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Topic: Equilibrium constant  (Read 1369 times)

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dummy_boy5

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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  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...

chenbeier

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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 »

dummy_boy5

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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?

chenbeier

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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.

dummy_boy5

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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...

chenbeier

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Re: Equilibrium constant
« Reply #5 on: November 05, 2020, 05:04:42 PM »
Yes its correct you got it.

dummy_boy5

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Re: Equilibrium constant
« Reply #6 on: November 05, 2020, 05:07:49 PM »
I really appreciate your help, thanks!

chenbeier

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Re: Equilibrium constant
« Reply #7 on: November 06, 2020, 02:53:06 AM »
One thing, the unit is missing.

Meter

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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.

chenbeier

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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.

Borek

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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 $\Delta G^0 = -RT \ln(K_p)$ 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

chenbeier

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Re: Equilibrium constant
« Reply #11 on: November 07, 2020, 05:43:53 AM »

Borek

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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.
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Meter

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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 »

chenbeier

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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 »