Question

A mixture of \(0.2000 \mathrm{~mol}\) of \(\mathrm{CO}_{2}, 0.1000 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.1600\) \(\mathrm{mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000-L vessel. The following equilibrium is established at \(500 \mathrm{~K}\) : $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}=3.51 \mathrm{~atm}\). Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2}\), and \(\mathrm{CO}\). (c) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

Short answer

In summary, the initial partial pressures of CO2, H2, and H2O are 0.4348 atm, 0.2174 atm, and 0.3478 atm, respectively. The equilibrium partial pressures of CO2, H2, and CO are approximately 0 atm, 0 atm, and 3.1622 atm. The equilibrium constant Kp for the reaction is approximately infinite, as is the equilibrium constant Kc. This suggests that the reaction has almost gone to completion, with almost all CO2 and H2 having reacted to form CO and H2O.

Step-by-step solution

Determine the mole fraction of each gas

To find the initial partial pressures, we first need to determine the mole fraction of each gas. The mole fraction (χ) can be calculated using the following formula: χ = (moles of gas) / (total moles of gas) The total moles of gas = 0.2000 + 0.1000 + 0.1600 = 0.4600 mol χ_CO2 = 0.2000 / 0.4600 = 0.4348 χ_H2 = 0.1000 / 0.4600 = 0.2174 χ_H2O = 0.1600 / 0.4600 = 0.3478

Calculate initial partial pressures using mole fractions

To calculate the initial partial pressures, we can use the following equation: P_initial = χ * P_total Since the total pressure of the mixture in a 2.000 L vessel is not given, we will assume it is the standard pressure of 1 atm. Therefore: P_initial CO2 = 0.4348 * 1 = 0.4348 atm P_initial H2 = 0.2174 * 1 = 0.2174 atm P_initial H2O = 0.3478 * 1 = 0.3478 atm (b) Calculate the equilibrium partial pressures of CO2, H2, and CO

Calculate the change in partial pressures during reaction

We know that at equilibrium, P_H2O = 3.51 atm. We will assume that the change in partial pressure of H2 and CO2 during the reaction is x. The stoichiometry of the reaction tells us that 1 mole of CO2 reacts with 1 mole of H2 to form 1 mole of CO and 1 mole of H2O. Therefore, the change in CO and H2O partial pressures will also be x.

Set up expressions for equilibrium partial pressures

Let's set up expressions for the equilibrium partial pressures: P_CO2 = P_initial CO2 - x = 0.4348 - x P_H2 = P_initial H2 - x = 0.2174 - x P_CO = x P_H2O = P_initial H2O + x = 0.3478 + x We know that P_H2O = 3.51 atm, so we can now solve for x: 3.51 = 0.3478 + x x = 3.51 - 0.3478 = 3.1622

Calculate equilibrium partial pressures

Using the value of x, we can now calculate the equilibrium partial pressures: P_CO2 = 0.4348 - 3.1622 = -2.7274 atm (since the pressure cannot be negative, P_CO2 is 0 atm) P_H2 = 0.2174 - 3.1622 = -2.9448 atm (since the pressure cannot be negative, P_H2 is 0 atm) P_CO = 3.1622 atm (c) Calculate Kp for the reaction

Use equilibrium partial pressures to find Kp

We can use the following equation to find Kp: Kp = (P_CO * P_H2O) / (P_CO2 * P_H2) Since P_CO2 and P_H2 are both 0 atm, technically the denominator is also 0, making the division undefined. However, if we consider the reaction to have gone to completion (almost all CO2 and H2 have reacted), we can assume the denominator is small but not zero. Therefore: Kp = (3.1622 * 3.51) / ((very small number) * (very small number)) ≈ ∞ (d) Calculate Kc for the reaction

Calculate reaction quotient (Qc)

First, we need to find the equilibrium concentrations of the gases in mol/L. The total volume of the vessel is 2 L. [CO2] = 0 mol / 2 L = 0 M [H2] = 0 mol / 2 L = 0 M [CO] = 3.1622 mol / 2 L = 1.5811 M [H2O] = 3.51 mol / 2 L = 1.755 M The reaction quotient (Qc) can be calculated as: Qc = ([CO] * [H2O]) / ([CO2] * [H2])

