Please help me in the question of attached image of Chemical Kinetics.
Riya , 5 Years ago
Grade 12th pass
Physical Chemistry> Please help me in the question of attache...
1 Answers
Mayank
Last Activity: 5 Years ago
first we have: let initial amount of N2O5 be “a”ml
2N2O5 
4NO2 + O2
a 0 0
at t=20min a-x 2x x/2
at t=very long time 0 2a a/2
at t=40min a-y 2y y/2
now solving for t=very long time we get a/2=10 so a=20ml
now solving for t=20min we get
x/2=5 so x=10
also k(rate constant) * t(time)=ln(a/(a-x)) // put t=20
now putting value of a and x in above equation we get k= ln(2)/20
now solving for t=40 min
using equation kt=ln(a/a-y)
here according to question we have to find y/2
so ln(2)*40/20=ln(a/a-y)
now solving we get equation as
ln(4)=ln(a/a-y) and now
put value for a=20 and hence we get y=15ml so ans is option 1 that is y/2=15/2=7.5ml
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