Problem 37 - Chemistry The Central Science In Si Units

Question

A mixture of 0.886 mol of \(\mathrm{CO}_{2}, 0.443 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and 0.713 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 10.00-L vessel. The following equilibrium is established at \(550 \mathrm{~K}\) : $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) (b) At equilibrium \(P_{\mathrm{H}_{2}}=182 \mathrm{kPa}\). Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{CO},\) and \(\mathrm{H}_{2} \mathrm{O}\). (c) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

Step-by-Step Solution

Answer

The initial pressures are CO₂: 406.1 kPa, H₂: 203.2 kPa, H₂O: 327.0 kPa. At equilibrium, CO₂: 384.9 kPa, CO: 21.2 kPa, H₂O: 348.2 kPa. Both \(K_p\) and \(K_c\) are 0.107.

Step 1: Calculate Initial Partial Pressures

For this step, use the ideal gas law \(PV = nRT\) to find the initial partial pressures. Since \(R = 8.314 \frac{J}{mol \cdot K}\) and temperature \(T = 550 K\), pressure \(P\) can be found using \(P = \frac{nRT}{V}\), where \(V = 10.00 \text{ L}\). Convert 10.00 L to 0.010 m³ for consistent units. - Pressure of \(\mathrm{CO}_{2}\): \(\frac{0.886 \times 8.314 \times 550}{0.010} = 406.1 \text{ kPa}\). - Pressure of \(\mathrm{H}_{2}\): \(\frac{0.443 \times 8.314 \times 550}{0.010} = 203.2 \text{ kPa}\). - Pressure of \(\mathrm{H}_{2} \mathrm{O}\): \(\frac{0.713 \times 8.314 \times 550}{0.010} = 327.0 \text{kPa}\).

Step 2: Apply Change in Reaction and Equilibrium Values

In this equilibrium system, \(x\) is the change in pressure of \(\mathrm{CO}\) formed. The stoichiometry gives that the decrease in \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\) will be \(x\), and the increase for \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be \(x\). The equilibrium pressure of \(\mathrm{H}_{2}\) is given as 182 kPa.i.e., \(203.2 - x = 182\) Solving for \(x\) gives:\(x = 203.2 - 182 = 21.2 \text{ kPa}\).

Step 3: Calculate Equilibrium Partial Pressures

Using the \(x\) value of 21.2 kPa established from the previous step:- \(\mathrm{CO}_{2}\): \(406.1 - x = 406.1 - 21.2 = 384.9\text{ kPa}\)- \(\mathrm{CO}\): \(0 + x = 21.2 \text{ kPa}\)- \(\mathrm{H}_{2} \mathrm{O} \): \(327.0 + x = 327.0 + 21.2 = 348.2 \text{ kPa}\)

Step 4: Calculate \(K_p\)

\(K_p\) is calculated using the equilibrium partial pressures in the expression for the equilibrium constant of a reaction with gaseous components.Formula: \[ K_p = \frac{P_{\text{CO}} \cdot P_{\text{H}_2O}}{P_{\text{CO}_2} \cdot P_{\text{H}_2} } \]\(\therefore K_p = \frac{(21.2)(348.2)}{(384.9)(182)} = 0.107\)

Step 5: Calculate \(K_c\)

To convert \(K_p\) to \(K_c\), use the relationship:\[ K_p = K_c (RT)^{\Delta n} \]Where \(\Delta n\) is the change in moles of gas which is 0 in this case (1 + 1 - 2 = 0). Therefore:\(K_p = K_c\)\( herefore K_c = 0.107\)

Key Concepts

Partial Pressure

When dealing with gas mixtures, it's important to understand the concept of partial pressure. Partial pressure refers to the pressure that a specific gas in a mixture would exert if it occupied the entire volume alone.
Each gas in a mixture contributes to the total pressure based on its proportion of the total moles of gas present. The sum of the partial pressures of all gases in the mixture equals the total pressure of the gas mixture.

• In the provided exercise, the partial pressures of carbon dioxide (\(\text{CO}_2\)), hydrogen (\(\text{H}_2\)), and water vapor (\(\text{H}_2O\)) are calculated using the initial amounts and the ideal gas law. This provides a snapshot of the gas situation before any reaction occurs.
• Using the ideal gas law, we determine the initial partial pressures with the formula: \(P = \frac{nRT}{V}\), where \(n\) represents the number of moles, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(V\) is volume in cubic meters.

Understanding and calculating partial pressures are foundational in predicting how gases in a system will behave when chemical reactions occur.

Equilibrium Constant

In chemical reactions involving gases, the equilibrium constant (\(K\)) plays a crucial role in understanding the system at equilibrium. The equilibrium constant can be expressed both in terms of partial pressures (\(K_p\)) and concentrations (\(K_c\)).
It provides a measure of the tendency of a chemical reaction to proceed in either direction under a set of conditions: towards the formation of products or reactants.
**The Role of the Equilibrium Constant:**

• The equilibrium constant is calculated using the partial pressures or concentrations of products and reactants at equilibrium. Specifically for gases, \(K_p\) is used: \[ K_p = \frac{P_{\text{products}}}{P_{\text{reactants}}} \]
• In simpler terms, a higher \(K\) value indicates a greater amount of products at equilibrium, indicating the reaction favors the forward direction. Conversely, a lower \(K\) value means the reaction leans towards the formation of reactants.

In the original exercise, \(K_p\) was calculated using the equilibrium partial pressures of the substances involved. The calculated \(K_p\) value of 0.107 suggests a balance between the products and reactants in the system.

Ideal Gas Law

The Ideal Gas Law is an essential equation in chemistry used to connect physical properties of gases. It relates the pressure, volume, temperature, and number of moles of a gas with the equation \(PV = nRT\). This equation assumes that gases behave ideally, meaning that they follow these relationships precisely under a wide range of conditions.

• In the model scenario, the Ideal Gas Law helps determine the initial conditions by calculating the partial pressures of each gas present in the 10.00 L container at a temperature of 550 K.
• Given that \(R\), the gas constant, is 8.314 \(\frac{J}{mol \cdot K}\), all parameters must be consistent in units, which often requires conversion (e.g., converting volume from liters to cubic meters).

By understanding the Ideal Gas Law, students can visualize how changes in one property, such as temperature or volume, can affect the others. It's a cornerstone principle for understanding and predicting the behavior of gases in chemical reactions, as demonstrated in the original exercise.