What could cause this reaction's equilibrium to shift to the left: 2SO2(g) + O2(g) 2SO3(g) (ΔH= -198 kJ/mol)

Question

Would Decreasing the volume of the system cause the equilibrium to shift to the left? Would decreasing the pressure? Would decreasing the temperature? Would adding SO2 to the system shift the equilibrium to the left? What about removing SO3?

Answer 1

2 + 1 → 2 ∆H -ve

2 moles + 1 mole → 2 mole

Decrease volume, increases pressure, equilibrium move to the side to partly reduce this increase, ie to the side with fewer moles, if that is possible. to the RIGHT

Decrease pressure, increased volume, equilibrium moves to the side to partly increase the reduced pressure, moves to the left

Decrease temperature ∆H -ve, heat is a "product" of the forward reaction. If the temperature is decreased, the reaction will move to the side that partially counteracts that - to the right

Add SO2, pushes to the right; take out SO3 pushes to the right also.

Apply Le Chatelier's principle - any change to an equilibrium will result in a movement in the direction of the equilibrium pair of reactions such that the change is partially reduced.

Answer 2

Remember that LeChatlier's Principle says that if you muck around with an equilibrium, the reaction will shift in the direction that relieves the strain you placed on the equilibrium.

If you think about this and each of your conditions, you should be able to figure out the answers. So, for the first one, of decreasing the volume: You have 3 moles of gas on the left and two on the right. Decreasing the volume will favor the side with the smaller number of moles of gas. So decreasing the volume should shift the equilibrium to the right.

Think about the others and see what you come up with...

Answer 3

Only decreasing the pressure would make it shift left. Since there are more moles of gases on the left, the system must create more moles of these gases if the pressure is decreased in order to restore the original pressure.