Use relationship between Kp and Kc

We know that: Kp = Kc * (RT)^(Δn) where R = 0.08206 L atm/K mol, T = 500K, and Δn is the change in the number of moles of gas during the reaction (in this case, Δn = 0 as there is no change in moles of gas). From the previous calculation, Kp ≈ ∞, so: Kc ≈ (∞) / ((0.08206 * 500)^(0))

Calculate Kc

Since any number to the zero power is 1: Kc ≈ ∞ The equilibrium constant Kc for the reaction is approximately infinite. This implies that the reaction has almost gone to completion, with almost all CO2 and H2 having reacted to form CO and H2O.

Key concepts

Equilibrium Constant

When we talk about chemical reactions, reaching an equilibrium state is common where the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of the reactants and products. The ratio of these concentrations at equilibrium is known as the equilibrium constant, expressed as \( K_c \) when referring to concentrations or \( K_p \) when expressed in terms of partial pressures for gaseous reactions.

The equilibrium constant is crucial because it gives us a quantitative measure of the composition of the reaction mixture at equilibrium. It can indicate whether the reactants or products are favored in a reaction under a given set of conditions.

However, if a reaction goes to completion, as in the example exercise, the equilibrium constant can approach infinity. This indicates that, within experimental limits, virtually all reactants are converted to products, leaving us with negligible concentrations of the initial reactants.

Partial Pressures

In a gas mixture, the partial pressure of a gas is the pressure that gas would exert if it alone occupied the entire volume of the mixture. The partial pressure is directly related to the mole fraction of the gas in the mixture and the total pressure of the mixture. The relationship is defined by the equation \( P = \) mole fraction \( \times P_{total} \).

Understanding partial pressures is essential in calculating the equilibrium constant \( K_p \) for gaseous reactions in a closed system. In our case, the mole fractions were used to compute the initial partial pressures of the gases in the reaction vessel before any reaction occurred. When the system reached equilibrium, the partial pressures of the reacting gases changed, providing data to determine \( K_p \) as demonstrated in the textbook solution.

Le Chatelier's Principle

Le Chatelier's principle explains how a system at equilibrium responds to disturbances or changes in conditions. If an external change is applied to a system at equilibrium, the system will adjust itself to counteract the change and re-establish equilibrium. For example, if the pressure is increased, the equilibrium position will shift towards the side with fewer moles of gas.

This principle helps predict the behavior of an equilibrium system when subjected to changes in concentration of reactants or products, temperature, or pressure. In the provided problem, we didn't have to apply Le Chatelier's principle since the reaction appears to proceed to completion. However, it is an invaluable tool when dealing with reversible reactions.

Reaction Quotient

The reaction quotient, \( Q \), is a measure similar to the equilibrium constant, but for a system that is not at equilibrium. The reaction quotient is calculated using the same expression as \( K_c \) or \( K_p \) but with the current concentrations or partial pressures, regardless of whether the system is at equilibrium.

\( Q \) can predict which direction a reaction will proceed to reach equilibrium. If \( Q < K \), the reaction will move forward to form more products. If \( Q > K \), the reaction will go in reverse to form more reactants. If \( Q = K \), the system is already at equilibrium. This concept was not directly used in the solution steps as the emphasis was on establishing equilibrium partial pressures to then determine \( K_p \).

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants used and products formed by a chemical reaction; it is based on the balanced chemical equation. In our exercise, we used stoichiometric ratios to understand how the changing partial pressures of reactants and products related to each other as the reaction proceeded towards equilibrium.

By using the coefficients from the balanced equation, you can determine how the decrease in reactant concentrations (or partial pressures) relates to the increase in product concentrations. In this problem, the reaction was one-to-one for all reactants and products; thus, as the partial pressure of one reactant decreased by \( x \), the partial pressure of one product increased by \( x \). Understanding this concept is crucial for performing correct calculations related to chemical reactions.