## 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)

1. Lena772

@Abhisar @Photon336

2. Photon336

You could work backwards

3. Photon336

wait those +/- thats the change in concentration right? @Lena772

4. Lena772

|dw:1442112283855:dw|

5. Lena772

Thats just showing the Tolerance for the answer will only be 0.1 above or below @Photon336

6. Lena772

I don't know how to solve for x working backwards @Photon336

7. Lena772

@abhisar

8. Photon336

So this was all the information they gave you?

9. Lena772

yep

10. 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.

11. Photon336

@Woodward thoughts? i'm not sure

12. 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!

13. anonymous

Whoops I said moles but this is concentration, let me straighten myself out before I say anymore lol

14. 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!

15. 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

16. anonymous

Yeah that's exactly what I got. :D

17. Photon336

@Woodward why didn't we need to use Kc in this case?

18. 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!

19. 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

20. Lena772

I got 5.5 and 2.5 and those were wrong

21. Lena772

@Woodward @Photon336