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A solution of N2O5 in CCl4 yields by decomposition at 45°C , 5 ml of O2, 20 minutes after the start of the experiment and 10 ml of O2 after a very long time. The decomposition obeys 1st order kinetics. What volume of O2 would have evolved, 40 minutes after the start? 1) 7.5 ml 2) 7.25 ml 3) 1.5 ml 4) 6 ml

A solution of N2O5 in CCl4 yields by decomposition at 45°C , 5 ml of O2, 20 minutes after the start of the experiment and 10 ml of O2 after a very long time. The decomposition obeys 1st order kinetics. What volume of O2 would have evolved, 40 minutes after the start?
1) 7.5 ml 2) 7.25 ml 3) 1.5 ml 4) 6 ml

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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