Question

A flask is charged with \(1.500\) atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(1.00\) atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\), and the following equilibrium is achieved: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(0.512 \mathrm{~atm}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) ? (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Short answer

The equilibrium partial pressure of N2O4 is 1.744 atm, the value of Kp for the reaction is 0.150, and the value of Kc for the reaction is 0.0112.

Step-by-step solution

Write the expressions for the partial pressures at equilibrium

Let x be the change in the amount of substances during the equilibrium process. The change in the partial pressure of N2O4 is -x, and the change in the partial pressure of NO2 is 2x. So, the equilibrium partial pressures of N2O4 and NO2 can be written as: \( P_{N2O4} = 1.500 - x\) \( P_{NO2} = 1.00 + 2x \)

Find the change in the amount of substances during the equilibrium process

At equilibrium, the partial pressure of NO2 is 0.512 atm. Using this information, we can find the value of x: \( 1.00 + 2x = 0.512 \) Solve for x: \( 2x = -0.488 \) \( x = -0.244 \)

Calculate the equilibrium partial pressure of N2O4

Now that we have the value of x, we can use the expression for the partial pressure of N2O4 at equilibrium to find its value: \( P_{N2O4} = 1.500 - x \) \( P_{N2O4} = 1.500 + 0.244 \) \( P_{N2O4} = 1.744 \space atm \)

Find the value of Kp

We can now use the expression for Kp to find its value: \( K_p = \frac{P_{NO2}^2}{P_{N2O4}} \) Substitute the equilibrium partial pressures: \( K_p = \frac{(0.512)^2}{1.744} \) Solve for Kp: \( K_p = 0.150 \)

Calculate the value of Kc

To find the value of Kc from Kp, we use the following equation: \( K_c = K_p \left(\frac{RT}{P}\right)^{-\Delta n} \) In this case, Δn = 2 - 1 = 1, R = 0.0821 L atm / mol K, T = 25°C = 298 K, and the total pressure P doesn't affect the equation in this case as we are working with partial pressures. So, \( K_c = K_p \left( \frac{0.0821 \times 298}{1} \right)^{-1} \) Substitute Kp: \( K_c = 0.150 \left( \frac{0.0821 \times 298}{1} \right)^{-1} \) Solve for Kc: \( K_c = 0.0112 \) #Results# (a) The equilibrium partial pressure of N2O4 is 1.744 atm. (b) The value of Kp for the reaction is 0.150. (c) The value of Kc for the reaction is 0.0112.

Key concepts

Equilibrium Partial Pressure

Equilibrium partial pressure is a term frequently encountered in chemical equilibrium problems involving gases. At a constant temperature, when a gas mixture's reactions reach a state where the concentrations of reactants and products remain stable over time, the system is said to be at equilibrium. Each gas in the mixture exerts a pressure, known as its partial pressure, which ideally depends on its mole fraction in the container and the total pressure.

In the exercise provided, the initial partial pressures of the gases \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) were given, and we had to determine the changes in their partial pressures when the system reached equilibrium. By setting up an equation reflecting these changes, the problem allowed us to solve for the equilibrium partial pressures. The steps effectively demonstrate how to consider the stoichiometry of the reaction when calculating changes in partial pressure due to the formation or consumption of gases.

Equilibrium Constant

The equilibrium constant of a chemical reaction, symbolized as \(K_p\) when expressed in terms of partial pressures and \(K_c\) when using molar concentrations, is a number that provides a measure of the position of the equilibrium. A larger equilibrium constant indicates a higher concentration (or partial pressure) of products relative to reactants at equilibrium. It's important to remember that \(K_p\) is used for gas-phase reactions under constant temperature.

The exercise illustrates how to calculate \(K_p\) using the relationship \( K_p = \frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \) which stems from the balanced equation of the reaction. Using the calculated equilibrium partial pressures, we were able to determine \(K_p\) and then relate it to \(K_c\) by considering the changes in the number of moles of gas during the reaction. This conversion between \(K_p\) and \(K_c\) is vital for comparing equilibrium constants obtained under different conditions and in different units.

Understanding the equilibrium constant deeply is essential for predicting the direction of the reaction and for numerous calculations in chemistry, including those related to reaction yields and concentrations.

Le Chatelier's Principle

Le Chatelier's principle offers significant insight into how a chemical system at equilibrium responds to various changes in conditions such as concentration, pressure, or temperature. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. It is a qualitative tool that explains the behavior of equilibrium systems without the need for numerical calculations.

In the context of the exercise, if additional \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) were added to the system or if the temperature were increased (assuming \(\mathrm{N}_{2}\mathrm{O}_{4}\) formation is endothermic), the equilibrium would shift to produce more \(\mathrm{NO}_{2}(g)\) to reduce the disturbance. Conversely, increasing the partial pressure of \(\mathrm{NO}_{2}(g)\) would shift the equilibrium towards the production of \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\).

Le Chatelier's principle is fundamental in predicting how changes in physical conditions can be used to maximize production in industrial chemical reactions and is also useful in understanding buffer systems in biology and other systems in equilibrium.