Equilibrium !!! Please Help??????

Question

I am in Grade 12 in Australia and I have a prac report due tomorrow ! I though I understood everything but then my friend saw this whole question from a different point of view and now I am confused. We had the solution : the * with number represents the charge of the ion (it is in equilibrium)
Fe *3+ (aq) + SCN* 1- (aq) ---> Fe(SCN)*2+
The colour of this is red. When enough water (H2O) is added to the solution to double the volume, the solution turns pale red. The question is: the direction in which the reaction must move to restore equilibrium?and why? And : Explain the observation for the test tube by referring to both collition theory and Le Chatelier's principle. I think that after the solution is diluted, there is an excess of products (because both the reactan concentrations would be halved - the are two reactants - and only one product's consentration would have to be halved). And therefore I think that the direction the reaction must move is to the left.

But today my friend suggested another theory about this - she said
that because there is no more volume, the particles won't collide as
frequently and before and therefore the reactants won't form such a big
amount of products and therefore there would be an excess of reactants and
the reaction would have to move to the right (to the product side).

I'm so confused now... Could you please help me and e-mail me as soon
as possible !!!

[e-mail]

no - it is an equilibrium question - the K value is doubled (if you sub in the numbers in the constant rule) even though the temp and pressure stayed the same. And therefore the reaction must either shift to the left or right to change K value and establish equilibrium. And I don't know if the solution has excess reactants or products............. my dilemma

Answer 1

Abby is right, and your friend was pointing you in the right direction, but maybe this will help clarify things.

The equilibrium expression can be written as

Q= [Fe(SCN)] / [Fe 3+]*[SCN-]

Now, remember at a given temp, Q is a constant called Keq. BUT, if you change the concentration of all three species by diluting , what happens to Q?

It goes UP because the denominator shrunk by more than the numerator. So equillibrium needs to be restored (you didn't change the temperature!) To do this, you get a backward reaction so that the numerator goes down, and the denominator goes up...Q returns to the value of Keq.

Since the Fe(SCN)2+ ion is responsible for the color, and its concentration goes down...the color gets more pale.

Answer 2

I was in yr12 last yr and got a band 6 in chemistry, this doesnt sound like a typical equillibrium question. I'm pretty sure adding water in this system would only act to dilute both the products and reactants but have no net effect on the posistion of the equillibrium. I think this is indicated by the pale red. Red > diluted > pale red....maybe i'm wrong but i doubt thy would ask something like this in the HSC.

Most of our problems concerning Le Chatiler(never mastered his name) concerned concentration, pressure and temperature.

Answer 3

A net BACKWARD RXN will occur. As Le Chatelier's Principle tells us that a equilibrium system would adjust itself partially to oppose the effects of any changes, and since the adding of water reduced the [ ] of all of the compounds in the system, it would result in a increase in K (halved, doubled, and doubled). Hence, to reduce K, the [reactant] must drop further while [products] increase. Hence a BACKWARD RXN.