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for the reactions, n205........>2no2+1/2o2 given -d[n205]/dt=k1[n2o5], d[no2]/dt=k2[n2o5] ,d[o2]/dt=k3[n2o5] the relation between k1, k2, k3 is

for the reactions, n205........>2no2+1/2o2 given -d[n205]/dt=k1[n2o5], d[no2]/dt=k2[n2o5] ,d[o2]/dt=k3[n2o5]  the relation between k1, k2, k3 is   

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
From the concepts of chemcial kinetcs we know,
-d(n205)/dt = (+1/2)d(no2)/dt = (+2)d(o2)/dt
and 
given -d[n205]/dt=k1[n2o5], d[no2]/dt=k2[n2o5] ,d[o2]/dt=k3[n2o5] .
Now put the values,
we get,
k1[n2o5] = 0.5k2[n2o5]
k1 = 0.5k2
and
(+1/2)k2[n2o5]  = (+2)k3[n2o5] .
k2 = 4k3
Thus,
k1/k2 = (0.5/4)(k2/k3)
k1*k3 = (k2^2)/8
is the final relation.

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