |

Step 3: Write the equilibrium constant expression

Write the equilibrium constant expression for the given reaction using the concentrations at equilibrium from the ICE table. The equation for the equilibrium constant (K) is: K = [CO]^2 * [O2] / [CO2]^2 Substitute the equilibrium concentrations from the ICE table: K = [(2x)]^2 * [(x)] / [(0.40 - 2x)]^2 Given K = 2.0 * 10^(-6), substitute K in the equation: 2.0 * 10^(-6) = (4x^2 * x) / (0.40 - 2x)^2

Step 4: Solve for x

To solve for x, the change in concentration from the initial values, we can first simplify the equation: 2.0 * 10^(-6) * (0.40 - 2x)^2 = 4x^3 Now we need to solve this equation for x. The equation is a bit complicated to solve algebraically, so we will use an iterative or trial and error method to find the approximate value of x.

Step 5: Find the equilibrium concentrations

After obtaining a value for x, we can find the equilibrium concentrations of CO2, CO, and O2 by substituting x back into the ICE table. CO2_equilibrium = 0.40 M - 2x CO_equilibrium = 0 M + 2x O2_equilibrium = 0 M + x You can compute the final equilibrium concentrations by substituting the numerical value you found for x in the previous step.

Key Concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol \( K \), is a crucial aspect of understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium in a reversible chemical reaction. For the reaction in question: \[2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) + \mathrm{O}_{2}(g),\]the equilibrium constant expression is written using the concentrations of the gases involved:

• \([CO]^2\) is the concentration of carbon monoxide squared,
• \([O_2]\) is the concentration of oxygen, and
• \([CO_2]^2\) is the concentration of carbon dioxide squared.

Thus, the expression for equilibrium constant \( K \) is:\[K = \frac{[CO]^2 [O_2]}{[CO_2]^2}.\]Introducing the initial values and solving this equation allows us to find the condition where the reaction reaches equilibrium.

ICE Table

The ICE table, which stands for Initial, Change, and Equilibrium, is an invaluable tool for systematically assessing the changes in concentration of reactants and products during a reaction. It consists of three rows that allow you to visualize:

• Initial concentrations of reactants and products before the reaction begins,
• Change in concentrations as the reaction progresses, which is usually represented by a variable, often \( x \), and
• Equilibrium concentrations after the reaction has reached equilibrium.

For our example reaction, the ICE table would initially show that \( [CO_2] \) is 0.40 M while \([CO]\) and \([O_2]\) start at zero. As the reaction proceeds, these concentrations change, depicted as

• Change in \( [CO_2] = -2x \),
• Change in \( [CO] = +2x \), and
• Change in \( [O_2] = +x \).

The table then allows substitution into the equilibrium expression to solve for \( x \), reflecting the change in concentrations.

Concentration

In the context of chemical reactions, concentration refers to the amount of a substance per unit volume of solution. It is usually expressed in molarity (M), which is moles per liter. Initial concentration is crucial for setting up the lever of changes that occur before a reaction reaches equilibrium.
For example, in the problem, the initial concentration of \( \mathrm{CO}_2 \) was calculated as:\[\text{Concentration} = \frac{\text{moles of } \mathrm{CO}_2}{\text{volume in liters}} = \frac{2.0 \text{ moles}}{5.0 \text{ L}} = 0.40 \text{ M}.\]Understanding concentration helps determine how changes will occur as the reaction shifts towards equilibrium, and it forms an integral part of the ICE table.
This starting point is invaluable for calculating other concentrations after incorporating \( x \) (the change in concentrations) into the equation derived from the ICE table.

Reaction Quotient

The reaction quotient, represented as \( Q \), serves a role similar to that of the equilibrium constant \( K \), but it applies to reactions not yet at equilibrium. It helps predict the direction in which a reaction will proceed to reach equilibrium.
The expression for \( Q \), like \( K \), is derived from the concentrations of the reactants and products:\[Q = \frac{[CO]^2 [O_2]}{[CO_2]^2}.\]Before equilibrium is reached:

• If \( Q < K \), the reaction will proceed forward, converting reactants into products.
• If \( Q > K \), it indicates the reaction will shift backward, converting products back into reactants.
• If \( Q = K \), the reaction is at equilibrium, and no further net change occurs.

This concept is pivotal as it determines how the system will adjust the concentrations of the reactants and products to achieve a state of equilibrium.