For Q1, if you increase the pressure, the system will try and reduce the pressure by shifting to the side with less moles of gas. There is only one of those equilibria which have less moles of gas on the right hand side and unfortunately it isn't D!
For Q2, I would have put a different answer (B) because when a system has reached equilibrium then the concentrations are constant, as the forward reaction and backward reaction occur at the same rate.
Q3 is quite long! You have to work out the moles of N
2O
4 that have become NO
2, by multiplying the concentration by the volume which gives 17.155 moles, so 47.345 moles of N
2O
4 have become NO
2. Then you multiply by 2, to give you the moles of NO
2 (1:2 ratio), which is 94.69.
Then find the concentration of this by dividing by 7.3, which gives 12.97123288...
And if you've written out the equation for K
c, then its [NO
2]
2 divided by [N
2O
4].
So you square the 12.97... to give 168.2528823...
Then as the equation says, you divide by the concentration of N
2O
4, which is 2.35, so you end up with 71.59697121...
Hence I think the answer is C. If there is a shorter way then somone please say!
You are on the right track for Q4! Just expand out the brackets and then solve to find x!