Calculating Equilibrium Pressures

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In summary, the conversation discusses a reaction involving NO2 and N2O4 with given values for ΔH°_{rxn} and ΔS°_{rxn}. The equilibrium pressures of NO2 and N2O4 are then calculated at a given temperature and under two different conditions: constant volume and constant pressure. The equilibrium constant is also discussed, with a focus on how it is expressed in terms of the partial pressures of the two gases. The equation used to find the equilibrium pressures is based on the fundamental thermodynamic equation, and the standard heat of reaction and entropy change are considered at constant pressure.
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Homework Statement



For the reaction

2 NO[itex]_{2}[/itex](g) [itex]\leftrightarrow[/itex] N[itex]_{2}[/itex]O[itex]_{4}[/itex](g)

ΔH°[itex]_{rxn}[/itex] = -57.2 kJ and ΔS°[itex]_{rxn}[/itex] = -175.8 [itex]J/K[/itex]

A 1 L container is initially filled with 1 atm of NO[itex]_{2}[/itex] at 298 K. Find the equilibrium pressures of NO[itex]_{2}[/itex] and N[itex]_{2}[/itex]O[itex]_{4}[/itex] at 298 K if

a. volume is held constant

b. pressure is held constant

Homework Equations



dG = VdP - SdT

The Attempt at a Solution



So using the fundamental thermodynamic equation above, I found that ΔG = -4811.6 J. SdT goes to zero since the temperature doesn't appear to change. VdP can be substituted withdH -TdS, with all three values available in the problem.

So using the following relation between K[itex]_{eq}[/itex] and free energy:

ΔG°[itex]_{rxn}[/itex] = -RTln(K[itex]_{eq}[/itex])

Solving for K[itex]_{eq}[/itex], I obtain 6.973

So from this point, how do I calculate the partial pressures of NO[itex]_{2}[/itex] and N[itex]_{2}[/itex]O[itex]_{4}[/itex]? Can I stop at 6.973?

As for the second question, I find that if pressure is held constant, the equilibrium constant is just 1, since ΔG[itex]_{rxn}[/itex] goes to zero.
 
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  • #2
How is the equilibrium constant expressed in terms of the partial pressures of NO2 and N2O4? Also note that the standard heat of reaction and standard entropy change of reaction are evaluated at constant pressure (1 atm), not constant volume. So the problem you have been working on is really problem 2. You have yet to address problem 1.
 

1. How do you calculate equilibrium pressure?

To calculate equilibrium pressure, you need to know the number of moles of each gas present in the reaction and their respective partial pressures. Then, use the ideal gas law (PV = nRT) to calculate the total pressure at equilibrium.

2. What is the significance of calculating equilibrium pressure?

Calculating equilibrium pressure allows us to understand the behavior of a chemical reaction and determine the optimal conditions for a reaction to reach equilibrium. It also helps us predict the direction in which a reaction will proceed.

3. Can equilibrium pressure be changed?

Yes, equilibrium pressure can be changed by altering the temperature, pressure, or concentration of the reactants or products. This will shift the equilibrium position and ultimately change the equilibrium pressure.

4. How does Le Chatelier's principle apply to calculating equilibrium pressure?

Le Chatelier's principle states that a system at equilibrium will respond to any changes by shifting the equilibrium position in a way that minimizes the effect of the change. This means that if the pressure is increased, the equilibrium will shift to the side with fewer moles of gas in order to decrease the pressure.

5. How does equilibrium pressure differ from initial pressure?

Equilibrium pressure is the total pressure of all gases present in a reaction at equilibrium, while initial pressure is the total pressure of all gases present at the beginning of the reaction. Equilibrium pressure takes into account the changes that occur as the reaction reaches equilibrium, while initial pressure does not.

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