Question

At 273 K and 1 atm, 10 litre of $\sf{N_{2}O_{2}}$ decomposes
⠀⠀⠀⠀
$\sf{N_{2}O_{2}\implies 2NO_{O}}$
⠀⠀⠀⠀
What is the degree of dissociation when the original volume is 25% less than that of existing volume?​

Answer 1

Explanation:

QUESTION :-

At 273 K and 1 atm, 10 litre of $\sf{N_{2}O_{2}}$ decomposes

⠀⠀⠀⠀

⠀⠀⠀⠀What is the degree of dissociation when the original volume is 25% less than that of existing volume?

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.

_____________________________________________

@TheStellar♥️