Question

Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 ∘C of 1.0×10−5s−1.

Calculate the partial pressure of O2 produced from 1.47 L of 0.605 M N2O5 solution at 45 ∘C over a period of 18.5 h if the gas is collected in a 11.4-L container. (Assume that the products do not dissolve in chloroform.)

Answer 1

Here we calculate the partial pressure of O2 produced from N2O5

Reaction:

2N2O5 --- > 4 NO2 + O2

Mole ration of N2O5 to O2 is 2 :1

So when two moles of N2 decomposed we get 1 mol of O2

Lets first find how many number of moles of N2O5 are decomposed.

Number of moles of N2O5 = Volume in L x molarity

= 1.47 L x 0.605 M

=0.88935 mol

Conversion of time into s

Time in s= 18.5 h x 3600 s / 1hr

= 66600 s

Since the reaction is first order and we are given its rate constant. We use integrated first law equation.

At = A exp (- kt)

Here [A]t is concentration or amount at time t, and [A] is initial amount. k and t are rate constant and time respectively.

Lets plug in…

At = 0.88935 mol x exp (- 1.0 E-5 x 66600 )

= 0.458 mol

Calculation of moles of O2

Moles of O2 produced = 0.458 mol N2O5 x 1 mol O2 / 2 mol N2O5

= 0.228 mol O2

Lets find partial pressure of O2

For this we use ideal gas law.

P = nRT/ V

P is in atm , n is mole , R is gas constant, T is temperature in K, V is volume in L

P = [0.228 x 0.08206 x 319.15/11.4 ] atm            ……..(we used temperature in K)

= 0.525 atm

So the partial pressure of O2 is 0.525 atm