Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started 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 3H_{2}+N_{2} :resonance: 2NH_{3}
Initial concentrations of N_{2} = H_{2} = 2 mol/L
Equilibrium concentration of NH_{3} is 0.5 mol/L
So, I have to somehow calculate the equilibrium constant, but don't know how to do it...

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

K = [NH_{3}]^{2}/[N_{2}][H_{2}]^{3}
I don't have the initial concentration of NH_{3}
You mean to calculate the remains concentrations I have to subtract from the initial concentrations like this:
[N_{2}] = 2  3×0.5 and [H_{2}] = 2  0.5 ?
Is it correct?

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.

I think it should be like that:
[N_{2}] = 2  1/2×0.5 = 1.75 and [H_{2}] = 2  3/2×0.5 = 1.25
and K would be: K = 0.5^{2} / (1.75 × 1.25^{3}) = 0.073
Its my intuition, but I think its not correct though...

Yes its correct you got it.

I really appreciate your help, thanks!

One thing, the unit is missing.

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.

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 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 K_{p} with unts?
Strict approach calls for unitless activities and for unitless equilibrium constant, what Meter wrote is perfectly correct.

That is what I learned.
https://www.toppr.com/ask/question/unitofequilibriumconstantis/

Just because the error is common doesn't mean the idea is right.

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*10^{5} Pa).
Directly from my textbook:
https://prnt.sc/vf4q7u
where c^{⦵} = 1 mol/L

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

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*10^{5} Pa for partial pressures.

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

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.