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:
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.
I'm not sure because I've never used a RICE table before, but here's what i did:
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.
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?