Equation for the dynamic equilibrium

[Janiceleong26]

1. Homework Statement

Homework Equations

The Attempt at a Solution

This is what I did:

Since they said at equilibrium, the total no. of moles is (2+x), so P and Q should have 1 mol each at equilibrium. So P and Q are in 1:1 ratio, but how do they know x is 1 ? Ans is B

 

[Borek]

So P and Q are in 1:1 ratio

Not necesarilly.

Write ICE table for each reaction separately and try to express total number of moles using equilibrium amounts (hint: some of xs will cancel out). The ICE table you wrote is wrong and doesn't fit any of the reactions. Change of P doesn't equal -1 in any of these reactions.

 

[epenguin]

It's at the limit of being a trick question. I guess the scientific insight it is testing for is to see it's a trick. I don't know what I can say according to rules, but don't want you to not waste time. Put it this way, if the reaction has gone to quantitative completion, what are the possibilities? Just a start that is.

(Sometimes after you have solved a problem like this you see a more efficient way of thinking for which you think doh you can see it immediately obvious. I haven't done this one long way or short way - but in this other recent problem https://www.physicsforums.com/threads/mendelians-epitasis-9-3-4-biology.838553/#post-5263793 a day or two after writing the table out I saw how it could have been immediately obvious without it. Student hasn't come back so is almost time to go into my black book ⇓⇓ though.)

 

[Janiceleong26]

Not necesarilly.

Write ICE table for each reaction separately and try to express total number of moles using equilibrium amounts (hint: some of xs will cancel out). The ICE table you wrote is wrong and doesn't fit any of the reactions. Change of P doesn't equal -1 in any of these reactions.

Oh yeah.. So if I let the change in mol of P be y ..

Is this right? So since the change in mol of P and Q are the same, i.e. y, and R is different, i.e. x, so that is why B is right?

 

[Janiceleong26]

It's at the limit of being a trick question. I guess the scientific insight it is testing for is to see it's a trick. I don't know what I can say according to rules, but don't want you to not waste time. Put it this way, if the reaction has gone to quantitative completion, what are the possibilities? Just a start that is.

(Sometimes after you have solved a problem like this you see a more efficient way of thinking for which you think doh you can see it immediately obvious. I haven't done this one long way or short way - but in this other recent problem https://www.physicsforums.com/threads/mendelians-epitasis-9-3-4-biology.838553/#post-5263793 a day or two after writing the table out I saw how it could have been immediately obvious without it. Student hasn't come back so is almost time to go into my black book ⇓⇓ though.)

Sorry, but what do you mean by the possibilities after a reaction has gone to quantitative completion? The total no. of moles in the mixture will always be the same?

 

[epenguin]

By gone to quantitative completion and similar phrases is meant equilibrium all over to one side, 100% conversion, nothing left of original reactant.

I've had my doh moment at lunch.

For your table, start again. Just have one variable.

 

[Borek]

Oh yeah.. So if I let the change in mol of P be y ..

View attachment 90686
Is this right? So since the change in mol of P and Q are the same, i.e. y, and R is different, i.e. x, so that is why B is right?

First - as epenguin said, one variable is enough.

Second - you can't write ICE table not explaining to which reaction it refers.

 

[epenguin]

Well tell us what you tried and got. On a third look I might have misinderstood this question, can you tell us where if comes from?
Not sure information enough for a numerical answer In general.

 

[Ygggdrasil]

Another approach to solve the question would be to choose a value of x (say 1 mol), then calculate the total moles at equilibrium in each of the four cases. In only one case will the total number of moles at equilibrium equal 2+x (it also works when using an ICE table with x, but sometimes its helpful to have actual numbers).

Well tell us what you tried and got. On a third look I might have misinderstood this question, can you tell us where if comes from?
Not sure information enough for a numerical answer In general.

Given the choices, there is a unique solution to the problem. However, there would not be a unique solution if it were not a multiple choice question.

 

[Janiceleong26]

By gone to quantitative completion and similar phrases is meant equilibrium all over to one side, 100% conversion, nothing left of original reactant.

I've had my doh moment at lunch.

For your table, start again. Just have one variable.

but the question states that P is "partly decomposed" this question is from Cambridge CIE AS level Chemistry.

First - as epenguin said, one variable is enough.

Second - you can't write ICE table not explaining to which reaction it refers.

If only one variable is used, then the change in moles of P and Q would be 2x each right? Judging by the answer i.e. Mol ratio in the equation in B. But if the question did not give us any options, how do I know the change would be 2x? The change in concentration for P and Q are the same and they can be ANY number, but not 2 as the question said P partly decomposes.

Can you give me an example or some hints on how to write ICE table for each reactions separately? Sorry but I've only been taught to write ICE table this way

 

[Janiceleong26]

Another approach to solve the question would be to choose a value of x (say 1 mol), then calculate the total moles at equilibrium in each of the four cases. In only one case will the total number of moles at equilibrium equal 2+x (it also works when using an ICE table with x, but sometimes its helpful to have actual numbers).
Given the choices, there is a unique solution to the problem. However, there would not be a unique solution if it were not a multiple choice question.

Oh right.. Only B would get a total mol of 3, if x=1. Thanks for the help :)

 

[Borek]

You are still ignoring the fact ICE table is built around the reaction equation. You can't say

the change in moles of P and Q would be 2x each

as if it was an universal rule. Is your statement correct if the reaction equation is

7P ⇔ 3Q + 9R

or

P ⇔ Q + R

or

P ⇔ Q

where R doesn't even exist?

You are given 4 different reaction equations, you have to write ICE table for each of them separately, express - for each reaction equation - final total amount of moles using equilibrium values (yes, it will contain some unknown x, but that's what is expected) and compare it to the information given in the problem.

I guess all the time you are writing ICE table for the first reaction - if so, please state so, otherwise it is not clear what you are doing.

 

[Janiceleong26]

You are still ignoring the fact ICE table is built around the reaction equation. You can't say
as if it was an universal rule. Is your statement correct if the reaction equation is

7P ⇔ 3Q + 9R

or

P ⇔ Q + R

or

P ⇔ Q

where R doesn't even exist?

You are given 4 different reaction equations, you have to write ICE table for each of them separately, express - for each reaction equation - final total amount of moles using equilibrium values (yes, it will contain some unknown x, but that's what is expected) and compare it to the information given in the problem.

I guess all the time you are writing ICE table for the first reaction - if so, please state so, otherwise it is not clear what you are doing.

Oh I see, I got it now, sorry for the trouble. Thanks !

 

[epenguin]

Janice, I asked in #8 where this question came from for a reason.Surely this is a bad question and it is unreasonable for you to have to guess the question that should have been asked.

Suppose the equilibrium is very unfavourable to formation of Q and R. You can have a little R and 2 moles of the rest combined for all equilibria represented by A, B, C, D. Extreme example to help think, but the equilibrium doesn't have to be extreme.

 

[Ygggdrasil]

Suppose the equilibrium is very unfavourable to formation of Q and R. You can have a little R and 2 moles of the rest combined for all equilibria represented by A, B, C, D. Extreme example to help think, but the equilibrium doesn't have to be extreme.

Unless x = 0 (implying an equilibrium constant of zero and an infinite free energy difference between products and reactants), the total number of moles will equal exactly 2+x only for one of the choices. For example, the total number of moles for reaction A is 2+2x. 2+2x ≠ 2+x for non-zero x even if x is very small.

 

[Borek]

I am with Ygg here, I think this is a perfectly nice question with a well defined answer.

 

[epenguin]

Oops silly me, I take it all back. For some reason I had been missing the first words of the question, that there were two moles to start with. Quite easy then.