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Equilibrium / Kc help please

2CO (g) + 2NO (g) ⇌ 2CO2 (g) + N2 (g) (delta)H = -788 kJMol-1

The Chemist mixes 0.46mol of CO with 0.45 mol of NO. The mixture is left to reach equilibrium at constant temperature.

The Student analyses the equilibrium mixture and finds that 0.25 mol NO remains. The total volume of the equilibrium mixture is 1.0dm^3.
Determine the value of Kc for this equilibrium mixture.
Show working.

So far for this I set out the table and now I require assistance,

Component....... CO.......NO........CO2.......N2
moles .............. 0.46......0.45.........0............0
Equilibrium.......0.26.......0.25.........?............?
Reacted...........0.2..........0.2..........?.............?

Not sure how to do the next steps but thanks!
Original post by harvingill
2CO (g) + 2NO (g) ⇌ 2CO2 (g) + N2 (g) (delta)H = -788 kJMol-1

The Chemist mixes 0.46mol of CO with 0.45 mol of NO. The mixture is left to reach equilibrium at constant temperature.

The Student analyses the equilibrium mixture and finds that 0.25 mol NO remains. The total volume of the equilibrium mixture is 1.0dm^3.
Determine the value of Kc for this equilibrium mixture.
Show working.

So far for this I set out the table and now I require assistance,

Component....... CO.......NO........CO2.......N2
moles .............. 0.46......0.45.........0............0
Equilibrium.......0.26.......0.25.........?............?
Reacted...........0.2..........0.2..........?.............?

Not sure how to do the next steps but thanks!

Same molar ratio between NO and CO2 so 0.2 moles of CO2 formed
Half molar ratio of N2 so 0.1 moles of N2 formed
Same amount of N2 (0.1) and CO2 (0.2) at equilibrium as are formed
Put the equilibrium concentrations (as its 1 mol dm3-) into the equilibrium constant expression, which is [CO2]squared x [N2] / [CO]squared x [NO] squared and you should get an answer?
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harvingill
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Original post by plasmaman
Same molar ratio between NO and CO2 so 0.2 moles of CO2 formed
Half molar ratio of N2 so 0.1 moles of N2 formed
Same amount of N2 (0.1) and CO2 (0.2) at equilibrium as are formed
Put the equilibrium concentrations (as its 1 mol dm3-) into the equilibrium constant expression, which is [CO2]squared x [N2] / [CO]squared x [NO] squared and you should get an answer?



when you meant 0.2 moles did you mean by mol reacted? I assume so, but thanks!
Original post by harvingill
when you meant 0.2 moles did you mean by mol reacted? I assume so, but thanks!


You could say reacted but technically because its a product I look at it as how much is formed. Did you put the numbers in and get the right answer? That's Ok, not sure if my answer was correct yet! :smile:
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Original post by plasmaman
You could say reacted but technically because its a product I look at it as how much is formed. Did you put the numbers in and get the right answer? That's Ok, not sure if my answer was correct yet! :smile:



Can I ask for help on a similar Question however it has two products so I need i'm a bit confused.
------------------------------------------------------------------------------------
CH3COOH + C2H5OH <---> CH3COOC2H5 + H20

2.00 moles of CH3COOH were mixed with 3.00moles of C2H50H and the mixture allowed to reach equilibrium when 0.43 moles of CH3COOH remained.

determine the concentration of everything and calculate Kc.

So far, I've:

component.......CH3COOH.......C2H50H.......CH3COOC2H5.......H20
mole.................2.......................3...................0..............................0
equilibrium......0.43...................1.57.............?...............................?
reacted............1.57...................1.43..............?...............................?

I'm not sure how to apply what you helped me with onto this Q, thanks beforehand!
Original post by harvingill
Can I ask for help on a similar Question however it has two products so I need i'm a bit confused.
------------------------------------------------------------------------------------
CH3COOH + C2H5OH <---> CH3COOC2H5 + H20

2.00 moles of CH3COOH were mixed with 3.00moles of C2H50H and the mixture allowed to reach equilibrium when 0.43 moles of CH3COOH remained.

determine the concentration of everything and calculate Kc.

So far, I've:

component.......CH3COOH.......C2H50H.......CH3COOC2H5.......H20
mole.................2.......................3...................0..............................0
equilibrium......0.43...................1.57.............?...............................?
reacted............1.57...................1.43..............?...............................?

I'm not sure how to apply what you helped me with onto this Q, thanks beforehand!

Did my explanation for the first question get the right answer?
Isnt the amount of ethanol that reacts the same as the amount of methanoic acid? Sorry, I actually cant answer this one. Hopefully somebody more qualified comes along to help us out. Have you tried typing the question word for word into google, you may find a mark scheme?
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harvingill
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Original post by plasmaman
Did my explanation for the first question get the right answer?
Isnt the amount of ethanol that reacts the same as the amount of methanoic acid? Sorry, I actually cant answer this one. Hopefully somebody more qualified comes along to help us out. Have you tried typing the question word for word into google, you may find a mark scheme?


You're answer to the first was correct, and the second question I have the answer but I don't understand how and why they values for products are what they actually are?!
Original post by harvingill
You're answer to the first was correct, and the second question I have the answer but I don't understand how and why they values for products are what they actually are?!

If you give me the full answer maybe I can work it out?
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harvingill
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Original post by plasmaman
If you give me the full answer maybe I can work it out?


CH3COOH =0.43
C2H50H = 1.43
CH3COOC2H5 = 1.57
H20 = 1.57

Kc = 4.01
Original post by harvingill
CH3COOH =0.43
C2H50H = 1.43
CH3COOC2H5 = 1.57
H20 = 1.57

Kc = 4.01


Ok, so C2H5OH is in excess because it has 3 moles compared to CH3COOH having 2 moles when the molar relationship is saying that they have the same amount of moles.
We then exclude C2H5OH for the minute as it is in excess
So we use the amount of moles of CH3COOH to work out how much of the products have been formed (or reacted in the table), same molar ratio as CH3COOH so the same amount, both 1.57, which is by default the same number of moles as there are at equilibrium

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