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Calculating the number of moles in an equilibrium mixture

Hi! :smile:

I've been doing some AQA Chem 4 papers over the last few weeks (as the exam is next week! :eek:), and there's one thing that comes up on most of them that I get completely wrong every time!

For example, the question says:

CH4(g) + 2H2O(g) <--> CO2(g) + 4H2(g)

Initially, 1.0 mol of methane and 2.0 mol of steam were placed in a flask and heated with a catalyst until equilibrium was established. The equilibrium mixture contained 0.25 mol of carbon dioxide.

a)i) Calculate the amounts, in moles, of methane, steam and hydrogen in the equilibrium mixture.

So, I was wondering how on earth you do this?! The mark scheme just gives the answer and no working, so would someone mind explaining to me how you would do/go about doing this question please?

I know it's normally only 1-2 marks, but it's annoying me as I cannot see any logical way to calculate this (other than multiplying the 0.25 by the ratio of moles, which is wrong!)!

Thanks :smile:

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I always start with writing down what I'm told first. So at the very start you have:

CH4 : 1.0 mol
H2O : 2.0 mol
CO2 : 0.0 mol
H2 : 0.0 mol

then you're told at equilibrium:

CH4 :
H2O :
CO2 : 0.25 mol
H2 :

We need to use the chemical equation to fill in the blanks above. SInce there was no CO2 at the start, it has been generated in the reaction of methane + water. You can see that the ratio of CH4 to CO2 is 1:1, therefore 0.25 mol of CH4 has reacted leaving 1 - 0.25 = 0.75 so now we know:

CH4 : 0.75 mol
H2O :
CO2 : 0.25 mol
H2 :

can you fill in the rest? :smile:
Reply 2
Avatar for maxjones2000
maxjones2000
OP
3
Original post by EierVonSatan
I always start with writing down what I'm told first. So at the very start you have:

CH4 : 1.0 mol
H2O : 2.0 mol
CO2 : 0.0 mol
H2 : 0.0 mol

then you're told at equilibrium:

CH4 :
H2O :
CO2 : 0.25 mol
H2 :

We need to use the chemical equation to fill in the blanks above. SInce there was no CO2 at the start, it has been generated in the reaction of methane + water. You can see that the ratio of CH4 to CO2 is 1:1, therefore 0.25 mol of CH4 has reacted leaving 1 - 0.25 = 0.75 so now we know:

CH4 : 0.75 mol
H2O :
CO2 : 0.25 mol
H2 :

can you fill in the rest? :smile:


Thanks very much for that! :smile:

I'm still a bit confused - I think I'm looking too deep into it to be honest!

This sounds really stupid, but how do you know to take it away from 1 (the initial moles of CH4), not 2 (the initial moles of H2O), as that 0.25 moles has been formed from both CH4 and H2O if you get what I mean?

Stupid q#2 - how would I go about doing it for H2O and H2, as there's more than one mole of each...? Sorry! :smile:
Original post by maxjones2000
This sounds really stupid, but how do you know to take it away from 1 (the initial moles of CH4), not 2 (the initial moles of H2O), as that 0.25 moles has been formed from both CH4 and H2O if you get what I mean?

Stupid q#2 - how would I go about doing it for H2O and H2, as there's more than one mole of each...? Sorry! :smile:


No need to apologise and they're not stupid questions :p:

An important thing to know is that the number of moles can change on either side of a reaction arrow, but the mass cannot.

To make one molecule (or mole, if you like) of CO2 it needs one CH4 and two H2O's. So regardless of what else is made if there are 15 molecules of CH4 and there were none before, 15 molecules of CO2 have been converted as well as 30 molecules of H2O (i.e. twice as much, according ot the ratio in the chemical equation).

So, going back to our table:

CH4 : 0.75 mol
H2O :
CO2 : 0.25 mol
H2 :

work out how much of the water has been used up and therefore what is left over at equilibrium? :smile:
Reply 4
Avatar for maxjones2000
maxjones2000
OP
3
aha that makes sense - thanks!

