Problem 69 - Chemistry

Question

For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

Step-by-Step Solution

Answer

The partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) at equilibrium are 1 atm and 4 atm, respectively.

Step 1: Set up the equilibrium expression

According to the given equilibrium constant, we can write the following equilibrium expression for this reaction: \[K_{\mathrm{p}} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_{2}}}\] where \(P_{\mathrm{CO_{2}}}\) and \(P_{\mathrm{CO}}\) represent the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\).

Step 2: Write expressions for partial pressures in terms of a single variable

Let's define the variable x as the partial pressure of \(\mathrm{CO}_{2}\) at equilibrium: \(x=P_{\mathrm{CO_{2}}}\) Since we know that the total pressure at equilibrium is 6.00 atm, and there are 2 moles of \(\mathrm{CO}\) formed for every mole of \(\mathrm{CO}_{2}\) consumed, we can express the partial pressure of \(\mathrm{CO}\) as: \(P_{\mathrm{CO}}=6.00 - 2x\)

Step 3: Substitute the expressions into the equilibrium equation

Now, we substitute the expressions for the partial pressures into the equilibrium expression: \[2.00 = K_{\mathrm{p}} = \frac{(6.00 - 2x)^2}{x}\]

Step 4: Solve for the variable and determine the partial pressures

To solve for x, we first multiply both sides of the equation by x: \[2.00x= (6.00-2x)^2\] Now, we can solve for x using either the quadratic formula or factoring (we'll use factoring): \[4x(3-x) = 2x(6-4x) = (6-2x)^2\] This equation simplifies to: \[6-4x = 2x\] Solving for x: \[6 = 6x\] \[x = 1\] Now, using the expressions from Step 2, we can determine the partial pressures at equilibrium: \(P_{\mathrm{CO}_{2}} = x = 1\,\mathrm{atm}\) \(P_{\mathrm{CO}}= 6 - 2x = 6 - 2(1) = 4\,\mathrm{atm}\) Therefore, the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) at equilibrium are 1 atm and 4 atm, respectively.

Key Concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant is a key concept that helps in quantifying how a reaction mixture reaches a state of balance. The equilibrium constant, denoted as \( K_{p} \) for reactions involving gases, is the ratio of the concentrations (or partial pressures) of the products to the reactants, each raised to the power of their stoichiometric coefficients. For instance, in the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g) \), the equilibrium constant expression is \( K_{p} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_{2}}} \).
Comparing this calculated ratio with the given \( K_{p} \) allows determination of reaction direction or extent, indicating how far a reaction has "gone forward" at a given temperature and pressure.
It's important to note that \( K_{p} \) is influenced only by temperature, not by changes in pressure or concentration. Thus, understanding the equilibrium constant helps in predicting the amounts of reactants and products during a chemical reaction at equilibrium.

Partial Pressure

Partial pressure describes the pressure exerted by an individual gas in a mixture of gases, each gas contributing to the total pressure independently. In a given reaction at equilibrium, the total pressure is the sum of the partial pressures of all gas components involved.
Let's consider our exercise where the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2\mathrm{CO}(g) \) reaches equilibrium with a total pressure of 6 atm. If \( P_{\mathrm{CO}_{2}} \) is the partial pressure of carbon dioxide and \( P_{\mathrm{CO}} \) is that of carbon monoxide, these can be expressed as \( x \) and \( 6-2x \) respectively, relying on stoichiometry where 2 moles of \( \mathrm{CO} \) are formed per mole of \( \mathrm{CO}_{2} \) consumed.
By solving the equilibrium expression with these variables, one can deduce the individual partial pressures, providing insights into the composition of the gas mixture at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept for predicting how a change in conditions can affect a system at equilibrium. It posits that if a stress is applied to a system in equilibrium, the system will adjust in a way that counteracts the stress in order to restore equilibrium.
This can involve changes like variations in concentration, pressure, volume, or temperature which affect the balance between reactants and products. For instance, increasing the pressure by decreasing volume in a gas reaction would favor the direction with fewer gas molecules.
While our specific example doesn't involve applying stress directly, understanding this principle helps determine how establishing or restoring equilibrium proceeds when conditions change. In practice, by comprehensively understanding Le Chatelier's Principle, chemists can manipulate conditions to favor the formation of desired products or minimize unwanted ones.