- #1

- 479

- 20

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter PFuser1232
- Start date

- #1

- 479

- 20

- #2

Borek

Mentor

- 28,702

- 3,190

- #3

- 479

- 20

- #4

Borek

Mentor

- 28,702

- 3,190

- #5

- 479

- 20

Does this apply even if we're given initial molar amounts of reactants/products?

Consider the Haber Process:

N_{2} + 3H_{2} <=> 2NH_{3}

K_{c} = [NH_{3}]^{2}/[N_{2}][H_{2}]^{3}

Suppose we're initially given 2 mol of N_{2} and 1 mol of H_{2}, in a system of unknown volume, V.

Now, I understand that change is going to be -x for Nitrogen, -3x for Hydrogen, and +2x for Ammonia (x has dimensions mol/L).

While computing K_{c} for this equilibrium, can we substitute (2-x) for [N_{2}] as though the initial concentration was 2 mol/L?

If we substitute the given data without glossing over any mathematical detail, we are left with:

K_{c} = (2x)^{2}/(2/V mol - x)(1/V mol - 3x)^{3}

If we do the binomial expansion, we would be left with Vs all over the place. I don't mean to be pedantic, it's just that I can't get my head around it and it's really bugging me.

Consider the Haber Process:

N

K

Suppose we're initially given 2 mol of N

Now, I understand that change is going to be -x for Nitrogen, -3x for Hydrogen, and +2x for Ammonia (x has dimensions mol/L).

While computing K

If we substitute the given data without glossing over any mathematical detail, we are left with:

K

If we do the binomial expansion, we would be left with Vs all over the place. I don't mean to be pedantic, it's just that I can't get my head around it and it's really bugging me.

Last edited:

- #6

Borek

Mentor

- 28,702

- 3,190

Suppose we're initially given 2 mol of N2 and 1 mol of H2, in a system of unknown volume, V.

If the V is unknown, and you are given Kc, problem can't be solved.

- #7

- 479

- 20

- #8

Borek

Mentor

- 28,702

- 3,190

Note you can always rescale the problem to V=1L, in which case volume again stops to matter.

Share: