Problem 71 - Chemistry: The Central Science

Question

A mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) is passed over a nickel catalyst at \(1000 \mathrm{~K}\). The emerging gas is collected in a \(5.00-\mathrm{L}\) flask and is found to contain \(8.62 \mathrm{~g}\) of \(\mathrm{CO}, 2.60 \mathrm{~g}\) of \(\mathrm{H}_{2}\), \(43.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\), and \(48.4 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). Assuming that equilibrium has been reached, calculate \(K_{c}\) and \(K_{p}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\)

Step-by-Step Solution

Answer

The short answer to calculate the equilibrium constants \(K_c\) and \(K_p\) for the given reaction is as follows: 1. Calculate the moles of each component at equilibrium using their molar masses and given masses. 2. Determine the equilibrium concentrations by dividing the moles by the volume. 3. Calculate \(K_c\) using the equilibrium expression: \(K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}]^3}{[\mathrm{CH}_{4}][\mathrm{H}_{2}\mathrm{O}]}\). 4. Convert \(K_c\) to \(K_p\) using the relationship: \(K_p = K_c(RT)^{\Delta{n}}\), where \(R = 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\), \(T = 1000 \mathrm{~K}\), and \(\Delta{n} = 2\). By following these steps, the calculated values for \(K_c\) and \(K_p\) represent the equilibrium constants for the given reaction.

Step 1: Using the molar masses, we need to find the moles of each gas component. Molar mass of \(\mathrm{CO} = 28.0 \mathrm{~g/mol}\), \(\mathrm{H}_{2} = 2.0 \mathrm{~g/mol}\), \(\mathrm{CH}_{4} = 16.0 \mathrm{~g/mol}\), and \(\mathrm{H}_{2} \mathrm{O} = 18.0 \mathrm{~g/mol}\) Moles of \(\mathrm{CO} = \frac{8.62 \mathrm{~g}}{28.0 \mathrm{~g/mol}}\), Moles of \(\mathrm{H}_{2} = \frac{2.60 \mathrm{~g}}{2.0 \mathrm{~g/mol}}\), Moles of \(\mathrm{CH}_{4} = \frac{43.0 \mathrm{~g}}{16.0 \mathrm{~g/mol}}\), and Moles of \(\mathrm{H}_{2} \mathrm{O} = \frac{48.4 \mathrm{~g}}{18.0 \mathrm{~g/mol}}\) #Step 2: Calculate equilibrium concentrations of each component#

Given that the volume of the flask is 5 L, we can calculate the equilibrium concentrations by dividing the moles by the volume. Equilibrium concentrations: \([\mathrm{CO}] = \frac{\text{moles of CO}}{5.00 \mathrm{~L}}\), \([\mathrm{H}_{2}] = \frac{\text{moles of H}_{2}}{5.00 \mathrm{~L}}\), \([\mathrm{CH}_{4}] = \frac{\text{moles of CH}_{4}}{5.00 \mathrm{~L}}\), and \([\mathrm{H}_{2}\mathrm{O}] = \frac{\text{moles of H}_{2} \mathrm{O}}{5.00 \mathrm{~L}}\) #Step 3: Calculate \(K_c\)#

Step 2: With equilibrium concentrations, we can now calculate \(K_c\) using the equilibrium expression: \(K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}]^3}{[\mathrm{CH}_{4}][\mathrm{H}_{2}\mathrm{O}]}\) Plug the equilibrium concentrations into the equation and calculate the value of \(K_c\). #Step 4: Calculate \(K_p\) using the relationship \(K_p = K_c(RT)^{\Delta{n}}\)#

To find \(K_p\), we need to convert \(K_c\) using the relationship: \(K_p = K_c(RT)^{\Delta{n}}\) where \(R = 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\), \(T = 1000 \mathrm{~K}\) and \(\Delta{n} =\) (moles of products – moles of reactants) = \((1 + 3) - (1 + 1) = 2\) Plug in the values and solve for \(K_p\). By following these steps, the values for \(K_c\) and \(K_p\) for the given equilibrium will be calculated.

Key Concepts

Understanding the Equilibrium Constant (Kc)

The equilibrium constant (\( K_c \) expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced equation. For the reaction \( \text{CH}_4(g) + \text{H}_2O(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g) \), the equilibrium constant is given as \( K_c = \frac{[CO][H_2]^3}{[CH_4][H_2O]} \).
The value of \( K_c \) is a unitless number that provides insight into the position of the equilibrium; a larger \( K_c \) indicates a greater concentration of products, while a smaller \( K_c \) signifies a higher concentration of reactants. It's crucial to understand that \( K_c \) does not change with changes in concentration or pressure, but it is temperature-dependent.
In the solved exercise, once the equilibrium concentrations for each substance are calculated, they are substituted into the \( K_c \) formula to determine the equilibrium constant for the reaction at 1000 K within a 5.00 L flask.

Calculating the Reaction Quotient (Kp)

The reaction quotient (\( K_p \) relates to the equilibrium constant, but it is used for gaseous reactions when we are interested in partial pressures instead of concentrations. The equation \( K_p = K_c(RT)^{\Delta n} \) incorporates the ideal gas law where \( R \) is the gas constant (\( 0.0821 \, \text{L \cdot atm} / \text{mol \cdot K} \)), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas between products and reactants.
When calculating \( K_p \) from \( K_c \) as shown in the exercise, it's important to accurately determine \( \Delta n \) from the balanced chemical equation. For every reaction, \( \Delta n \) takes into account the stoichiometry of the gaseous reactants and products, often leading to a different value than \( K_c \) due to the impact of temperature and the change in the number of moles of gas.

The Mole Concept and Its Role in Chemistry

The mole concept is a fundamental pillar in chemistry, providing a bridge between the mass of substances and the number of particles. Avogadro's number (\( 6.022 \times 10^{23} \)) is the foundation for this concept, representing the number of atoms or molecules in one mole of a substance.
Understanding how to convert between grams and moles is crucial. This conversion involves using the molar mass, which is the mass in grams of one mole of a substance. In the worked example, determining the number of moles of each gas (\( \text{CO}, \text{H}_2, \text{CH}_4, \text{H}_2O \)) from their respective masses, using their molar masses, was the first step. This process is pivotal for further calculations related to equilibrium since reactions occur on a molecular level, and knowing the amounts in moles is necessary for stoichiometric calculations.

Determining Equilibrium Concentrations in Chemical Reactions

In a chemical equilibrium situation, the equilibrium concentrations represent the concentrations of reactants and products when the rates of the forward and reverse reactions are equal. These are the concentrations that remain constant over time at a given temperature and pressure.
To find these concentrations, like in the given problem, first calculate the number of moles for each species involved in the reaction. Then, using the volume of the reaction container, we can find the concentration (\( [C] = \frac{\text{moles}}{\text{volume}} \) in moles per liter). As demonstrated in the textbook solution, equilibrium concentrations are essential for calculating the equilibrium constant \( K_c \) as they represent the 'snapshot' of the reaction's dynamic balance point. These values are specifically important in predicting the direction of the reaction, and in designing chemical processes and experiments.