Question

Can anyone help me please? I have to find equilibrium constant but I am only given mole ratio of substances not volume or concentrations..

Answer 1

The following system was allowed to come to equilibrium at 1500°C


2NO(g) ↔ N2(g) + O2(g)


At equilibrium, the system contains 0.0035 moles of NO(g), 0.040 moles of N2(g) and 0.040 moles of O2(g). What is the value of Kc for this system.


There is no volume to calculate the concentration of each of these mole ratios, I think I must be overlooking something?

Answer 2

It wouldn't make a difference. in principle for 100% hard core chemists (like me), the equilibrium does not have a unit and since it is the total volume, they will all cancel in the end. therefore you can base you equilibrium calculation on mole instead of concentration :)

Answer 3

oh ok cool. Thank you :-)

Answer 4

No problem at all. But also another thing I might want to add :)
Notice all the molecules are gasses :) then we usually use partial pressure instead of concentration.

Can you write the equilibrium expression for the reaction? :)

Answer 5

I had Kc=products over reactants. Kc= (N2)(O2) /(NO)2

Gave me a Kc value of 1.31x10^4

Answer 6

Would it change the value if I were to use partial pressures?

Answer 7

Perfect and then let me show you why I did as I did! (this works both for partial pressure and volume) :D

\[\large Kc=\frac{ \left[ N_2 \right] \times \left[ O_2 \right] }{ \left[ NO \right]^2 }=\frac{ \frac{ n(N_2) }{ V } \times \frac{ n(O_2) }{ V } }{ \frac{ n(NO) }{ V } \times \frac{ n(NO) }{ V } }=\frac{ n(N_2) \times n(O_2) }{ n(NO)^{2} }\]

as you can see the volume cancel just as predicted.

Answer 8

partial pressure and concentration* sorry writing wrong all the time.

Answer 9

:-) Thanks. We can cancel as the volume would be the same through the entire equation?

Answer 10

Yes, as it is the total volume of the solution or in this case the container.

Answer 11

Would you please be able to help me with another question? I have not come across combining to equilibrium constants and have a question on it.

Answer 12

Then you can ask your self: what would be the case if it was A -> B + C Then we would get:
\[Kc=\frac{ n(B) \times n(C) }{ n(A) } \]
That expression would have the unit "mol"... but I claimed the thing to be unitless, so usually we put in a "normalization factor" to cancel it:
\[Kc=\frac{ n(B) \times n(C) }{ n(A) \times n^{\Theta} }\]
\(n^{\Theta}\) = 1 mol so the units cancel. However this step is usually canceled unless it is hardcore chemistry. :)

Answer 13

Yes I recognise that equation from my textbook but it isn't used in any of the examples

Answer 14

Because it is a good help to do it usually (you'll see why at some point), it is only when you combine thermodynamics into it, it is important the equilibrium constant is unit less otherwise it become a horrible mess.

About your second question I have a beautiful rule:
add 2 reactions = multiple 2 equilibrium contacts
subtract 2 reactions = divide 2 equlibrium contacts with each other

This is the fast way, but if you want we can do it the long way to establish that conclusion?

Answer 15

You said you had never done this kind of question before, if that is the case we take the long way and then make some tricks as we go.

Answer 16

long way please, still a little lost

Answer 17

Alright:
So we notice first of all we notice that the final reaction does not contain oxygen at all aka the must cancel in the reaction, we can do this by doing the reversed reaction like so:
\(\sf PCl_3 + \frac{1}{2}O_2 \rightleftharpoons POCl_3\)
and
\(\sf NO_2 \rightleftharpoons NO + \frac{1}{2}O_2\)

If we add these two reactions we get (add reactants with reactants and products with products):
\(\sf PCl_3 + \frac{1}{2}O_2 + NO_2 \rightleftharpoons POCl_3 + NO + \frac{1}{2}O_2 \)

Notice the oxygen cancel against each other so the reaction is:
\(\sf PCl_3 + NO_2 \rightleftharpoons POCl_3 + NO \)

exactly like we wanted!

with me so far?

Answer 18

Sorry for the poor writing, a little tired, but the math and reactions are at least correct :)

Answer 19

good so far :)

Answer 20

Alright, lets just for now keep oxygen in the reaction to make it simple. The equilibrium constant expression would be:
\[\large K3= \sf \frac{\left| POCl_3 \right| \times \left| NO \right| \times \left| O_2 \right|^{\frac{ 1 }{ 2 }}}{ \left| PCl_3 \right| \times \left| NO_2 \right| \times \left| O_2 \right|^{\frac{ 1 }{ 2 }} }\]
But we remember that we got equilibrium constant 1 and 2 given by:
\[K1=\sf \frac{ \left| POCl \right| }{ \left| PCl \right| \times \left| O_2 \right|\frac{ 1 }{ 2 } }\] and\[K2=\sf \frac{ \left| NO_2 \right| }{ \left| NO \right| \times \left| O_2 \right|^{\frac{ 1 }{ 2 }} }\]

If you look closely you can see K3 is made from K1 and K2 giving us:
\[K3=\frac{ K1 }{ K2 }\]

Try substitute and check it is correct! :)

Answer 21

I can see K1 but why is K2 flipped (NO2 on bottom instead of on top?) because we reversed it in the beginning?

Answer 22

Exactly.
If you have a reaction A + B -> C + D with the equilibrium constant K, the reverse reaction C + D -> A + B is given by the reciprocal equilibrium constant (1/K).

Answer 23

As we demonstrated from by adding the reactions together to get the wanted reaction

Answer 24

great. thank you :-)

Answer 25

But you understand it perfectly? and would be able to do it if you had a similar problem? :)

Answer 26

I will practice the concept a few times to ensure :) But makes sense here.

Answer 27

my baby just woke up. I imagine I will have other questions when I come back. If you are still on in a little while I might talk to you then. Thank you so much for your help. Studying these units in correspondence is hard.

Answer 28

No problem at all, and you are more than welcome to tag me or message me :)

Answer 29

In the mean time I will write a little trick associated to this question that may in some cases make it more easy and also save a lot of time.

Answer 30

Lets try and start backwards, meaning we look at the wanted reaction with the unknown equilibrium constant. The expression is:
\(\large K3= \sf \frac{\left| POCl_3 \right| \times \left| NO \right| }{ \left| PCl_3 \right| \times \left| NO_2 \right| }\)

We multiply the expression with:
\(\sf 1=\frac{\left| O_2 \right|^{\frac{ 1 }{ 2 }}}{\left| O_2 \right|^{\frac{ 1 }{ 2 }}}\)

as \(1 \times x =x, \forall x\).

We then get:
\(\large K3= \sf \frac{\left| POCl_3 \right| \times \left| NO \right| \times \left| O_2 \right|^{\frac{ 1 }{ 2 }}}{ \left| PCl_3 \right| \times \left| NO_2 \right| \times \left| O_2 \right|^{\frac{ 1 }{ 2 }} }\)

which is equal to:
\(\large K3=\frac{ K1 }{ K2 }\)

This method should only be done if you got the surplus. :)

Answer 31

Obviously in the whole question I should use brackets not the absolute as I did, my apologies for that.

Answer 32

Yes. Thank you :)

Answer 33

No problem at all.

Answer 34

I didn't think you were online. I have opened another question if you are free?

Answer 35

Well I just came online now, but I'll take a look, where the other questions connect btw? :)