Can you please explain if it would be possible to calculate the K value if more than one product or reactant in a chemical equation is a gas?

Answer 1

I'm not sure where you're headed with this but I'm guessing that K for liquids where M is used for the concentration is no problem for you but if a gas is involved you don't know how to handle it. For example, in the reaction
A(aq) + B(aq) ==> C(aq),
Keq = (C)/(A)(B). We substitute M at equilibrium for A, B, and C and there is no problem calculation K. If the reaction is
Na2CO3(s) --> Na2O(s) + CO2(g)and
Kp = pCO2 or
Kc = (CO2)
Finally, if we have a mixed equation such as
Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) then Keq =
(ZnCl2)*pH2/(HCl)
If I've interpreted your question incorrectly explain a little more fully (perhaps an example) and what you don't understand and I'll try again.

Answer 2

Thanks but not exactly, the question is asking what would happen if you had more than one product or reactant that was a gas

Answer 3

What are you studying? The following answer assumes you are studying reasons for a reaction to go to completion.
When one or more products are a gas the reaction is shifted to the right and it goes to completion.
Reactions go to completion for one of three reasons:
1. A gas is formed.
2. A ppt is formed.
3. A slightly ionized product is formed.

Answer 4

Ièm studying grade 12 chemistry and this is the Chemical Systems and Equilibrium

Answer 5

Thanks. The last answer is the one you want. When a gas is evolved in a closed system, the system eventually reaches equilibrium so the forward and the reverse reactions occur but the system as a whole does not change from the equilibrium point. When the system is open, the gas escapes into the surroundings, the system cannot reach equilibrium and Le Chatelier's Principle applies and the entire system shifts to the right trying to reach equilibrium. The end result is that the reaction goes to completion.