# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: dummy_boy5 on November 05, 2020, 03:23:00 PM

Title: Equilibrium constant
Post by: dummy_boy5 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...
Title: Re: Equilibrium constant
Post by: chenbeier 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
Title: Re: Equilibrium constant
Post by: dummy_boy5 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?
Title: Re: Equilibrium constant
Post by: chenbeier 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.
Title: Re: Equilibrium constant
Post by: dummy_boy5 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...
Title: Re: Equilibrium constant
Post by: chenbeier on November 05, 2020, 05:04:42 PM
Yes its correct you got it.
Title: Re: Equilibrium constant
Post by: dummy_boy5 on November 05, 2020, 05:07:49 PM
I really appreciate your help, thanks!
Title: Re: Equilibrium constant
Post by: chenbeier on November 06, 2020, 02:53:06 AM
One thing, the unit is missing.
Title: Re: Equilibrium constant
Post by: Meter 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.
Title: Re: Equilibrium constant
Post by: chenbeier 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.

Title: Re: Equilibrium constant
Post by: Borek 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.
Title: Re: Equilibrium constant
Post by: chenbeier on November 07, 2020, 05:43:53 AM
That is what I learned.

Title: Re: Equilibrium constant
Post by: Borek on November 07, 2020, 06:22:13 AM
Just because the error is common doesn't mean the idea is right.
Title: Re: Equilibrium constant
Post by: Meter 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
Title: Re: Equilibrium constant
Post by: chenbeier 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

Title: Re: Equilibrium constant
Post by: Meter on November 07, 2020, 09:49:23 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.
Because quantities like the reaction quotient, pH, etc. are a measure for the activity of a certain physical property. For instance, pH measures the activity of hydrogen ions in a solution. Likewise, a reaction quotient measures the activity of molecules in a solution. All activities are dimensionless, and depending on what type of activity, your quantities have to be divided by some conversion factor, like 1 mol/L for concentrations or 1*105 Pa for partial pressures.
Title: Re: Equilibrium constant
Post by: chenbeier on November 07, 2020, 09:53:51 AM
Thanks for the explanation.

Activities and concentration are not the same like fugacities and pressure are also not the same.

Title: Re: Equilibrium constant
Post by: Borek on November 07, 2020, 10:16:16 AM
You ignore the units part. log (nor any other function) doesn't miraculously remove units from the quantity it acts on, it can be used only on unitles quantities.

40 years ago I went in university i also never heard it.

That was definitely part of my course back in eighties, and we used books written decade or two earlier. As I already wrote - this a lousy chemistry, sadly, it is often taught without a second thought and deeper understanding of why it is wrong and inconsistent.

Just because it is popular doesn't mean it is correct.