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Analogy to describe pressure in a reversible reaction? watch

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1. Er...can anynone help me out?

Do u have an analogy to describe the effect of Pressure in a reversible reaction, Like the Haber process?

i know that an increase in pressure for the forward reaction (exothermic)means an increase in yeild but i dont' really UNDERSTAND how if u know wot i mean? i feel like im just memorising pointless rubbish
lol. i know it's very simple stuff, but any suggestions?
2. In gases the volume of a gas is proportional to the number of moles. Thus, if you look at the number of moles on each side, you should be able to tell which side of the reaction will have the highest volume.

High pressure will favour the side of the reaction with the lowest volume, since effectively the pressure is trying to reduce the volume. So in the Haber process, where 4 moles (N2 and 3H2) go to 2 moles (2NH3), high pressure favours the forward reaction and so you increase the yield with higher pressure.
3. ahh.... I need to better understand the concept of pressure methinks and what effects it has when it increases and decreases in general.
ta muchly. that is clearer, i'll consolidate all that info and read the textbook page agen. Honestly, half of wot our chem teacher says is all in her head and does'nt translate to us [well..me ]
4. When you increase the pressure you push the particles closer together increasing the concentrations.

The rate of both the forward and backward reactions increases as there are now more collisions.

However, the side with the most moles in the equation must increase its reaction rate more so than the side with fewer moles (as there are more particles involved in the former case).

The forward and backward rates are temporarily now NOT equal, the equilibrium is disturbed and the reaction moves in the direction of fewer moles until a new equiibrium is established.

This can be simply seen by using an equiibrium in which two moles goes to one mole such as:

N2O4 <==> 2NO2

Double the pressure and you halve the volume.

The concentration of each component doubles. On the left hand side the N2O4 concentration moves from 1 to 2, but on the right hand side the concentration moves from 2 to 4, i.e. an increase of 2 (double the increase of the LHS). To cope with this change in concentration the reaction moves to the LHS to make more N2O4

Mathematically:

Let the equilibrium concentrations of N2O4 and NO2 be x and y respectively.

the equilibrium law says that k= y2/x

Double the pressure and you double the concentration of both - the ratio becomes (2y)2/2x = 4y2/2x = 2y/x (which is exactly double the original ratio value of x/y)

In other words the increase in concentration has now made the ratio NOT EQUAL to K (originally y/x).

The reaction responds by making more of component x (the N2O4 ) until the value of K is restored and equilibrium established once again.
5. woah-Thanks for that detailed explaination,
It has helped my understanding, the concept of pressure to me is not so 'flaky'.Will stick it in my excersise book and make it my mantra=DDD

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Updated: March 12, 2008
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