How will the following system at equilibrium shift in each of the following cases?

Question

2 SO3(g) = 2 SO2(g) + O2(g) H= 197kj

a)SO2(g) is added
b)the pressure is decreased by increasing the volume of the container
c)the pressure is increased by added Ne(g)
d)the temperature is decreased
e) O2(g) is removed

Answer 1

A) More SO2 will push the system to the left, increasing SO3 and decreasing O2. There is more SO2 for the O2 to react with and this direction will counter the change.
B) Reducing pressure will move equilibrium right because this will compensate for the reduction in pressure. There are 3 volumes of gas on the RHS and only 2 on the LHS.
C) The increase in pressure will be countered by moving the equilibrium to the left. This reduces pressure by reducing the volume of gasses in the system. (note I have had to correct some teachers of chemistry on this one; it makes NO difference how the pressure is increased!!!)
D) Your equation suggests that it takes energy to convert SO2 and O2 to SO3 so heating the system will push the equilibrium left to reduce the temperature.
E) Removing O2 will push the reaction right to replace the lost oxygen.
The rule is that the system will react to reduce the effect of the change; whatever that is.

Answer 2

I do find these questions misleading. Changing concentrations does not change the equilibrium constant (Kp). So if you add SO2 then more SO3 will be produced to keep the equilibrium constant constant.

As in this case the Kp has pressure in the numerator so increasing the pressure will reduce the pressure of the O2 and SO2 to again keep Kp constant.
As for temperature /\G = -RT ln(Kp) so T decrease will cause Kp to become smaller so more SO3 will be formed.

Answer 3

a). More SO3(g) is formed
b). Nothing
c) Don't know
d). More SO3(g) forms
e). Less SO3(g) is formed

Mostly guesses

Answer 4

a) shift to the left
b & c &d) not a lot
e) shift to the right