ausm8 said:
Hi,
I have a question about equilbirum.
2SO2 + O2 --> 2SO3 Delta H = -99kj/MOL
How does temperature, pressure and concentration affect this reaction of equilibrium.
The reaction is exothermic I believe.
Cheers
okay, well if the reaction is 2SO2 + O2 --> 2SO3 then i guess none of those factors will affect it cuz the reaction isnt in equilibrium.
but lets assume that it is
since you say the reaction has an enthalpy change of -99KJ/Mol it IS exothermic in the forward reaction so:
2SO
2 + O
2 <--> 2SO
3 + heat
temperature: if there is a decrease in temperature, then le chateliers principle states that equilibrium will favour the reaction that will produce more heat to compensate for the heat lost.
hence, equilibrium will shift to the right to produce more heat.
if temperature was to increase the equilibrium will shift to the left (it will favour the reverse reaction)
pressure: if there is an increase in pressure, for example the space of the system is decreasing; compacting in size, then this will cause the system to want to favour a reaction that will produce fewer moles. (more moles take up more space, but since space is limited, the system will want to decrease it's volume as well)
hence, for 2SO
2 + O
2 <--> 2SO
3 equilibrium will favour the forward reaction.
it will shift to the right, since there are fewer moles on the RHS of the equation.
the converse applies for this also
concentration: this kinda depends on what substance is being affected. say for example more SO2 is being added then equilibrium will favour the forward reaction (or equilibrium will shift to the right) since, according to le chatelires principle -> system will react to counter balence change, and to try to make it how the system used to be, and the only way to do this is by making 'added SO2' react with O2 to reduce concentration of SO2.
and so forth.
hope that explains man. cuz ai....its the most ive ever written