Problem 102 - Chemistry

Question

An equilibrium mixture contains 0.60 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 \(K_{\mathrm{p}}\) for the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \Longrightarrow 2 \mathrm{CO}(g)\)

Step-by-Step Solution

Answer

The equilibrium constant (Kₚ) for the reaction C(s) + CO₂(g) ⇌ 2 CO(g) can be calculated by substituting the given partial pressures into the Kₚ expression: \(K_{\text{p}}=\frac{(\text{partial pressure of CO})^2}{(\text{partial pressure of CO}_2)}\). Plugging in the given values, we obtain \(K_{\mathrm{p}}=3.212\,\text{atm}\).

Step 1: Write the expression for Kₚ

For a given reaction at equilibrium, the expression for Kₚ is: \[ K_{\text{p}}=\frac{(\text{partial pressure of products})^{\mathrm{n}}}{(\text{partial pressure of reactants})^{\mathrm{m}}} \] where n and m are the stoichiometric coefficients of the products and reactants, respectively. For the given reaction: \[ \text{C}(s)+\text{CO}_2(g) \Longleftrightarrow 2 \text{CO}(g) \] The expression for Kₚ will be: \[ K_{\text{p}}=\frac{(\text{partial pressure of CO})^2}{(\text{partial pressure of CO}_2)} \]

Step 2: Substitute the given values and solve for Kₚ

We are given the partial pressures of CO (2.89 atm) and CO₂ (2.60 atm). Let's plug these values into the expression for Kₚ: \[ K_{\text{p}}=\frac{(\text{2.89 atm})^2}{(\text{2.60 atm})} \] Now, we can calculate the value of Kₚ: \[ K_{\text{p}}=\frac{(2.89\,\text{atm})^2}{(2.60\,\text{atm})}= \frac{8.3521\,\text{atm}}{2.60\,\text{atm}} \] \[ K_{\mathrm{p}}=3.212\,\text{atm} \] The equilibrium constant, \(K_{\mathrm{p}}\), for the given reaction is \(3.212\,\text{atm}\).

Key Concepts

Understanding Partial Pressure

Partial pressure is a crucial concept in the study of gases within any chemical reaction at equilibrium. It represents the pressure that each gas in a mixture would exert if it alone occupied the entire volume. In simpler terms, it tells us how much a specific gas contributes to the total pressure in a mixture.

In the context of our exercise, we are dealing with gases such as carbon dioxide and carbon monoxide. The partial pressures of these gases are essential because they are directly used in calculating the equilibrium constant, \(K_{\text{p}}\).

Here’s what you need to remember about partial pressure:

• Each gas in a mixture contributes to the total pressure based on its own 'partial' pressure.
• The total pressure is the sum of the partial pressures of all gases in the mixture.
• Partial pressures are often given in units of atmosphere (atm).

In calculations regarding chemical equilibrium, like finding \(K_{\text{p}}\), you’ll often use these pressures directly in your formulas. It's crucial to ensure that all partial pressures are measured accurately, as any error will affect your equilibrium calculations.

Reaction Stoichiometry in Equilibrium

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It is essential for understanding how substances transform from reactants to products and in what proportions.

In the context of achieving chemical equilibrium, stoichiometry helps determine\(n\) and\(m\), the stoichiometric coefficients, which are vital in the equilibrium constant expression. For the reaction \( \text{C}(s) + \text{CO}_{2}(g) \Longrightarrow 2 \text{CO}(g) \), these coefficients are unique to each specific reaction and determine the ratio of reactants to products.

Consider these points about stoichiometry in a chemical equilibrium:

• The coefficients in a balanced chemical equation indicate the proportional amounts of reactants and products.
• These proportions are crucial in calculating the equilibrium constants, whether \(K_c\) or \(K_{\text{p}}\).
• In our equation, the number of moles of each species corresponds to the coefficients, where there's 1 mole of CO₂ reacting with solid carbon to produce 2 moles of CO.

Understanding stoichiometry is often the first step towards solving any equilibrium problem because it helps us plug in the correct values into our \(K_{\text{p}}\) expression.

Basics of Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal. This balance results in no net change in the amounts of reactants and products.

For the given reaction \( \text{C}(s) + \text{CO}_{2}(g) \Longrightarrow 2 \text{CO}(g) \), equilibrium is achieved when the formation and consumption of CO reach a point where their rates are equal.

Key aspects of chemical equilibrium include:

• At equilibrium, the concentrations of reactants and products remain constant, although they are not necessarily equal.
• Equilibrium is dynamic, meaning the reactions are still occurring, but the overall concentrations don't change.
• The equilibrium position can be affected by changes in pressure, temperature, and concentration.

The equilibrium constant, \(K_{\text{p}}\), quantifies the ratio of product pressures to reactant pressures once equilibrium is reached. It provides insight into the favorability of product formation under given conditions. In our exercise, we calculated \(K_{\text{p}}\) to understand this balance better in terms of partial pressures.