Hey, orgoclear, thanks for the reply
I'm still a bit confused...
You say the [H2] increases.
But does the the change in concentration (H2) affect anything?
c=n/v
If V increases, and n (H2) decreases (since the reverse reaction is favored), shouldn't the concentration decrease. 8-o
c (H2) = n (decrease)/V (increase)
The only reason I'm asking is in some questions like this, they ignore the affect of volume on the concentration calculation, so
I'm just seeking clarity.
Thanks for your help again !