posted by Sammy on .
How would increasing the volume of the reaction vessel affect these equilibria. You don't have to tell me the answers, but can you please explain to me how I get the answer.
a) NH4Cl(solid) <===> NH3(gas) + HCL(gas)
b) N2(gas) + O2(gas) <===> 2NO(gas)
If the volume increases, the pressure must decrease AND the concn must decrease (moles/greater volume = smaller concn).
If Kc = (NH3)(HCl). Then making (NH3) and (HCl) smaller means the product becomes less than Kc and the reaction must shift to increase it to the original value of Kc. The only way it can do that is to shift the equilibrium to the right. That's the long approach. The shorter approach is to say (for a gas reaction) that increasing P will shift the equilibrium to the side with the smaller number of moles. So increase P and we see a shift to left. Decrease P and it goes to the right. Same answer but shorter.
For b), changing the pressure has no effect BECAUSE the moles of gas is the same on both sides. But you can get the same answer if we want to go the concn route. Kc = (NO)^2/(N2)(O2)
If we increase the volume, then concn is moles/larger V = smaller number.
(moles NO/V)(moles NO/V)/(moles N2/V)(moles O2/V) and you notice V cancels so it makes no difference what we choose for a volume (or pressure).