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# A-level Chemistry Revision Squad! watch

A mixture of hydrogen of concentration 10moldm^-3 and CO2 of concentration 90moldm^-3 was heated at 1200K so that steam and CO was produced. At equilibrium only 80.53moldm^-3 od CO2 remained in the mixture. Calculate Kc for the reaction.

So far I have the expression for it: Kc = [CO][H2O] / [H2][CO2]
I'm not sure if I did this right but I did: 90 - 80.53 to give me 9.47 and I think this means that only 9.47 of CO2 was reacted with H2. Not sure if thats correct?

I have no clue what to do after this? I know that I have to find the concetration for bot hof the products formed but how.. ://

Help appreciated
2. Why doesn't Kc change if you change any concentrations or pressure at equilibrium? It only changes with temperature change.
surely concentrations of products and reactants change so Kc would change?
3. (Original post by olivia7001)
Why doesn't Kc change if you change any concentrations or pressure at equilibrium? It only changes with temperature change.
surely concentrations of products and reactants change so Kc would change?
If we change the concentration or pressure, Kc/Kp initially change because according to Chatelier principle, the P.O.E will shift to the side that oppose the change to restore equilibrium, so when we shift to a side, the new Kc/Kp value is smaller or bigger depending to which side was shifted then when equilibrium is restored, the value of Kc/Kp is restored

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4. (Original post by olivia7001)
Why doesn't Kc change if you change any concentrations or pressure at equilibrium? It only changes with temperature change.
surely concentrations of products and reactants change so Kc would change?

Kc is a constant, right? And the general equation is:
$K_c=\frac{[a]^A[b]^B}{[c]^C[d]^D} \\for: \\Aa+Bb\rightleftharpoons Cc+Dd$
Let's say that the conc. of c increases. Because the left hand side of the equation must be constant (kc has to be kept the same), the top of the fraction must change in order to account for the change in conc of c. Therefore, even though the conc. of c has changed, the conc. of the other reactants will change, rather than kc changing.

Pressure is just the conc. in a given volume. so it works by the same principle stated above.

Have a look at the mindmap for more kinetics stuff. Hope that helps.
Attached Images
5. Untitled page.pdf (974.3 KB, 64 views)
6. (Original post by PlayerBB)
If we change the concentration or pressure, Kc/Kp initially change because according to Chatelier principle, the P.O.E will shift to the side that oppose the change to restore equilibrium, so when we shift to a side, the new Kc/Kp value is smaller or bigger depending to which side was shifted then when equilibrium is restored, the value of Kc/Kp is restored

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What is Kp ?
7. (Original post by FemaleBo55)
A mixture of hydrogen of concentration 10moldm^-3 and CO2 of concentration 90moldm^-3 was heated at 1200K so that steam and CO was produced. At equilibrium only 80.53moldm^-3 od CO2 remained in the mixture. Calculate Kc for the reaction.

So far I have the expression for it: Kc = [CO][H2O] / [H2][CO2]
I'm not sure if I did this right but I did: 90 - 80.53 to give me 9.47 and I think this means that only 9.47 of CO2 was reacted with H2. Not sure if thats correct?

I have no clue what to do after this? I know that I have to find the concetration for bot hof the products formed but how.. ://

Help appreciated
Hi- I've given it a shot apologies if there are any mistakes...
Here goes:

So yes, I did the same thing I did 90-80.53= 9.47 then (using the RICE) method I did the same with (H2). So I did 10-9.47= 0.43moldm-3.
Then on the other side of the equation we are not told the concentration of CO or H20 so I assumed it was 0 for both. So the conc of both CO and H20 would be 9.47. (0+9.47=9.47)
Then using the Kc equation I multiplied (9.47 x 9.47)/ (0.53 x 80.53)= 2.1

Kc = 2.1 (i think)
8. (Original post by P____P)

Kc is a constant, right? And the general equation is:
$K_c=\frac{[a]^A[b]^B}{[c]^C[d]^D} \\for: \\Aa+Bb\rightleftharpoons Cc+Dd$
Let's say that the conc. of c increases. Because the left hand side of the equation must be constant (kc has to be kept the same), the top of the fraction must change in order to account for the change in conc of c. Therefore, even though the conc. of c has changed, the conc. of the other reactants will change, rather than kc changing.

Pressure is just the conc. in a given volume. so it works by the same principle stated above.

Have a look at the mindmap for more kinetics stuff. Hope that helps.
So is it that proportionally everything changes by the same amount, so the factor that everything changed by ends up cancelling out in the fraction?
9. (Original post by haj101)
Hi- I've given it a shot apologies if there are any mistakes...
Here goes:

So yes, I did the same thing I did 90-80.53= 9.47 then (using the RICE) method I did the same with (H2). So I did 10-9.47= 0.43moldm-3.
Then on the other side of the equation we are not told the concentration of CO or H20 so I assumed it was 0 for both. So the conc of both CO and H20 would be 9.47. (0+9.47=9.47)
Then using the Kc equation I multiplied (9.47 x 9.47)/ (0.53 x 80.53)= 2.1

Kc = 2.1 (i think)
Thanks I get that now! But is the RICE method? I've never heard of it LOL -.-
10. (Original post by FemaleBo55)
Thanks I get that now! But is the RICE method? I've never heard of it LOL -.-
It's basically:

