Lena772
  • Lena772
A 2.0 L vessel contains HI and H2. The following equilibrium is established: H2 (g) + I2 (g) ⇌ 2 HI (g) At equilibrium [H2] = 2.0 M [I2] = 1.5 M [HI] = 2.5 M The initial concentration of I2 was not mentioned so you must assume [I2]o = 0. Calculate the initial moles of: HI (Tol: ± 0.1) H2 (g) (Tol: ± 0.1)
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Lena772
  • Lena772
@Abhisar @Photon336
Photon336
  • Photon336
You could work backwards
Photon336
  • Photon336
wait those +/- thats the change in concentration right? @Lena772

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Lena772
  • Lena772
|dw:1442112283855:dw|
Lena772
  • Lena772
Thats just showing the Tolerance for the answer will only be 0.1 above or below @Photon336
Lena772
  • Lena772
I don't know how to solve for x working backwards @Photon336
Lena772
  • Lena772
@abhisar
Photon336
  • Photon336
So this was all the information they gave you?
Lena772
  • Lena772
yep
Lena772
  • Lena772
I know M= moles solute/L of soln but i dont know if i could manipulate that equation with what I'm given here.
Photon336
  • Photon336
@Woodward thoughts? i'm not sure
anonymous
  • anonymous
Yeah I am just working through this myself, I haven't done these in a while but I think I can help you. For instance, your ICE table doesn't seem quite right, you're starting with 0 moles of \(I_2\) and then below you are losing x amount to get 1.5 moles! |dw:1442114000906:dw| This doesn't make sense, the equilibrium will shift the other way!
anonymous
  • anonymous
Whoops I said moles but this is concentration, let me straighten myself out before I say anymore lol
anonymous
  • anonymous
Ok, so here's how I solved it, write out the ICE table but we don't need to use the equilibrium constant expression! We just simply label our unknowns in Molarity: |dw:1442114537041:dw| So we get three equations: \[[H_2]_0 + x = 2\] \[0+x = 1.5 \] \[[HI]_0 -2x = 2.5\] That middle one tells us that x=1.5, so the rest will just be algebra! Once you find these concentrations, don't forget to multiply by 2.0L to get the number of moles. I think that does it!
Photon336
  • Photon336
@Woodward interesting; well if the change in x must arguably be the same for both hydrogen gas and Iodine, does this mean that we can easily find out what the initial concentration of HI is. For HI it's just 5.5 M for Iodine initial concentration is just 0 And for H2 it's 0.5 M
anonymous
  • anonymous
Yeah that's exactly what I got. :D
Photon336
  • Photon336
@Woodward why didn't we need to use Kc in this case?
anonymous
  • anonymous
We don't need to use Kc because the information we have is essentially Kc. Calculate Kc for fun to see if you can!
Photon336
  • Photon336
Oh I see, because we're given the equilibrium concentrations already that's basically Kc, so all we need to do is to just find the initial concentrations. thanks for clarifying that
Lena772
  • Lena772
I got 5.5 and 2.5 and those were wrong
Lena772
  • Lena772
@Woodward @Photon336

Looking for something else?

Not the answer you are looking for? Search for more explanations.