# Question about equilibrium position

• etotheipi
In summary, the equilibrium position refers to the state where the forces acting on an object are balanced, resulting in no net acceleration. This position can change depending on various factors, such as external forces or changes in the system itself. Additionally, the equilibrium position can be either stable, where the object returns to its original position after being displaced, or unstable, where the object moves away from its original position when displaced. Understanding equilibrium position is crucial in analyzing and predicting the behavior of systems in physics and chemistry.
etotheipi
Suppose we have a reversible reaction $$aA + bB <=> cC + dD$$ which is exothermic in the forward direction. If the temperature of the system decreases, the rate of the forward reaction will increase so that the quotient Q increases to the new (higher) value of Kc. This I think I am ok with (I am also taking Q to be a measure of the position of equilibrium).

Now consider reducing the concentration of C at a constant temperature. This will initially reduce the quotient, however the rate of forward reaction will then increase to restore the quotient to the original value of Kc. My question is that many sources would say that the position of equilibrium shifts to the right, when in fact it seems that the position of equilibrium has moved left then right again to its original position.

What am I missing here? Thanks :)

Last edited by a moderator:
etotheipi said:
If the temperature of the system decreases, the rate of the forward reaction will increase
No, it will not. The rates of both forward and backward reactions decrease with decreasing temperature, but the back reaction more so.

Q is not a measure of the position of equilibrium. K is. Q = K when the reaction is at equilibrium. Reducing the quotient (by reducing [C]) does not move the equilibrium to the left; it moves the position of the reaction to the left, away from the equilibrium condition. The rate of the forward reaction will not increase; the rate of the reverse reaction decreases because [C] is decreased.

mjc123 said:
No, it will not. The rates of both forward and backward reactions decrease with decreasing temperature, but the back reaction more so.

Yeah sorry you're right, I should have said the net forward reaction.

mjc123 said:
Reducing the quotient (by reducing [C]) does not move the equilibrium to the left; it moves the position of the reaction to the left, away from the equilibrium condition. The rate of the forward reaction will not increase; the rate of the reverse reaction decreases because [C] is decreased.

This makes a lot of sense. However why would we say that the position of equilibrium shifts to the right if we reduce [C] ?

The reaction shifts to the right, i.e. there is a net conversion of A and B to C and D. The position of equilibrium as defined by K does not change, Q = K at equilibrium, but the individual concentrations will be different from before.

etotheipi
mjc123 said:
The reaction shifts to the right, i.e. there is a net conversion of A and B to C and D. The position of equilibrium as defined by K does not change, Q = K at equilibrium, but the individual concentrations will be different from before.

Thank you, this clears up the confusion I had.

Just as a follow up, I got the following from chemguide; what is the position of equilibrium actually referring to? It seems like it should actually be the position of reaction like you mentioned instead.

"The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.

Suppose you have an equilibrium established between four substances A, B, C and D.

According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again."

I would say it means the position of reaction at equilibrium - whatever that means! The trouble is that "position of reaction" and "position of equilibrium" are not well-defined terms.
The equilibrium constant K is constant at constant temperature, but changes with change of temperature. At equilibrium the reaction quotient Q is equal to K. If you change the concentrations, e.g. by removing C, then Q is no longer equal to K. If Q > K, the reaction will shift to the left to reduce Q. If Q < K, the reaction will shift to the right to increase Q. When equilibrium is re-established, Q = K again, but the composition of the mixture is different from the previous equilibrium state. That is why, depending on your definition, you could say the "position of equilibrium" is either the same or different!

etotheipi

## What is equilibrium position?

Equilibrium position refers to the point at which the forces acting on a system are balanced, resulting in no net change in the system's position or state.

## What factors affect the equilibrium position?

The equilibrium position is affected by factors such as temperature, pressure, concentration, and the nature of the reactants and products involved in the system.

## How is equilibrium position calculated?

The equilibrium position can be calculated using the equilibrium constant, which is a ratio of the concentrations of the products and reactants at equilibrium.

## What happens if the equilibrium position is disturbed?

If the equilibrium position is disturbed, the system will shift in a direction that minimizes the disturbance and restores the equilibrium position. This is known as Le Chatelier's principle.

## Can the equilibrium position ever be reached?

In theory, the equilibrium position can be reached, but in practice, it may take a very long time for a system to reach equilibrium. Factors such as reaction rate and external conditions can affect the time it takes for equilibrium to be reached.

Replies
3
Views
1K
Replies
1
Views
1K
Replies
9
Views
3K
Replies
9
Views
27K
Replies
3
Views
1K
Replies
14
Views
2K
Replies
1
Views
1K
Replies
2
Views
3K
Replies
131
Views
6K
Replies
5
Views
3K