Question

An equilibrium mixture contains \(0.60 \mathrm{~g}\) solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of \(2.60\) atm and \(2.89\) atm, respectively. Calculate the value of \(K_{\mathrm{e}}\) for the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\)

Short answer

The equilibrium constant for the reaction \(C(s) + CO_2(g) \rightleftharpoons 2 CO(g)\) considering the given partial pressures for CO\(_2\) and CO gases is \(K_p \approx 3.21\).

Step-by-step solution

Write the expression for \(K_p\)

First, we need to write the expression for the equilibrium constant in terms of partial pressures. For the given reaction: \[C(s) + CO_2(g) \rightleftharpoons 2 CO(g)\] The \(K_p\) expression is: \[K_p = \frac{(P_{CO})^2}{P_{CO2}}\]

Convert the given partial pressures

We are provided with the partial pressures of CO\(_2\) and CO gases in units of atm. Use these values directly in the equilibrium expression: \(P_{CO2} = 2.60\) atm \(P_{CO} = 2.89\) atm

Calculate the value of \(K_p\)

Now, substitute the partial pressures into the \(K_p\) expression: \[ K_p = \frac{(P_{CO})^2}{P_{CO2}} = \frac{(2.89)^2}{2.60} \] Calculate the equilibrium constant: \[ K_p = \frac{8.3521}{2.60} = 3.212 \approx 3.21 \] Thus, the equilibrium constant for the given reaction is \(K_p \approx 3.21\).

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes a state in a reversible reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. It's essential to grasp that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their concentrations have stabilized and remain constant.

When a reaction reaches equilibrium, it's not static but dynamic. This means that the reactants are still converting to products, and vice versa, just at a rate that balances out. Think of it as a busy crossroad where the number of cars—or molecules in this case—coming in equals to the number leaving, maintaining a constant traffic—or concentration—flow.

For the reaction \[C(s) + CO_2(g) \rightleftharpoons 2 CO(g)\],the solid carbon (C) does not appear in the equilibrium constant expression as its concentration remains constant being a pure solid. Only the gaseous components are included, as they are the ones changing in concentration when the reaction proceeds toward equilibrium.

The Role of Partial Pressure in Equilibrium

In the context of gas-phase reactions, partial pressure is a crucial concept. It is the pressure a gas would exert if it alone occupied the volume of the mixture at the same temperature. In a mixture of gases, each gas exerts a pressure independently. This pressure is proportional to its concentration, making it a useful proxy when calculating the equilibrium constant, especially for reactions involving gases, known as \(K_p\).

The given exercise asks us to use the partial pressures of carbon dioxide (\(CO_2\)) and carbon monoxide (\(CO\)) to calculate \(K_p\) for the reaction. Here's a refined step that aids understanding:

• Identify the equilibrium concentrations or partial pressures as given or calculated.
• Apply these values into the equilibrium expression that relates the concentrations or pressures of the products and reactants.

In our case, we directly used the provided partial pressures of the gases to calculate \(K_p\), demonstrating the quantifiable relationship between gas concentration and its effect on the reaction's equilibrium.

Applying Le Chatelier's Principle

Le Chatelier's principle is a guiding framework for predicting how a change in conditions (like pressure, temperature, or concentration of reactants/products) affects the position of equilibrium. It states that if an external stress is applied to a system in equilibrium, the system adjusts to partially offset the stress and establishes a new equilibrium state.

For instance, increasing the pressure of a gas reaction mixture by decreasing volume causes the equilibrium to shift toward the side with fewer molecules of gas to reduce pressure. Similarly, adding more \(CO_2\) to our reaction would shift the equilibrium to produce more \(CO\) to reestablish the equilibrium condition, as per Le Chatelier's principle.

The concept is practical not only in academic exercises but also in industrial processes where yield optimization is crucial. If the principle was applied to the reaction in our exercise, we might adjust the conditions to favor the production of either carbon dioxide or carbon monoxide depending on the required output, showcasing the principle’s utility in real-world applications.