Question

At 273k and 1 atm 10 l of n2o4 decomposes to 2no2. what is degree of dissociation when original volume is 25%less than that of existiong volume? Plz answer this..

Answer 1

The degree of dissociation will be 0.33.

Explanation:

Let the initial volume of N₂O₄ be x, and initial volume of NO₂ is 0

If the degree of dissociation is a, then final volume of N₂O₄ is x(1-a) and that of NO₂ is 2ax

                          N₂O₄------------------> 2NO₂

initial                     x                              0

At equilibrium      x(1-a)                        2ax

Hence total initial volume = x + 0 = x

Total Final volume = x(1-a) + 2ax = x + ax = x(1+a)

It is given that the initial volume is 25% less than the final volume

=> x = 0.75* x(1+a)

=> 1+a = 1.33

=> a = 0.33

Hence the degree of dissociation will be 0.33.

Answer 2

Answer:0.33

Explanation: