Chapter 14: Problem 104
At \(25^{\circ} \mathrm{C},\) the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.15 atm and 0.20 atm, respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.
Short Answer
Expert verified
The partial pressures when a new equilibrium is established are: N2O4 = 0.32 atm, NO2 = 0 atm.
Step by step solution
01
Define the Chemical Reaction
The given chemical reaction is: 2 NO2 ↔ N2O4
02
Initial Conditions
The partial pressure of NO2 is 0.15 atm and that of N2O4 is 0.20 atm before the volume change. Let's denote the change in pressure of NO2 as \( \Delta p_{NO2} \) in atm and the change in pressure of N2O4 as \( \Delta p_{N2O4} \) in atm when the volume is halved.
03
Le'Chatelier's Principle
Now according to Le'Chatelier's principle, the reaction will shift towards the side with the less number of moles of gas, which is the N2O4 side. Thus, \( \Delta p_{NO2} = -2 \Delta p_{N2O4} \).
04
Expressing change
As Boyle's law states that pressure and volume are inversely proportional, halving the volume doubles the pressure. Therefore, the total pressure after volume change will be 2 times the initial total pressure.
05
Calculating Change
Let's denote the total initial pressure as \( p_{i} \) and the total pressure after the change as \( p_{f} \). So, \( p_{f} = 2 * p_{i} \) or \( p_{i} = p_{f}/2 \). Now, the initial total pressure \( p_{i} \) is the sum of initial partial pressures of N2O4 and NO2, i.e., \( p_{i} = 0.20 + 0.15 = 0.35 \) atm. Therefore, the total pressure after volume change \( p_{f} \) is \( p_{f} = 2 * 0.35 = 0.70 \) atm.
06
Calculating New Partial Pressures
The total pressure after the change \( p_{f} \) is the sum of the pressures after the change of NO2 and N2O4. Therefore, 0.70 = (0.15 - \( \Delta p_{NO2} \)) + (0.20 + \( \Delta p_{N2O4} \)). By substituting \( \Delta p_{NO2} = -2 \Delta p_{N2O4} \), we can solve for \( \Delta p_{N2O4} \): \( \Delta p_{N2O4} = (0.70 - 0.15 - 0.20) / 3 = 0.12 \) atm and \( \Delta p_{NO2} = -2 * 0.12 = -0.24 \) atm.
07
Final Partial Pressures
The final partial pressures will be: N2O4 = 0.20 + 0.12 = 0.32 atm and NO2 = 0.15 - 0.24 = -0.09 atm. However, since pressure can't be negative, the partial pressure of NO2 is 0 atm, implying all NO2 is converted to N2O4 when the volume is halved.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Le Chatelier's Principle
Understanding how chemical reactions respond to changes in their environment is crucial, and that's where Le Chatelier's principle comes into play. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of the equilibrium will shift to counteract that change.
In the context of the given exercise, when the volume of the container is doubled, the system originally at equilibrium finds itself with too much space and not enough pressure. According to Le Chatelier's principle, the reaction will attempt to increase pressure by shifting towards the side with more moles of gas. In the case of the reaction 2 NO2 ↔ N2O4, there are initially more moles of NO2 (two moles of gas) than N2O4 (one mole of gas). So, upon doubling the volume, the reaction shifts to produce more NO2 from N2O4 to increase the pressure, directly demonstrating Le Chatelier's principle in action.
In the context of the given exercise, when the volume of the container is doubled, the system originally at equilibrium finds itself with too much space and not enough pressure. According to Le Chatelier's principle, the reaction will attempt to increase pressure by shifting towards the side with more moles of gas. In the case of the reaction 2 NO2 ↔ N2O4, there are initially more moles of NO2 (two moles of gas) than N2O4 (one mole of gas). So, upon doubling the volume, the reaction shifts to produce more NO2 from N2O4 to increase the pressure, directly demonstrating Le Chatelier's principle in action.
Partial Pressure
The term partial pressure refers to the individual pressure exerted by a single gas within a mixture of gases. It can also be seen as the contribution that gas makes to the total pressure of the gas mixture. Understanding partial pressures is necessary to solve equilibrium problems in gaseous systems, such as in the given exercise.
When the volume is doubled at constant temperature, the partial pressure of each gas would theoretically be halved if the system were not at equilibrium. However, because the system seeks to re-establish equilibrium as Le Chatelier's principle suggests, the new partial pressures will not simply be half of the initial values. Determining these new partial pressures involves considering how the equilibrium will shift to compensate for the volume change.
When the volume is doubled at constant temperature, the partial pressure of each gas would theoretically be halved if the system were not at equilibrium. However, because the system seeks to re-establish equilibrium as Le Chatelier's principle suggests, the new partial pressures will not simply be half of the initial values. Determining these new partial pressures involves considering how the equilibrium will shift to compensate for the volume change.
Equilibrium Constant
The equilibrium constant, denoted as K, is a number that expresses the ratio of product concentrations to reactant concentrations at equilibrium. Each reaction at a given temperature has its own constant value. In the case of gaseous reactions, like the one presented in the exercise, the equilibrium constant is often expressed in terms of partial pressures, known as Kp.
It's important to note that while Le Chatelier's principle tells us the direction in which the equilibrium will shift, the equilibrium constant itself does not change with a change in volume or pressure. This is because it is only dependent on temperature. This principle helps to understand how the ratios of the partial pressures change as the reaction reaches a new equilibrium after a disturbance, such as the volume change in our exercise.
It's important to note that while Le Chatelier's principle tells us the direction in which the equilibrium will shift, the equilibrium constant itself does not change with a change in volume or pressure. This is because it is only dependent on temperature. This principle helps to understand how the ratios of the partial pressures change as the reaction reaches a new equilibrium after a disturbance, such as the volume change in our exercise.