Problem 90 - Chemistry The Science in Context

Question

At \(2000^{\circ} \mathrm{C},\) the value of \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ is \(1.0 .\) What is the ratio of \([\mathrm{CO}]\) to \(\left[\mathrm{CO}_{2}\right]\) at \(2000^{\circ} \mathrm{C}\) in an atmosphere in which \(\left[\mathrm{O}_{2}\right]=0.0045 M\) at equilibrium?

Step-by-Step Solution

Answer

Based on the given reaction and equilibrium constant, the ratio of the concentration of CO to CO2 at 2000°C in an atmosphere with [O2] = 0.0045 M at equilibrium is approximately 222.2.

Step 1: Write the expression for the equilibrium constant \(K_c\)

The expression for equilibrium constant \(K_c\) is given by the concentrations of products divided by the concentrations of reactants, raised to the power of their stoichiometric coefficients. For the given reaction $$ 2 CO(g) + O_2(g) \rightleftharpoons 2 CO_2(g), $$ the expression of the equilibrium constant \(K_c\) is: $$ K_c = \frac{[CO_2]^2}{[CO]^2 [O_2]}. $$

Step 2: Set up the ICE (Initial, Change, Equilibrium) table

In the ICE table, we list the initial concentrations of the substances, the change in their concentrations as the reaction proceeds, and the final concentrations at equilibrium. Let the initial concentrations of CO and CO2 be \(x\) and \(y\), respectively. As the reaction goes to equilibrium, the concentration of CO decreases by \(2\) moles and that of CO2 increases by \(2\) moles. The equilibrium concentrations are: | | [CO] | [O2] | [CO2] | |----------|------------|------------|------------| | Initial | x | 0.0045 | y | | Change | -2a | -a | 2a | | Equilibrium | x-2a | 0.0045-a | y+2a | where \(a\) is the change in concentration.

Step 3: Substitute ICE table values into \(K_c\) expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium expression and plug in the known value of \(K_c = 1.0\): $$ 1.0 = \frac{[(y+2a)]^2}{[(x-2a)]^2 (0.0045-a)}. $$

Step 4: Express the ratio

We are interested in finding the ratio of \([CO]\) to \([CO2]\): \(\frac{[CO]}{[CO2]} = \frac{x-2a}{y+2a}\).

Step 5: Solve for the ratio

By rearranging the equation in step 3, we can simplify it to $$ (y+2a)^2 = (x-2a)^2 (0.0045-a). $$ Now consider the ratio \(\frac{[CO]}{[CO2]} = \frac{x-2a}{y+2a}\). If we square both sides, we obtain $$ \frac{(x-2a)^2}{(y+2a)^2} = \frac{(x-2a)^2}{(y+2a)^2 (0.0045-a)}. $$ Since \((y+2a)^2\) cannot be zero, the ratio \(\frac{[CO]}{[CO2]} = \frac{1}{0.0045-a}\).

Step 6: Calculate the value of the ratio

To find the ratio, we first need to determine the value of \(a\). From the given data, we know that [O2] at equilibrium is \(0.0045 - a\). Since it is unlikely for the concentration to become negative, we can approximate \(a\) by assuming [O2] does not change significantly. Thus, \(a \approx 0.0045\). Now we can calculate the ratio: $$ \frac{[CO]}{[CO2]} = \frac{1}{0.0045 - a} \approx \frac{1}{0.0045 - 0.0045} \approx 222.2. $$ So the ratio of [CO] to [CO2] at \(2000^\circ \mathrm{C}\) in an atmosphere in which \(\left[\mathrm{O}_{2}\right]=0.0045 M\) at equilibrium is approximately \(222.2\).

Key Concepts

chemical_equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the rates of the forward and reverse reactions are equal, ensuring that the concentrations of reactants and products remain constant over time. This dynamic state implies that although individual molecules react over time, the overall ratio of individual species does not change.
In the exercise described, we have a reaction involving carbon monoxide and oxygen gas producing carbon dioxide. At equilibrium, the rate of the formation of carbon dioxide from carbon monoxide and oxygen is equal to the rate of the decomposition of carbon dioxide back into carbon monoxide and oxygen. This balance can be expressed via the equilibrium constant, denoted as \( K_c \).
The value of \( K_c \) is derived from the concentration of the products over the reactants, each raised to the power of their stoichiometric numbers from the balanced chemical equation:

• The equation for \( K_c \) in this reaction is \( \frac{[CO_2]^2}{[CO]^2 [O_2]} \).
• It serves as a measure of the extent of a reaction at equilibrium. A constant \( K_c \) value at a given temperature implies that the proportions of reactants and products are fixed, even though molecules continue reacting.

Understanding chemical equilibrium helps predict how certain changes, like shifts in concentration or pressure, can impact the concentrations of reactants and products at equilibrium.

Le Chatelier's principle

Le Chatelier's principle is a fundamental concept in chemistry that helps us predict the response of a system in equilibrium to external changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change, establishing a new equilibrium state.
In the context of our chemical reaction, if we were to alter concentrations, pressure, or temperature, the system would try to offset those changes to restore equilibrium. Let's explore a few scenarios:

• If more oxygen gas \([O_2]\) is added to the system, the principle predicts the reaction will shift to the right, favoring the production of more carbon dioxide \([CO_2]\), since it tries to "remove" the added \([O_2]\).
• Conversely, removing some \([O_2]\) would shift the equilibrium to the left, increasing \([CO]\) and reducing \([CO_2]\) as the system attempts to "replace" the lost \([O_2]\).
• An increase in pressure would generally favor the side of the reaction with fewer gas molecules, shifting equilibrium accordingly, but in this instance, the number of moles of gas is the same on both sides.

Le Chatelier's principle explains how systems can adapt to changes, maintaining stability and allowing chemists to predict outcomes when manipulating conditions.

reaction_quotient

The reaction quotient, denoted by \(Q\), is a snapshot of a reaction's progress towards equilibrium. It is calculated using the same expression as the equilibrium constant \(K_c\), but at any given point during a reaction, not only at equilibrium.
By comparing the value of \(Q\) to \(K_c\), we can determine whether a reaction mixture is at equilibrium or which direction it must shift to achieve equilibrium.

• If \(Q = K_c\), the system is at equilibrium.
• If \(Q < K_c\), the reaction will proceed in the forward direction, producing more products to reach equilibrium.
• If \(Q > K_c\), the reaction will proceed in the reverse direction, producing more reactants.

In the context of the exercise above, the reaction quotient would help us evaluate whether any given set of concentrations of \([CO]\), \([O_2]\), and \([CO_2]\) corresponds to equilibrium or if the system needs adjustment. By determining \(Q\), we can understand how changes in concentrations affect the direction in which the reaction proceeds to achieve equilibrium.