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Please help me in the question of attached image of Chemical Kinetics.

Please help me in the question of attached image of Chemical Kinetics.

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Grade:12th pass

1 Answers

Mayank
368 Points
4 years ago
 
first we have: let initial amount of  N2O5   be “a”ml
                                                  2N2O5                         \leftrightharpoons                    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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