Problem 33 - Chemistry: The Central Science

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 CO. (c) Calculate \(K_{p}\) for the reaction.

Step-by-Step Solution

Answer

(a) Initial partial pressures: \(P_{CO_{2}} = 4.57 \mathrm{~atm}\), \(P_{H_{2}} = 2.29 \mathrm{~atm}\), \(P_{H_{2}O} = 3.66 \mathrm{~atm}\) (b) Equilibrium partial pressures: \(P_{CO_{2}} = 0.81 \mathrm{~atm}\), \(P_{H_{2}} = 1.53 \mathrm{~atm}\), \(P_{CO} = 3.76 \mathrm{~atm}\), and \(P_{H_{2}O} = 3.51 \mathrm{~atm}\) (c) The equilibrium constant, \(K_{p} = 8.61\)

Step 1: Calculate mole fraction of each gas

First, we'll calculate the mole fraction of each gas by dividing the moles of each gas by the total moles of the mixture. Mole fraction of CO₂ = \(\frac{0.2000 \, mol}{0.2000 + 0.1000 + 0.1600} = \frac{0.2000}{0.4600}\) Mole fraction of H₂ = \(\frac{0.1000 \, mol}{0.4600}\) Mole fraction of H₂O = \(\frac{0.1600 \, mol}{0.4600}\)

Step 2: Calculate initial partial pressures

Now, we need to find the total pressure of the system. The Ideal Gas Law gives us: PV = nRT where P is the pressure, V is the volume, n is the moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. Given the total number of moles, volume, and temperature, we can now solve for the total pressure: P = \(\frac{nRT}{V} = \frac{0.4600 \times 0.0821 \times 500}{2.000}\) Now, we can calculate the initial partial pressures of each gas by multiplying the mole fraction by the total initial pressure: \(P_{CO_{2}} = P \times \text{Mole fraction of CO₂}\) \(P_{H_{2}} = P \times \text{Mole fraction of H₂}\) \(P_{H_{2}O} = P \times \text{Mole fraction of H₂O}\) Step 2: Calculate the equilibrium partial pressures

Step 3: Determine the change in partial pressures

In the equilibrium state, the partial pressure of H₂O is given to be 3.51 atm. We can assume a change of x in the partial pressures of CO₂, H₂, and CO to represent the shifting of the equilibrium. This gives us: \(P_{CO_{2}} - x\) \(P_{CO} = x\) \(P_{H_{2}} - x\) \(P_{H_{2}O} = 3.51\) The final equilibrium partial pressures can be solved using the information provided. Step 3: Calculate Kₚ for the reaction

Step 4: Calculate Kₚ

The equilibrium expression for the given reaction is: \(K_{p} = \frac{P_{CO} \times P_{H_{2}O}}{P_{CO_{2}} \times P_{H_{2}}}\) Using the equilibrium partial pressures obtained in Step 2, substitute the values into the Kₚ expression and solve for Kₚ.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for anyone studying chemistry. Put simply, when a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, it has achieved equilibrium.

At this point, the concentrations of reactants and products remain constant over time, not because the reactions have ceased, but because they are occurring at the same rate and balance each other out.

For the reaction \( \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2O(g) \), a dynamic equilibrium is established in a closed container when the production of \( \text{CO} \) and \( \text{H}_2O \) is balanced with the formation of \( \text{CO}_2 \) and \( \text{H}_2 \) from the reverse reaction.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and amount of moles of an ideal gas. The law is commonly expressed as \( PV = nRT \), where \( P \) represents the pressure, \( V \) is the volume, \( n \) signifies the moles of gas, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature in Kelvin.

This law assumes that all gases behave ideally, which means they have perfectly elastic collisions and do not experience intermolecular forces. In reality, no gas is truly ideal, but many gases approximate this behavior at certain conditions of temperature and pressure.

It's essential to apply this law to find the total pressure of a gaseous system and then work out the initial partial pressure of each component in a mixture by using their respective mole fractions. This step is pivotal in the approach to solving for equilibrium conditions in a chemical reaction involving gases.

Equilibrium Constant (Kp)

The equilibrium constant for gaseous reactions, denoted as \( K_p \), is a ratio of the partial pressures of the products over the reactants, each raised to the power of their stoichiometric coefficients.

When a reaction has reached equilibrium at a particular temperature, \( K_p \) remains constant and is given by the expression: \[ K_{p} = \frac{P_{products}^{coefficients}}{P_{reactants}^{coefficients}} \]
In our reaction, \( K_{p} = \frac{P_{CO} \times P_{H_{2}O}}{P_{CO_{2}} \times P_{H_{2}}} \), where \( P \) denotes partial pressure. The value of \( K_p \) is a reflection of the position of the equilibrium; a larger \( K_p \) indicates a greater concentration of products, whereas a smaller \( K_p \) favors the reactants.

Calculating \( K_p \) provides insight into the behavior of a system under equilibrium conditions and allows chemists to predict how the system will respond to changes in conditions such as pressure and temperature.