What happens to the concentration of HI(g) when the total pressure on the equilibrium reaction

2HCl(g) + I2(s) <--> 2HI(g) + Cl2(g)
is increased (by compression)?
1. increases
2. Unable to determine
3. decreases
4. remains the same

For an increase in pressure the reaction shifts to the side with the fewer mols. In this reaction the number of mol gas is the same on both side; therefore, there will be no shift due to pressure changes.

I meant to take out number 4 because it was an incorrect answer, I just don't understand why.

I just reread the question and my answer is wrong. Give me a second.

2HCl(g) + I2(s) <--> 2HI(g) + Cl2(g)

I tried to delete my first response before you read it but I didn't make it so I put in the oops message. Here is the scoop. (I was alerted by another of your questions below.)
My initial statement concerning pressure changes is correct; i.e., an increase in pressure will shift the reaction to the side with the fewer mols gas (NOTE gas). In this reaction, you have 3 mols gas on the right and only two mols gas on the left (note that I2 is a solid and not a gas); therefore, the reaction will shift to the left which will decrease the concn of HI.

It decreases

What happens to the concentration of NO(g) when the total pressure on the equilibrium reaction 3NO2(g) + H2O(l) = 2 HNO3(aq) + NO(g)

What happens to the concentration of NO(g) when the total pressure on the equilibrium reaction 3NO2(g) + H2O(l) = 2 HNO3(aq) + NO(g) is increased by compression