# How Does Halving the Volume Affect Gas Equilibrium in a Closed System?

• Qube
In summary, the conversation discussed the equilibrium constant of a reaction in a cylinder filled with three gases - ammonia, nitrogen, and hydrogen. It was determined that there is only 1 mole of each gas at equilibrium, and if the volume is halved at constant temperature, the reverse reaction will be favored. The new number of moles of each gas can be calculated, and the equilibrium constant remains the same. The conversation also addressed some questions about the logic and understanding of the problem, and clarified that it is important to be critical of sources but also open to different perspectives.
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## Homework Statement

There is a constant volume in a cylinder filled with three gases - ammonia, nitrogen, and hydrogen. The volume is 1.0 L. The equilibrium constant of the reaction of nitrogen and hydrogen to form ammonia is 1.0. Find the number of moles of each gas at equilibrium, and what happens if we halve the volume of the cylinder at constant temperature?

## Homework Equations

2 H3N ⇌ N2 + 3 H2

K = [non-solid products]/[non-solid reactants] = 1 = ([N2]

## The Attempt at a Solution

The equilibrium constant is 1, and the volume is one. This greatly simplifies calculations. We can easily figure out that there is only 1 mole of each gas. 1 times 1 all divided by 1 is still 1.

Because the volume is 1, volume does not factor into the equilibrium constant calculation. We can therefore just plug moles into the equilibrium constant equation. If we have 1 mole of N2, H2, and ammonia, then that causes the equilibrium constant to equal 1, as given.

What happens when we halve the volume of the cylinder?

We're not changing the temperature, only the pressure, so K, the equilibrium constant, should remain the same.

Since we're disturbing the system, the system will attempt to return to equilibrium.

First we must consider which way the reaction will proceed to return to equilibrium. I did this two ways.

1) Decreasing the volume increases the pressure. Upon consideration of the ammonia synthesis equation, there are fewer reactant moles than product moles. Therefore, the reverse reaction will be favored. It is "easier" for fewer moles of a gas to exist at a high pressure than for more moles to exist.

2) I also considered Q. Because volume was halved, molarities were doubled. The molarities of the gases were all originally 1, since the original volume was 1 and the number of moles of each was was 1. Therefore, Q =

(2 * 2^3) / (2^2) = 4.

Because Q > K, Q must decrease to equal K. For Q to decrease, the numerator must decrease in value, and the denominator must increase in value. Because the expression of Q, like K, places non-solid products over non-solid reactants, this implies that the amount of product must decrease, and the amount of reactant must increase. Therefore, the reverse reaction is favored; ammonia is going to be produced as the system attempts to return to equilibrium following the disruption.

So now we know the reverse reaction is favored. We are going to be producing ammonia at the "expense" of nitrogen and hydrogen.

Therefore, the new number of moles of each molecule are:

1 + 2x moles of ammonia.
1 - 3x moles of hydrogen.
1 - x moles of nitrogen.

These moles can be converted to molarities by dividing by the new volume (0.5 L). This effectively doubles the number of moles of each. New molarities:

2 + 4x M ammonia
2 - 6x M hydrogen
2 - 2x M nitrogen.

Now we can finally restore these values to the equilibrium constant equation. K = 1 still holds because the disruption to the system occurred at constant temperature.

1 = ([2-2x][2-6x]^3 )/ [2+4x]^2

Now, this is a rather messy rational function to deal with, so I'll first figure out the constraints of x.

2 - 6x > 0. We cannot have molar concentrations equal to 0, or negative. There wouldn't be much of a reaction if one concentration were 0, and it makes no sense to have a negative concentration. From this inequality we can see that x < 0.34.

Also, x > 0. We can't be working with a negative number of moles of anything.

So I guess x = 0.1. I get 0.85. No cigar. This is significantly less than the equilibrium constant of 1. This implies I must increase the numerator and decrease the denominator.

To this effect I guess x = 0.09.

I get 1.017. Too big. I must decrease the numerator and increase the denominator.

x = 0.091

I get 0.999. Close enough.

Questions:

1) I'm actually pre-learning general chemistry II for next semester. I've been following a prof's lecture videos. Could one please look over my train of thought above and tell me if there are any logical holes, misunderstandings, etc.? It would be very, very much appreciated. This prof does his own thing with chemistry and has kicked a student's general chemistry book across the classroom, denouncing it as a piece of trash and instead promoting his own typed up chemistry "textbook" (more like a bunch of bound notes). One of his pet peeve about mass-produced books is that they fail to write the "correct" formulas for chemical substances, such as writing NH3 instead of H3N, or OH- instead of HO-. Hence my usage of H3N above.

2) Only temperature changes affect the equilibrium constant K, correct? Had the problem said that volume was halved but did not state that temperature was kept constant, then I would have to assume that temperature was kept constant to be able to use the same equilibrium constant, right?

3) Also, how come no equilibrium with respect to solute exists in a solution of 1 molar CaCl2? Is it because all the calcium chloride dissolves and that there is no calcium chloride that exists? The lack of a solute would make equilibrium with respect to solute impossible, right?

Last edited:

Hi there,

Thank you for your detailed response to the forum post. I am always happy to see individuals taking the time to think critically and analyze scientific concepts.

After reviewing your thought process, I do not see any major logical holes or misunderstandings. Your approach to the problem is sound and your calculations seem to be correct.

Here are some additional thoughts and clarifications:

1) It is always important to be critical of information and sources, especially in the scientific field. However, it is also important to be open to different perspectives and to consider multiple sources of information. While your professor may have his own opinions on certain textbooks and formulas, it is important to also refer to reputable sources and textbooks for accurate information. In this case, both NH3 and H3N are acceptable notations for ammonia, and OH- and HO- are both used to represent hydroxide ions.

2) You are correct, temperature is the only factor that affects the equilibrium constant K. If the volume was halved and the temperature was not specified, it would be safe to assume that the temperature remained constant.

3) In a solution of 1 molar CaCl2, there is no equilibrium with respect to solute because all of the calcium chloride has dissociated into its constituent ions (Ca2+ and 2Cl-). Therefore, there is no solute present to establish an equilibrium with. This is different from a solution where only a portion of the solute has dissociated, in which case an equilibrium with respect to solute would exist.

I hope this helps clarify any questions you may have had. Keep up the good work in your studies and always stay curious and open-minded.

## What is Le Chatelier's Principle?

Le Chatelier's Principle is a fundamental principle in chemistry that states when a system at equilibrium is subjected to a stress, it will shift in a way that minimizes the effect of the stress.

## What are the three types of stresses that can affect a system at equilibrium?

The three types of stresses are changes in temperature, pressure, and concentration. These stresses can cause the equilibrium to shift in order to maintain balance.

## How does Le Chatelier's Principle apply to exothermic and endothermic reactions?

In an exothermic reaction, an increase in temperature will cause the equilibrium to shift to the left, favoring the reactants. In an endothermic reaction, an increase in temperature will cause the equilibrium to shift to the right, favoring the products.

## What is the effect of a change in pressure on a system at equilibrium?

If the number of moles of gas are different on the reactant and product side, a change in pressure will affect the equilibrium. An increase in pressure will shift the equilibrium to the side with fewer moles of gas, while a decrease in pressure will shift the equilibrium to the side with more moles of gas.

## How does Le Chatelier's Principle apply to changes in concentration?

If the concentration of a reactant or product is changed, the equilibrium will shift in the opposite direction to minimize the stress. An increase in concentration of a reactant will cause the equilibrium to shift to the right, while an increase in concentration of a product will cause the equilibrium to shift to the left.

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