A reaction takes place in various steps- The rate const- for 1st- 2nd- 3rd and 5th steps are k1-k2-k3 and k5 resp- The overall rate constant is given by k-k2k3k1k51-2- If activation energy are 40- 60- 50 and 10 kJ-mole resp- The overall energy of activation kJ-mole is-

The correct option is D 25 The correct option is C .Given, K= k _ 2 k _ 3 ( k _ 1 k _ 5 ) ^ 1 2 #160; Taking log both side we get, ln ...

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Standard XII

Chemistry

Rate Constant

A reaction ta...

Question

A reaction takes place in various steps. The rate const, for $1^{s t}, 2^{n d}, 3^{r d}$ and $5^{t h}$ steps are $k_{1}, k_{2}, k_{3}$ and $k_{5}$ resp. The overall rate constant is given by $k = \frac{k_{2}}{k_{3}} {(\frac{k_{1}}{k_{5}})}^{1 / 2}$ . If activation energy are $40, 60, 50$ and $10 k J / m o l e$ resp. The overall energy of activation $(k J / m o l e)$ is:

10

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20

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25

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none of these

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Solution

The correct option is D $25$
The correct option is $C$ .
Given,
$K = \frac{k_{2}}{k_{3}} (\frac{k_{1}}{k_{5}})^{\frac{1}{2}}$
Taking log both side we get,
$l n K = l n k_{2} - {l n k}_{3} + \frac{1}{2} ({l n k}_{1} - {l n k}_{5})$
We know that, $K = A e^{\frac{- E a}{R T}}$ ; $l n K = l n A e - \frac{E a}{R T}$
Let every step has the same arrhenius constant and E1,E2,E3 and E5 are the corresponding activation energy of 1st,2nd,3rd and 5th step respectively.
So, $K = (l n A - \frac{E a 2}{R T}) - (l n A - \frac{E a 3}{R T}) + \frac{1}{2} [(l n A - \frac{E a 1}{R t}) - (l n A - \frac{E a 5}{R T})$
$K = 60 - 50 + \frac{1}{2} (40 - 10) = 10 + 15 = 25$

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A reaction takes place in various steps. The rate const, for 1st,2nd,3rd and 5th steps are k1,k2,k3 and k5 resp. The overall rate constant is given by k=k2k3(k1k5)1/2. If activation energy are 40,60,50 and 10 kJ/mole resp. The overall energy of activation (kJ/mole) is: