September 21, 2021, 12:48:15 PM
Forum Rules: Read This Before Posting


Topic: Equilibrium constant  (Read 938 times)

0 Members and 1 Guest are viewing this topic.

Offline dummy_boy5

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Equilibrium constant
« on: November 05, 2020, 03:23:00 PM »
Hey, guys!
Can someone explain how to solve this problem?

We have a reaction 3H2+N2  :resonance: 2NH3

Initial concentrations  of N2 = H2 = 2 mol/L
Equilibrium concentration of NH3 is 0.5 mol/L

So, I have to somehow calculate the equilibrium constant, but don't know how to do it...

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #1 on: November 05, 2020, 04:26:52 PM »
1. Write the equation of the mass action law.
2. From the given equilibrium NH3 concentration read how much nitrogen and hydrogen was used.
3. Calculate the remaining concentration of nitrogen and hydrogen
4. Fill in the values in the mass action law

Now its your work to show
« Last Edit: November 05, 2020, 04:50:47 PM by chenbeier »

Offline dummy_boy5

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Equilibrium constant
« Reply #2 on: November 05, 2020, 04:37:02 PM »
K = [NH3]2/[N2][H2]3
I don't have the initial concentration of NH3
You mean to calculate the remains concentrations I have to subtract from the initial concentrations like this:
[N2] = 2 - 3×0.5 and [H2] = 2 - 0.5 ?
Is it correct?

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #3 on: November 05, 2020, 04:47:49 PM »
Mass action law is correct.
Of course you don't have initial concentration of ammonia because its 0.
Figure out how much hydrogen and nitrogen was used to get 0,5 mol NH3.
If you have that then subtract each from the initial values.
The results fill in mass action law.

Offline dummy_boy5

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Equilibrium constant
« Reply #4 on: November 05, 2020, 05:01:23 PM »
I think it should be like that:
[N2] = 2 - 1/2×0.5 = 1.75 and [H2] = 2 - 3/2×0.5 = 1.25

and K would be: K = 0.52 / (1.75 × 1.253) = 0.073

Its my intuition, but I think its not correct though...

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #5 on: November 05, 2020, 05:04:42 PM »
Yes its correct you got it.

Offline dummy_boy5

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Equilibrium constant
« Reply #6 on: November 05, 2020, 05:07:49 PM »
I really appreciate your help, thanks!

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #7 on: November 06, 2020, 02:53:06 AM »
One thing, the unit is missing.

Offline Meter

  • Full Member
  • ****
  • Posts: 229
  • Mole Snacks: +13/-3
  • Take what I say with a grain of salt
Re: Equilibrium constant
« Reply #8 on: November 06, 2020, 05:17:20 PM »
One thing, the unit is missing.
The equilibrium constant is unitless as it is implicitly understood that each concentration is divided by the standard concentration (1 mol/L), removing any units.

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #9 on: November 07, 2020, 03:13:18 AM »
That is wrong. The unit depends on the equation.

In this case.

K = [NH3]2/[N2][H2]3

The unit is (l/mol)2.


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26749
  • Mole Snacks: +1734/-403
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Equilibrium constant
« Reply #10 on: November 07, 2020, 04:25:01 AM »
That is wrong. The unit depends on the equation.

In this case.

K = [NH3]2/[N2][H2]3

The unit is (l/mol)2.

That's a lousy chemistry. How are you going to calculate [itex]\Delta G^0 = -RT \ln(K_p)[/itex] using Kp with unts?

Strict approach calls for unitless activities and for unitless equilibrium constant, what Meter wrote is perfectly correct.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #11 on: November 07, 2020, 05:43:53 AM »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26749
  • Mole Snacks: +1734/-403
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Equilibrium constant
« Reply #12 on: November 07, 2020, 06:22:13 AM »
Just because the error is common doesn't mean the idea is right.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Meter

  • Full Member
  • ****
  • Posts: 229
  • Mole Snacks: +13/-3
  • Take what I say with a grain of salt
Re: Equilibrium constant
« Reply #13 on: November 07, 2020, 07:29:54 AM »
That is wrong. The unit depends on the equation.

In this case.

K = [NH3]2/[N2][H2]3

The unit is (l/mol)2.
You do the same thing when working with gases, where the partial pressures are all divided by the standard pressure (1*105 Pa).

Directly from my textbook:

https://prnt.sc/vf4q7u

where c = 1 mol/L
« Last Edit: November 07, 2020, 07:43:51 AM by Meter »

Offline chenbeier

  • Sr. Member
  • *****
  • Posts: 1348
  • Mole Snacks: +102/-22
  • Gender: Male
Re: Equilibrium constant
« Reply #14 on: November 07, 2020, 09:30:39 AM »
I understand it. But why I can do so, to devide by 1 mol/ l or 1 bar, if pressure is used. A lot of websides dont consider it. And 40 years ago I went in university i also never heard it.
Its like a price is 100 $ and we devide by 1$ and say its 100. Make this sense? But I have an open ear to learn new things, if they are explained.
Sentences like :"Just because the error is common doesn't mean the idea is right" dont help.
Explanation is given here

https://en.m.wikipedia.org/wiki/Equilibrium_constant

« Last Edit: November 07, 2020, 09:47:43 AM by chenbeier »

Sponsored Links