I've got there being 1.5 moles of H2O at equilibrium by doing 2-2(0.25)=1.5. Is that right?

I'm still stuck with the hydrogen though, as what do I do about the 4?
Would it be 4-4(0.75)=1?

Thanks so much for all your help! :smile:

Give me a pure-evil calculus question and I'm fine, but give me a basic 2-mark chemistry one.... :tongue:
(edited 11 years ago)
Original post by maxjones2000
aha that makes sense - thanks!

I've got there being 1.5 moles of H2O at equilibrium by doing 2-2(0.25)=1.5. Is that right?


Yeah :awesome:

I'm still stuck with the hydrogen though, as what do I do about the 4?
Would it be 4-4(0.75)=1?


You start with no hydrogen, you're not taking away :smile:

Thanks so much for all your help! :smile:

Give me a pure-evil calculus question and I'm fine, but give me a basic 2-mark chemistry one.... :tongue:


No worries and hah, plenty of calculus in chemistry if you;re looking for it :teeth:
Reply 6
Avatar for maxjones2000
maxjones2000
OP
3
woo :cool:

I think I may have got it now:
To work out the moles of hydrogen, would you do (1-0.75)+(2-1.5)-0.25=0.5?
As the rest of the reactants have to have been turned into hydrogen? Or am I being completely thick?! :tongue:

Original post by EierVonSatan
Yeah :awesome:
No worries and hah, plenty of calculus in chemistry if you;re looking for it :teeth:


Yay, the worlds a better place with calculus.. :biggrin: Just out of interest, where does it come into it? :smile:
Original post by maxjones2000
woo :cool:

I think I may have got it now:
To work out the moles of hydrogen, would you do (1-0.75)+(2-1.5)-0.25=0.5?
As the rest of the reactants have to have been turned into hydrogen? Or am I being completely thick?! :tongue:


You're over thinking it :yes:

If you have 1 molecule of CO2, you have four times as much H2.

Yay, the worlds a better place with calculus.. :biggrin: Just out of interest, where does it come into it? :smile:


University level stuff :smile:
Reply 8
Avatar for maxjones2000
maxjones2000
OP
3
Original post by EierVonSatan
You're over thinking it :yes:

If you have 1 molecule of CO2, you have four times as much H2.



Arghh! :tongue: so would it just be 4x0.25=1?

I really am missing the total obvious here, aren't I?!



University level stuff :smile:

Oh right, I'll pass on that one then... :tongue:

Sorry to be so 'thick'! :smile:
Original post by maxjones2000
Arghh! :tongue: so would it just be 4x0.25=1?

I really am missing the total obvious here, aren't I?!


You got it :yes:

Oh right, I'll pass on that one then... :tongue:

Sorry to be so 'thick'! :smile:


It's all just practice, you'll get it :yep:
Reply 10
Avatar for maxjones2000
maxjones2000
OP
3
Original post by EierVonSatan
You got it :yes:



It's all just practice, you'll get it :yep:


Oh! :eek: It really is that simple then?!! :biggrin:

Thanks very much for all your help - I really appreciate it :smile:
Reply 11
Avatar for mar junior
mar junior
Original post by EierVonSatan
I always start with writing down what I'm told first. So at the very start you have:

CH4 : 1.0 mol
H2O : 2.0 mol
CO2 : 0.0 mol
H2 : 0.0 mol

then you're told at equilibrium:

CH4 :
H2O :
CO2 : 0.25 mol
H2 :

We need to use the chemical equation to fill in the blanks above. SInce there was no CO2 at the start, it has been generated in the reaction of methane + water. You can see that the ratio of CH4 to CO2 is 1:1, therefore 0.25 mol of CH4 has reacted leaving 1 - 0.25 = 0.75 so now we know:

