The equilibrium reaction below is endothermic, which change wil increase the amount of NO2 at equilibrium?

Equation:
N2O4(g) yields 2NO2(g)

1. adding a catalyst
2. decreasing the temperature
3. increasing the volume of the container
4. adding an inert gas to increase the pressure

Answer:
2
LE CHATLIER'S PRINCIPLE-
..............IF A SYSTEM IS SUBJECTED TO CHANGE IN CONCENTRATION , TEMPERATURE , OR PRESSURE THEN SYSTEM TENDS TO MOVE SO AS TO UNDO THE EFFECT OF CHANGE.

SO TO INCREASE THE AMOUNT OF NO2,

IF WE INCREASE THE VOLUME OF CONTAINER,
THEN
OVERALL PRESSURE WOULD DECREASE
SO
MORE MOLECULES ARE REQUIRED TO RESTORE THE PRESSURE.
HENCE ANSWER IS .................."(3.)"
Le Chatlier's principle allows us to predict the direction a reaction will take when we perturb the equilibrium by changing the pressure, volume, temperature, or component concentrations.

Simply stated, the principle says that if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize that stress.

A good non-chemical analogy is two people on a see-saw. If their masses are equal then the see-saw balances. If we stress the system by adding weight to one side, the only way we can return to balance is by having the heavier person move closer to the fulcrum.

For a general equilibrium such as this:

A + B equil. arrow C + D

If we add more of compound A to the system, the system will try to get rid of the excess A. The only way it can do that is to make the forward reaction (i.e. left to right) favored, producing more C and D as some of the excess A is consumed. Notice we can't convert all the excess A to C and D because the reverse reaction tries to get rid of the "new" C and D in the same fashion.

Likewise, if we took away some A, the system would try to make more A by consuming C and D (favoring the reverse reaction, i.e. from right to left).
Let's consider what happens when we increase the pressure on this reaction, something that we can also do by decreasing the volume

N2O4 (g) equil. arrow 2 NO2 (g)

In the forward reaction, the N-N bond of N2O4 breaks and the two NO2 pieces fly apart in space. To go in the opposite direction, two NO2 molecules must not only come close together, but they also be aligned properly to form a new N-N bond. This will be important a little later, so keep it in mind.
From Le Chatlier, we know that if we increase the pressure, the system will do what it can to counteract that change. Notice in this equation that the right side has two moles of gas but the left side has only one (Deltan = +1). Therefore, one way to reduce the pressure would be to convert some of the NO2 (g) to N2O4 (g). In other words, we predict the reaction will shift to the left under these conditions.

We could express this in terms of the reaction quotient, Qc if we wished:

QC = [NO2]2/[N2O4]

Notice that as both concentrations increase, the value of QC becomes larger (because the top one is squared). Therefore, Qc > Kc, meaning there is an excess of products over reactants. Therefore, the system will shift to the left (consume the excess products) until QC = KC.

Whatever method we use to analyze the problem, it should become obvious that an increase in pressure (or decrease in volume) favors the net reactions that DECREASES the TOTAL number of moles of gases.
There are two fairly easy ways to shift the equilibrium, the first makes use of the exothermicity of the reaction. As written, the reaction releases releases 58 kJ of energy ( DHo = -58 kJ). Since energy is released, adding more energy via heating shift the equilibrium to the left and removing energy via cooling will shift the equilibrium to the right. The second method utilizes changes in pressure to shift the equilibrium. This will be covered in another demonstration.

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