Calculate the mole percentage of N2O5 that will be decomposed at equilibrium.?

Calculate the mole percentage of N2O5 that will be decomposed at equilibrium if the volume is increased by a factor of 0.53.





Consider the decomposition equilibrium for dinitrogen pentaoxide.

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



At a certain temperature and a total pressure of 1.00 atm, the N2O5 if 0.53% decomposed (by moles) at equilibrium.Calculate the mole percentage of N2O5 that will be decomposed at equilibrium.?
Assume ideal gas:

p路V = N路R路T



For constant pressure and temperature the volume is proportional to the total amount of substance

V = (R路T/p) 路 N = constant 路 N



Initially we had pure dinitrogen pentoxide let say N鈧€

Let x be the amount of N鈧侽鈧?which had reacted away in equilibrium. From reaction equation you find that per moles N鈧侽鈧? 4/2=2 moles of NO鈧? and 1/2 moles of O鈧?are formed. Hence the amount of substance of the components in equilibrium is:

N(N鈧侽鈧? = N鈧€ - x

N(NO鈧? = 2x

N(O鈧? = (1/2)路x

The total amount of substance in equilibrium is

N_e = N(N鈧侽鈧? + N(NO鈧? + N(O鈧? = N鈧€ + (3/2)路x



We know that the volume in equilibrium has increased by 53%

V_e/V鈧€ = 1.53

because volume and amount of substance are proportional

=%26gt;

N_e/N鈧€ = 1.53

%26lt;=%26gt;

(N鈧€ + (3/2)路x) / N鈧€ = 1.53

%26lt;=%26gt;

1 + (3/2)路(x/N鈧€) = 1.53

Hence the fraction of dinitrogen pentoxide which had reacted away is

(x/N鈧€) = (2/3)路0.53 = 35.3%