Change in concentration in equilibrium caused by addition

[TT0]

Homework Statement

Cu²⁺ ions react with Fe²⁺ ions according to the following reaction.

Cu²⁺ + 2Fe²⁺ --> Cu + 2Fe³⁺

At equilibrium, the concentration of Cu²⁺ ions is not changed by the addition of

1. A) Cu²⁺+
2. B) Fe²⁺
3. C) Cu
4. D) Fe³⁺.
   

Homework Equations

The Attempt at a Solution

I chose A through elimination.

B won't work because the equilibrium will shift to the right if Fe²⁺ is added.

C and D won't work because the equilibrium will shift to the right.

Therefore A seems the most correct. However, if Cu₂₊ is added, then the equilibrium will also shift to the right. This means that it shifts to the right until the Cu²⁺ concentration is back to what it was before which seems impossible. Could someone shed some light on this?

 

[Borek]

Can you write formula for the equilibrium constant for this problem?

 

[TT0]

$$\frac {[Cu][Fe^{3+}]^2} {[Cu^{2+}][Fe^{2+}]^2}$$

This is the equilibrium constant. (took me a while to get the latex working)

 

[Borek]

Not exactly. Solid copper is not dissolved, so it doesn't have a concentration.

 

[TT0]

I see, so since it isn't dissolved, adding solid copper wouldn't change it's concentration. But won't it will still increase the backwards rate of reaction so it will still affect the Cu²⁺ concentration? This is the best I could come up with from the comment.

 

[Borek]

Changes in the amount of solid change both forward and backward reaction speed in exactly the same way, the net effect cancels out.

That's why there are no solids in the equilibrium formula (we assume activity of the solids equals exactly 1).

 

[TT0]

I see, so as a rule, is this correct?

If solid reactants or products is added to a system already in equilibrium, the concentrations of all chemicals will stay the same; the reaction rate will increase equally in both directions.

Thank you very much!

 

[Borek]

If solid reactants or products is added to a system already in equilibrium, the concentrations of all chemicals will stay the same; the reaction rate will increase equally in both directions.

And the equilibrium won't shift in any direction.

 

[TT0]

Cheers! This is useful knowledge!