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Kp equilibrium constant q

Hi!
I was hoping someone could just check my working for this question! Any help if it's incorrect is much appreciated
Capture.PNG

Here's my working:
Pressure of NO2 = 2 - 1.1 = 0.9atm
Kp = 0.9^2 / 2 = 0.405atm

Thanks for your help in advance :smile:
Original post by kswales1
Hi!
I was hoping someone could just check my working for this question! Any help if it's incorrect is much appreciated
Capture.PNG

Here's my working:
Pressure of NO2 = 2 - 1.1 = 0.9atm
Kp = 0.9^2 / 2 = 0.405atm

Thanks for your help in advance :smile:

Try drawing a rice table. What do you think the pressure of N2O4 at equilibrium will be?
Reply 2
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kswales1
OP
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Original post by Hellllpppp
Try drawing a rice table. What do you think the pressure of N2O4 at equilibrium will be?

I'm not sure because I've never used a RICE table before, but here's what i did:
Capture.PNG
Since we know the pressure of NO2 dropped by 1.1atm, that means it's partial pressure is 2 - 1.1 = 0.9. Using the RICE table, this means that 2x = 0.9, so x=0.45. The pressure of N2O4 at equib. is 2-x, so 2 - 0.45 = 1.55? Once again i've never used a rice table before so im not even sure if i've used it correctly.
Original post by kswales1
I'm not sure because I've never used a RICE table before, but here's what i did:
Capture.PNG
Since we know the pressure of NO2 dropped by 1.1atm, that means it's partial pressure is 2 - 1.1 = 0.9. Using the RICE table, this means that 2x = 0.9, so x=0.45. The pressure of N2O4 at equib. is 2-x, so 2 - 0.45 = 1.55? Once again i've never used a rice table before so im not even sure if i've used it correctly.

You’ve got the right idea. C stands for change and its -1.1 for NO2. You’ve done the right thing dividing C by 2 and switching the sign for N2O4. What you’ll need to do is 2 + 0.55. Then you put this into the Kc equation.
(edited 2 years ago)
Reply 4
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kswales1
OP
8
Original post by Hellllpppp
You’ve got the right idea. C stands for change and its -1.1 for NO2. You’ve done the right thing dividing C by 2 and switching the sign for N2O4. What you’ll need to do is 2 + 0.55. Then you put this into the Kc equation.

I'm sorry for replying so late!! I'm confused though, where did the 0.55 come from?
Original post by kswales1
I'm sorry for replying so late!! I'm confused though, where did the 0.55 come from?

-1.1 divided by 2 then flip the sign from negative to positive because it’s on the other side of the reaction
Reply 6
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kswales1
OP
8
Original post by Hellllpppp
-1.1 divided by 2 then flip the sign from negative to positive because it’s on the other side of the reaction

Ohh ok. So does that mean the partial pressure of the N2O4 is 2 - 0.55?
Original post by kswales1
Ohh ok. So does that mean the partial pressure of the N2O4 is 2 - 0.55?

It’s 2 + 0.55 because the partial pressure of NO2 falls so the pressure of N2O4 must rise

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