Unit 4 Questions MEGATHREAD!

[HarveyD]

N2O4 <===> 2NO2
where NO2 is brown and N2O4 is clear

Keeping temp constant, the plunger of the syringe is pushed in to halve the volume. Equilibrium is then re-established. Is the brown colour of the mixture more intense or less intense than before volume is halved?

Don't understand why answer says more intense...

[luken93]

If the volume is halved, then it will favour the side that has few particles, that being the left. Since N2O4 is clear, it will be less intense.

See below.

[b^3]

If the volume is halved, then it will favour the side that has few particles, that being the left. Since N2O4 is clear, it will be less intense.

Yes but since the volume is halfed, the concentration of of NO2 increases (this factor is greater than the shift of the equilibrium), making the intensity of the brown color increase.

[tony3272]

Initially when the volume is pressed down, it will be more intense and become much darker. Then it will re-establish equilibrium and by Le Chatalier's principle it will only partially oppose the changes made to it. Hence the concentration on your concentration-time graph will never go back down to or lower than its original concentration. At this new equilibrium it will be more intense than it was originally

[b^3]

Well explained.

[luken93]

Woops, forgot to see that equilibrium was re-established :/

Can't wait until UMAT is over 

[HarveyD]

oh how but can you tell? does that always happen during dilution or..

[tony3272]

This isn't a dilution but rather a reduction of volume. When the syringe is initially pushed down, technically, the reactants and products will have not changed as the reaction to offset the change has not yet occurred. At this instant in time, the amount of both reactants and products will be the same as they were before, but the volume will be halved. Hence, the concentrations of both will be greater and the solution will become more intense. Then, in order to offset the change, the reaction much proceed backwards in order to reduce the number of particles and reduce the pressure change. Once it is back in equilibrium the concentrations of both will have changed, but they will not become lower than the concentration before the syringe was pressed down. Because of this, it will become more intense.

[Vincezor]

oh how but can you tell? does that always happen during dilution or..

By looking at the colour intensity. We did a prac of this last week at school. It is a bit hard to see, but after a few attempts you'll be able to to see the colour shift to a more intense colour when plunger is pushed in, and then it was become a bit paler afterwards (however it is darker than it was originally).

If you have Heinemann Chemistry 2 textbook, have a look at page 274! Hopefully that sure clear up any confusion.

[HarveyD]

yeahh i saw the graph and i understand how it works
just a bit confused how you can tell the colour intensity i.e. how it would be still darker than orignal after the offset to change.

so would it be correct to say that if there was an increase in volume instead
the end product would be paler than the original even after offset to change?
like does this apply to all such reactions?

[generalkorn12]

Is there any tips or ideas on drawing/interpreting concentration-time graphs, I'm having trouble understanding them.

[tony3272]

I found the best way to do this is to understand how the concentration changes after certain actions have been performed. \

For example, when a dilution occurs, the concentration of everything will initially fall (as there is greater volume), but then the reaction will need to go forwards or backwards to reestablish equilibrium. One side's concentration will decrease further whilst the other side will increase in concentration.

[luken93]

Might look for a site with all of these graphs in 1 place, could be very helpful!

[dehoqu]

For the reaction:
H_2(g) + I_2(g) <---> 2HI _(g)
Why doesn't a change in pressure/volume affect the equilibrium position of the reactants/products? Is there no Le Chatelier's response?

[pi]

For the reaction:
H_2(g) + I_2(g) <---> 2HI _(g)
Why doesn't a change in pressure/volume affect the equilibrium position of the reactants/products? Is there no Le Chatelier's response?

Equal mole (or particle) ratio of reactants and products on each side, therefore no change.