Ratio
Initial moles
Change
Equilibrium moles

(It allows me to find the equilibrium moles a lot quicker and easier!)
11. (Original post by P____P)
What is Kp ?
Pressure equilibrium constant
12. (Original post by thatcooldude2.0)
What uni are you at? How are you finding the course in general and what is the most interesting topic you've done so far?
I'm at uea doing their mchem degree. Honestly, its great, the lecturers are all really nice, they clearly know their stuff and put the students first. The first year really is just to get everyone on the same level, which means the content doesnt come as heavy as you might think, they are also a bit more lenient about the quality of lab reports and things in the first year. Come second year they expect you to pull your socks up and start throwing a hell of a lot of knowledge your way, especially in physical chemistry, but that just adds to the excitement I feel. My favourite topic so far has been frustrated lewis pairs, the general concept of which is you take a strong lewis acid (eg.a borane) And a strong lewis base (an amine) and stearically hinder them so they cant react. You can then pass small molecules such as hydrogen gas through the mixture, and the force between the lewis pair rips the H2 molecule apart, allowing the extraction of the electrons into a battery of some form.
I hope this helped, sorry about grammar, im typing this on a tablet
13. (Original post by Spudgunhimself)
I'm at uea doing their mchem degree. Honestly, its great, the lecturers are all really nice, they clearly know their stuff and put the students first. The first year really is just to get everyone on the same level, which means the content doesnt come as heavy as you might think, they are also a bit more lenient about the quality of lab reports and things in the first year. Come second year they expect you to pull your socks up and start throwing a hell of a lot of knowledge your way, especially in physical chemistry, but that just adds to the excitement I feel. My favourite topic so far has been frustrated lewis pairs, the general concept of which is you take a strong lewis acid (eg.a borane) And a strong lewis base (an amine) and stearically hinder them so they cant react. You can then pass small molecules such as hydrogen gas through the mixture, and the force between the lewis pair rips the H2 molecule apart, allowing the extraction of the electrons into a battery of some form.
I hope this helped, sorry about grammar, im typing this on a tablet
Thanks for the reply, that sounds really interesting! Do you like the uni itself? Also what are you thinking of doing when you finish your degree?
14. (Original post by olivia7001)
So is it that proportionally everything changes by the same amount, so the factor that everything changed by ends up cancelling out in the fraction?
Yup, that's one (mathematical) way of looking at it. Also remember that chemically 'the equilibrium position shifts to opposite side, to oppose the change in pressure/volume or conc.'

That phrase works basically every time in AS and A2 on AQA spec.
15. (Original post by haj101)
It's basically:

Ratio
Initial moles
Change
Equilibrium moles

(It allows me to find the equilibrium moles a lot quicker and easier!)
Okay so for my question lets see if I understand it using your method lol:

Ratio for reactants is 1:1 and for the products is also 1:1
Initial moles: they haven't given me any moles so I'm just gunna use concentration, is that okay? so initial conc of: H2 = 10moldm-3, CO2 = 90moldm-3 and for the products is 0 because the reaction hasnt started yet?
Change: from the question we know that 80.53 of CO2 remained so from 90 only 9.47 of reacted with H2. As the ratio is one to one same the amount of H2 must have been reacted too hence 10 - 9.47 = 0.53?
WIth regards to the prodcuts as both are 1:1 ration with respect to product:reactants the product must have increased by 9.47 if the reactants decreased by 9.47? Therefore both H2O and CO are 9.47moldm-3
Equilibrium mole: not sure what you write here? LOL

Then for Kc you just plug them in the formula?
16. (Original post by P____P)

Kc is a constant, right? And the general equation is:
$K_c=\frac{[a]^A[b]^B}{[c]^C[d]^D} \\for: \\Aa+Bb\rightleftharpoons Cc+Dd$
Let's say that the conc. of c increases. Because the left hand side of the equation must be constant (kc has to be kept the same), the top of the fraction must change in order to account for the change in conc of c. Therefore, even though the conc. of c has changed, the conc. of the other reactants will change, rather than kc changing.

Pressure is just the conc. in a given volume. so it works by the same principle stated above.

Have a look at the mindmap for more kinetics stuff. Hope that helps.
Lol I really like your mind maps, what other ones have you made for chemistry?

Is (C2H5)2C(OH)CH3 2-ethyl-butan-2-ol
18. (Original post by FemaleBo55)
Okay so for my question lets see if I understand it using your method lol:

Ratio for reactants is 1:1 and for the products is also 1:1
Initial moles: they haven't given me any moles so I'm just gunna use concentration, is that okay? so initial conc of: H2 = 10moldm-3, CO2 = 90moldm-3 and for the products is 0 because the reaction hasnt started yet?
Change: from the question we know that 80.53 of CO2 remained so from 90 only 9.47 of reacted with H2. As the ratio is one to one same the amount of H2 must have been reacted too hence 10 - 9.47 = 0.53?
WIth regards to the prodcuts as both are 1:1 ration with respect to product:reactants the product must have increased by 9.47 if the reactants decreased by 9.47? Therefore both H2O and CO are 9.47moldm-3
Equilibrium mole: not sure what you write here? LOL

Then for Kc you just plug them in the formula?
Yes!! We can apply it to conc questions... For the equilibrium mole we put the calculated conc (from the change) and the equlibrium conc/moles is used in the equation.
Aww Im happy you understand!
19. Anyone doing edexcel?

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20. (Original post by kiiten)

Is (C2H5)2C(OH)CH3 2-ethyl-butan-2-ol
Yes you are correct.
21. (Original post by jyyl)
Yes you are correct.
Thanks - I wasn't sure if it was ethyl or methyl

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