Problem 33 - Chemistry: The Central Science in SI Units, Global Edition

Question

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(550 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\), and \(35.46 \mathrm{kPa}\) for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively.(a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L},\) calculate \(K_{c}\) at this temperature.

Step-by-Step Solution

Answer

The values of Kp and Kc for the given reaction are: (a) \( K_p = 3.059 \) (b) \( K_c = 0.376 \)

Step 1: Calculate Kp using the given equilibrium partial pressures

We are given the partial pressures of the reactants and products at equilibrium. Using the equilibrium expression for Kp, we can calculate its value as follows: The equilibrium expression is \( K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_{2}}} \) Now, substitute the given partial pressures into the expression: \( P_{NO} = 10.13 kPa \) \( P_{Cl_{2}} = 20.27 kPa \) \( P_{NOCl} = 35.46 kPa \) \( K_p = \frac{(35.46)^2}{(10.13)^2 \times 20.27} \) Calculate the value of Kp: \( K_p = 3.059 \)

Step 2: Convert Kp to Kc using the relationship between Kp and Kc

Given that the volume of the vessel is 5.00 L and the temperature is 500.0 K, we can use the relationship between Kp and Kc to find Kc. For a reaction with a change in the number of moles of gas (Δn), the relationship between Kp and Kc is given by: \( K_p = K_c(RT)^{Δn} \) Where R is the gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin, and Δn is the change in moles of gas between reactants and products. For the given reaction: \( Δn = (2 - 2 - 1) = -1 \) Now, solve for Kc: \( K_c = K_p / (RT)^{Δn} \) Since we are given the pressures in kPa, we will use the value of R in L kPa / mol K, which is R = 8.314 L kPa / mol K, and the given temperature to find Kc: \( K_c = \frac{3.059}{(8.314 \times 500.0)^{-1}} \) Calculate the value of Kc: \( K_c = 0.376 \) So the values of Kp and Kc for the given reaction are as follows: (a) \( K_p = 3.059 \) (b) \( K_c = 0.376 \)

Key Concepts

Equilibrium Constant (Kp and Kc)

The equilibrium constant is essential to understanding chemical equilibrium. It tells us how the concentrations or pressures of reactants and products relate when a system reaches equilibrium. There are two main types of equilibrium constants: \( K_p \) and \( K_c \).

\( K_p \) is used when dealing with gases, as it focuses on the partial pressures of the gases involved. To find \( K_p \), you use the equation that relates the pressure of the products to the reactants. For example, in the reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2\mathrm{NOCl}(g)\), the equation becomes \( K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_{2}}} \).

On the other hand, \( K_c \) is utilized when concentrations in a solution are involved, typically measured in molarity (moles per liter). Both \( K_p \) and \( K_c \) are related by the equation \( K_p = K_c(RT)^{Δn} \), where \(Δn\) is the difference in moles of products and reactants in the gas phase.

• Both constants may reveal whether the equilibrium favors products or reactants.
• They vary with temperature changes and not with pressure or concentration changes.

If you know \( K_p \), you can find \( K_c \), and vice versa, provided the temperature and \( Δn \) are known.

Partial Pressure

Partial pressures play a vital role when working with gas-phase equilibriums. Each gas in a mixture exerts its own pressure, called partial pressure. In the context of equilibrium, partial pressures help us calculate \( K_p \).

Taking the example equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2\mathrm{NOCl}(g)\), the partial pressures might be \(10.13\, \mathrm{kPa}\), \(20.27\, \mathrm{kPa}\), and \(35.46\, \mathrm{kPa}\) for \(\mathrm{NO}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{NOCl}\), respectively. These values help compute \( K_p \) using the equation \( K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_{2}}} \).

In a gas mixture:

• Each gas contributes to the total pressure based on its mole fraction and temperature.
• The sum of all partial pressures equals the total pressure of the gas mixture.

Understanding partial pressures allows us to predict how gases will behave under different conditions, especially important for reactions at dynamic equilibrium.

Gas Constant

The gas constant, often denoted as \( R \), is a crucial component in the equations relating thermodynamic quantities. Depending on the units you are working with, \( R \) has different numerical values. For instance:

• In the equation \( K_p = K_c(RT)^{Δn} \), \( R = 8.314\, \mathrm{L \cdot kPa / mol \cdot K} \) when pressure is in kPa.
• If pressure were in atmospheres, you would use \( R = 0.0821\, \mathrm{L \cdot atm / mol \cdot K} \).

Understanding \( R \) is fundamental when converting between \( K_p \) and \( K_c \), as it ties in the effect of temperature with the change in moles of gas, \( Δn \).

\( R \) helps to:

• Factor in the role of temperature in shifts in equilibrium.
• Bridge the relationship between pressure and concentration-based equilibrium expressions.

Always ensure that your units of \( R \) align with the units used in your calculations, otherwise, it might lead to errors in interpreting the equilibrium constants.