Methanol can be produced by the following reaction
CO + 2H2---CH3OH
under the conditions used all are gaseous and concentrations measured in dm-3
1. write equation for K(equilibrium constant) and show units for K
2. explain how the equilibrium yield will be affeced by: - increase in
pressue, increase in temperature, increase in concentration of CO &
the presence of a catalyst

Hello Paulinej

1) The equilibrium constant can be given the following equation:

K = [CH3OH] / ([CO][H2]^2)

If the units are dm-3, that would give units of dm6 for K.  Are you
sure the units are correct here?  dm-3 usually indicates a volume, I
would expect something more like mol dm-3 (in which case the answer
would be mol2dm6).

This is done by examining the units thus:

Units of K = dm-3 / dm-3 x (dm-3)(dm-3)
Units of K = 1 / dm-6 which is equivalent to dm6

2) Increases in pressure favour the side of the reaction with fewer
molecules.  On the left hand side of the equation we have 3 mol (one
of CO and two of H2), on the right hand side we have 1 mol of CH3OH. 
Increasing the pressure will therefore increase the yield of CH3OH as
the reaction will want to go to the right.

Chatelier's Principle states that a reaction will proceed to
counteract any change.  Therefore an exothermic reaction (one that
gives out heat) will favour reactants and endothermic reactions (one
that requires heat to start) will favour products when INcreasing the
temperature.  As your reaction is exothermic the yield will decrease
when temperature is increased.

An increase in concentration of one of the reactants will will yield
more product, this is the process of equilibrium.  Therefore
increasing the CO concentration should increase the yield of CH3OH.

The presence of a catalyst will not effect the yield of the reaction. 
A catalyst ONLY effects the speed at which equilibrium will be reached
(a trick question!).