61] At 273 K and 1 atm, 10 litre of N204decomposes to NO2 according to

1.	61] At 273 K and 1 atm, 10 litre of N204decomposes to NO2 according to equationNO 2 NO2(e)What is degree of dissociation (a) when theoriginal volume is 25% less than that of existingvolume?1) 0.252) 0.333) 0.664) 0.5
Answer» 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. this question the answer of 0.33

61] At 273 K and 1 atm, 10 litre of N204decomposes to NO2 according to equationNO 2 NO2(e)What is degree of dissociation (a) when theoriginal volume is 25% less than that of existingvolume?1) 0.252) 0.333) 0.664) 0.5

Answer»

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.

this question the answer of 0.33