Problem 37 - Chemistry The Central Science

Question

A mixture of 0.2000 mol of \(\mathrm{CO}_{2}, 0.1000 \mathrm{mol}\) of \(\mathrm{H}_{2},\) and 0.1600 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 \(P_{\mathrm{H}_{2} \mathrm{O}}\) \(=3.51\) atm. Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .(\mathbf{c})\) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

Step-by-Step Solution

Answer

\(P_{CO_{2}}^{initial} = 4.082~atm\), \(P_{H_{2}}^{initial} = 2.041~atm\), and \(P_{H_{2}O}^{initial} = 3.282~atm\). At equilibrium, \(P_{CO_{2}} = 2.806~atm\), \(P_{H_{2}} = 0.765~atm\), and \(P_{CO} = 1.276~atm\). The equilibrium constants are \(K_{p} = K_{c} = 0.897\).

Step 1: Calculate the initial partial pressures of CO2, H2, and H2O.

To calculate the initial partial pressures of the gases, we use the ideal gas law: \( PV = nRT \) Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Rearranging, we get: \( P = \frac{nRT}{V} \) First, calculate the total pressure: \( P_{total} = \frac{(0.2000~mol + 0.1000~mol + 0.1600~mol)(0.0821~L~atm~mol^{-1}~K^{-1})(500~K)}{2.000~L} \) Next, we'll compute the mole fraction of each gas and multiply it by the total pressure to obtain the initial partial pressure of each gas.

Step 2: Set up the reaction and the equilibrium expression.

We have the following equilibrium reaction: \( CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2}O(g) \) Let's write down the equilibrium expression using partial pressures: \( K_{p} = \frac{P_{CO}P_{H_{2}O}}{P_{CO_{2}}P_{H_{2}}} \)

Step 3: Calculate the equilibrium partial pressures of CO2, H2, and CO.

To calculate the equilibrium partial pressures, we need to relate the initial partial pressures to the changes during the reaction. We are given that the equilibrium partial pressure of H2O is 3.51 atm. Using the stoichiometry of the reaction, we can calculate the amounts of CO2, H2, and CO at equilibrium. Assume that x atm of CO2 and H2 reacted to produce x atm of CO and H2O. Then the equilibrium pressures would be as follows: \(P_{CO_{2}} = P_{CO_{2}}^{initial} - x \) \(P_{H_{2}} = P_{H_{2}}^{initial} - x \) \(P_{CO} = x \) \(P_{H_{2}O} = P_{H_{2}O}^{initial} + x \) We know that the final partial pressure of H2O is 3.51 atm, so we can solve for x and calculate the equilibrium pressures for the other gases.

Step 4: Calculate Kp and Kc for the reaction.

Using the equilibrium partial pressures obtained in step 3, we can calculate the equilibrium constant Kp using the expression: \( K_{p} = \frac{P_{CO}P_{H_{2}O}}{P_{CO_{2}}P_{H_{2}}} \) Now we'll convert Kp to Kc using the relationship between them: \( K_{p} = K_{c}(RT)^{\Delta{n}} \) where \(\Delta{n}\) is the change in the number of moles of gas in the reaction, which is zero in this case. So in this case, Kp = Kc. Finally, we have calculated the initial partial pressures, equilibrium partial pressures, and the equilibrium constants Kp and Kc for the given reaction.

Key Concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. In a gaseous mixture within a fixed volume at a given temperature, each gas contributes to the total pressure. This contribution is known as the partial pressure. Understanding how each gas's partial pressure plays into the total pressure in a system is crucial for analyzing chemical reactions and equilibria.

To calculate partial pressure, you can use the Ideal Gas Law equation rearranged for pressure:

• \( P = \frac{nRT}{V} \) where \( n \) is the number of moles of the gas, \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( V \) is the volume.

Given different gases, each gas's partial pressure is proportional to its mole fraction in the mixture.

• The mole fraction is the ratio between the moles of the gas and the total moles of all gases combined.

Once you know the mole fraction, you multiply it by the total pressure to get the gas's partial pressure. This approach forms the basis for understanding dynamic changes that occur as reactions reach equilibrium.

Equilibrium Constant (Kp and Kc)

In chemical reactions, the equilibrium constant is a key concept used to express the ratio of concentrations or partial pressures of products to reactants at equilibrium. For gases, we often use two types of equilibrium constants: \( K_p \) and \( K_c \). They provide insight into how favorably a reaction produces products under set conditions.

- \( K_p \) is the equilibrium constant based on the partial pressures of the gases involved. It is determined from the expression: \[ K_{p} = \frac{P_{CO}P_{H_{2}O}}{P_{CO_{2}}P_{H_{2}}} \]- \( K_c \) is the equilibrium constant based on concentration (mol/L).

The relationship between \( K_p \) and \( K_c \) is given by the equation:

• \( K_{p} = K_{c}(RT)^{\Delta{n}} \) where \( \Delta{n} \) is the difference in moles between gaseous products and reactants.
• In cases where \( \Delta{n}=0 \), as in this reaction, \( K_p \) and \( K_c \) are equal.

Knowing these constants helps predict the direction of the reaction and the position of equilibrium.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the four main properties of gases—pressure (P), volume (V), temperature (T), and number of moles (n). It captures how these variables affect the behavior of gases under ideal conditions, serving as a reliable model in many practical situations.

The equation is structured as:

• \( PV = nRT \) where \( R \) stands for the universal gas constant.

This law helps in calculating one variable if the others are known. For example, to determine the partial pressure of a gas, you can rearrange it as:

• \( P = \frac{nRT}{V} \).

In chemical equilibrium problems like the one we faced, the Ideal Gas Law is critical for finding initial pressures set by introducing specific moles of gases into a defined volume. This calculation is a stepping stone towards further determining the equilibrium state, partial pressures, and associated constants for any dynamic system involving gases.