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Lena772
 one year ago
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)
Lena772
 one year ago
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)

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1You could work backwards

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1wait those +/ thats the change in concentration right? @Lena772

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Thats just showing the Tolerance for the answer will only be 0.1 above or below @Photon336

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0I don't know how to solve for x working backwards @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1So this was all the information they gave you?

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0I know M= moles solute/L of soln but i dont know if i could manipulate that equation with what I'm given here.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@Woodward thoughts? i'm not sure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0Whoops I said moles but this is concentration, let me straighten myself out before I say anymore lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that's exactly what I got. :D

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@Woodward why didn't we need to use Kc in this case?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We don't need to use Kc because the information we have is essentially Kc. Calculate Kc for fun to see if you can!

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0I got 5.5 and 2.5 and those were wrong
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