[msij280]

2N2O5 (g) ----> 4NO2 (g) + O2 (g)

Derive the relationship between the initial pressure of N2O5, total pressure and the current partial pressure of N2O5.

Please explain how this is done.

Thanks.

[Borek]

You have to combine mole balance and stoichiometry with partial pressures. Something similar to ICE table.

[Bryan Sanctuary]

I am guessing that you start with pure N₂O₅(g) at a given P and then let it go to equilibrium.  At equilibrium, you want the relationship between the starting pressure of N₂O₅ and the partial pressures of the others.

Let's start with one mole of pure N₂O₅, so the initial pressure is, assuming ideal gases, P_initial = RT/V.

               2N₂O₅(g) ----> 4NO₂ (g) + O₂ (g)

initial         1  mole              0              0

equilibrium   1-2α               4α             α

At equilibrium, the total number of moles is:  1+3α

If we assume an ideal gas, then the equilibrium partial pressures are related to the mole fractions:

P_i/P_total =X_i

Therefore the mole fractions of each are 

P_N2O5/P_total = (1-2α)/(1+3α)
P_NO2/P_total = 4α/(1+3α)
P_O2/Ptotal = α/(1+3α)

and the total pressure at equilibrium is

P_total = P_N2O5 +P_NO2 + P_O2

Note that each partial pressure obeys its own ideal gas law:

P_NO2=n_NO2RT/V   etc.

The total pressure at equilibrium is:

P_total=(1+3α)RT/V = P_initial+3αRT/V

which is the relationship between the initial pressure of pure N₂O₅ and the final total pressure.

This is all Dalton's Law of Partial Pressures and is covered in section 1.9 of my physical chemistry text book which you can get for free:

http://www.mchmultimedia.com/store/register.php?action=try&cat=64