CH4 : 0.75 mol
H2O :
CO2 : 0.25 mol
H2 :

can you fill in the rest? :smile:

i have a few questions that i really want you to answer

1: What do you mean by 'the chemical equation" because i have the same problem

2:Please, i really need to kinow, explain to me how you came to the conclusion that CH4 moles was 0.75 at equilibrium, in addition how do you know that the amount of CO2 Made was completely dependant on the ch4 concentration and not both the h20 and co2 concentrations, i dont understand how it is simply a case of minusing one from the other. if it was 1 product to 2 reactants i would understand

3: even if this was right, why wouldnt the concentration of h2 be 1 mole if the co2 was 0.25, as 0.25 x 4 is 1

Thank you so much.
(edited 11 years ago)
Original post by mar junior
i have a few questions that i really want you to answer

1: What do you mean by 'the chemical equation" because i have the same problem


CH4(g) + 2H2O(g) <--> CO2(g) + 4H2(g)

2:Please, i really need to kinow, explain to me how you came to the conclusion that CH4 moles was 0.75 at equilibrium, in addition how do you know that the amount of CO2 Made was completely dependant on the ch4 concentration and not both the h20 and co2 concentrations, i dont understand how it is simply a case of minusing one from the other. if it was 1 product to 2 reactants i would understand


Firstly, there are no concentrations in this problem, only amounts in moles :yes: This is an important distinction.

Okay, so we know that there is no CO2 at the start of the reaction. We are also told that there is 0.25 mol of CO2 at equilibrium.

This means that it has to have been generated by the reaction of methane with hydrogen - happy with this?

The chemical equation (above) shows us that when 1 mole of CH4 reacts with 2 moles of H2O it produces 1 mole of CO2 and 4 moles of H2. This is the stoichiometric ratio.

As there is 0.25 moles of CO2 this means that 0.25 moles of CH4 must have reacted. Make sense?

So, if you start with 1 mole and use up 0.25 moles, you're left with 0.75 moles :smile:

The amount of CO2 (a product) is dependant on how much CH4 and H2O there is (the reactants). You can't make something from nothing, after all. Again, this is not related to concentrations - just amount.

3: even if this was right, why wouldnt the concentration of h2 be 1 mole if the co2 was 0.25, as 0.25 x 4 is 1

Thank you so much.


Did you read the thread? :p: It is 1 mole :yes:
(edited 11 years ago)
Reply 13
Avatar for mar junior
mar junior
Thank you SO MUCH, for taking your time to explain what went on, Your help and patience with a guy like me is invaluable here.

is it okay if i pm you for help with other examples, because i sometimes struggle with what to do when the molar ratios are scrambled.

I need your knowledge one more time,

so if you have REVERSIBLE the equation:

A + 3B <-----> 2C + 4D (1.2 moles of A and 0.991 moles of B are in the mix initially and make 0.4 moles of C are in equilibrium mixture)

What will the number of moles of all chemicals be at equilibrium and tell me why? im going to use your reasons to make my notes

Thanks in advance.
(edited 11 years ago)
Original post by mar junior
Thank you SO MUCH, for taking your time to explain what went on, Your help and patience with a guy like me is invaluable here.

is it okay if i pm you for help with other examples, because i sometimes struggle with what to do when the molar ratios are scrambled.


The forum is best as lots of people can see it, instead of just muggins here :p:

I need your knowledge one more time,

so if you have REVERSIBLE the equation:

A + 3B <-----> 2C + 4D (1.2 moles of A and 0.991 moles of B are in the mix initially and make 0.4 moles of C are in equilibrium mixture)

What will the number of moles of all chemicals be at equilibrium and tell me why? im going to use your reasons to make my notes

Thanks in advance.


Like above determine what you have at the start and at equilibrium which is already given to you. Initially:

A = 1.2
B = 0.991
C = 0
D = 0

Equilibrium:

A =
B =
C = 0.4
D =

As before since there is no C or D present at the start the forward reaction is responsible for all of amount of C and D at eqilibrium...so if you have 0.4 moles of C then there must be twice as much of D. This is because there are 4D for every 2C in the chemical equation :smile:

Have a go at determining A and B at equilibrium :yep:
Reply 15
Avatar for mar junior
mar junior
Original post by EierVonSatan
The forum is best as lots of people can see it, instead of just muggins here :p:


Equilibrium:

A =
B =
C = 0.4
D =

As before since there is no C or D present at the start the forward reaction is responsible for all of amount of C and D at eqilibrium...so if you have 0.4 moles of C then there must be twice as much of D. This is because there are 4D for every 2C in the chemical equation :smile:

Have a go at determining A and B at equilibrium :yep:


Correct me if im wrong but i still cant make this out, i purposely chose violent
looking equations

D = 0.8, Because the ratio of that to C is 1:2, 0.4 moles of C must mean 0.8 moles of D

I really for the life of me cannot find any way of deducing the amounts of A and B, because i cant see any relationship or multipliers to use to get their amount, Even if 0.4 and 0.8 moles are made, i cant see how that relates to the original concentration

to try and work it out I tried making 0.4 2x (As theres 2 moles of C) and 0.8 4x, and so X = 0.2, and subtracting numbers as the products had to come from somewhere, Am i overthinking this and is there a simpler method

A = 0.991 - 3X (Because its 3 moles) = 0.391
B = 1.2 - X (because its 1 mole) = 1mole

Also is the number of moles or the concentration (if liquid) or volume (if gas) used when using the equilibrium constant equation?
If im not right please explain why not because this topic is proving very tricky so far, thanks so much.
(edited 11 years ago)
Original post by mar junior
Correct me if im wrong but i still cant make this out, i purposely chose violent
looking equations

D = 0.8, Because the ratio of that to C is 1:2, 0.4 moles of C must mean 0.8 moles of D


Indeed :yy:

to try and work it out I tried making 0.4 2x (As theres 2 moles of C) and 0.8 4x, and so X = 0.2, and subtracting numbers as the products had to come from somewhere, Am i overthinking this and is there a simpler method


Yeah, so if 2C = 0.4 then C = 0.2 and A = 0.2, B = 0.6

The A and B quantities are what have been used up and need to be subtracted from the initial amounts...

A = 0.991 - 3X (Because its 3 moles) = 0.391
B = 1.2 - X (because its 1 mole) = 1mole


You've got the A and B mixed up:

A = 1.2 -0.2 = 1.0 at equilibrium
B = 0.991 - 0.6 = 0.391 at equilibrium

Also is the number of moles or the concentration (if liquid) or volume (if gas) used when using the equilibrium constant equation?


Depends on the exact nature of the expression :yes:
(edited 11 years ago)
Reply 17
Avatar for mar junior
mar junior
phew, music to my ears i thought i had got everything wrong im happy i was on the right track, i really wish i could give you something valuable for your help and patience with me, Were you previously a teacher? But before you go one more question,

You said i had A and B mixed up, care to explain how you worked that out and why that was the case?? i need to know so that i can adapt it for other mole questions.
(edited 11 years ago)
Original post by mar junior
phew, music to my ears i thought i had got everything wrong im happy i was on the right track, i really wish i could give you something valuable for your help and patience with me, Were you previously a teacher? But before you go one more question


Not a school teacher :no: :p:

You said i had A and B mixed up, care to explain how you worked that out and why that was the case?? i need to know so that i can adapt it for other mole questions.


A(equilibrium) = A(Initial) - A(reacted)

so A(initial) = 1.2 mol and A(reacted) = 0.2 mol

You just had A and B mixed up so you were doing:

A(equilibrium) = B(Initial) - B(reacted)
Reply 19
Avatar for DomJay24
DomJay24
HAS ANYBODY GOT A YOUTUBE LINK TO A VIDEO EXPLAINING HOW TO DO A QUESTION LIKE THIS??!!?!

I'm sorry but reading the solution just makes it even more complicated

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