Equilibrium constant is K=3.4*10^90 for the reaction: 2K+Cl2 2KCl Is Cl2 concentration very large or very small at equilibrium?

Question

Equilibrium constant is K=3.4*10^90 for the reaction: 2K+Cl2 <--> 2KCl
Is Cl2 concentration very large or very small at equilibrium?

somy · Answer

what do u think? any idea?

anonymous · Answer

I really have no idea where to begin, I have several questions like this one that I have to answer though. @Somy

somy · Answer

hmmm 
 2K+Cl2 <--> 2KCl
what is Kc formula?

anonymous · Answer

Uhh. maybe [K+]^2[Cl-]/[K+]^2[Cl2-]

anonymous · Answer

or am I completely lost

somy · Answer

wrond

somy · Answer

$$Kc = \frac{ product }{ reactant }$$
$$Kc = \frac{  [KCl]^2 }{ [K]^2 [Cl_2] }$$

anonymous · Answer

Ok would this be high? since ^90 and it would be far to the right… @Somy

somy · Answer

Kc is pretty a lot 
so u can assume that KCl's concentration is pretty high

somy · Answer

to get such big number that would mean that K and Cl's product concentrations should be a pretty less
K is to the power of 2 so it must be pretty bigger the Cl
so i would expect Cl's concentration to be Less/ small

somy · Answer

i think, im not really sure tbh

anonymous · Answer

@Somy ok thanks. So if I have K = 0.274*10^42 for the reaction:
C2H2+2H2<-->C2H6     Would C2H6 also be small or would this be large since I think the products are favored.

anonymous · Answer

I just don't want to get confused

somy · Answer

by product of concentrations i meant K and Cl 
not KCl

somy · Answer

C2H6 i think would be larger