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..
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Answered by
40
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.
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3
Answer:0.33
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