Name: Mechanism of a Chemical Reaction
Uploaded: 2019-07-19
Duration: 5 min 23 s
Description: An energy profile diagram theoretically represents a single energetic pathway for a chemical reaction. Learn the reaction mechanism with an energy diagram.

Question

An organic compound dissociates into

n

parallel first-order reactions simultaneously and produces

n

different products

P_{1}, P_{2}, P_{3}, \dots P_{n}

having rate constants

k, 2 k, 3 k, \dots nk

and activation energies

E, 2 E, 3 E, \dots nE

, respectively. Calculate the overall energy of activation of the compound

P

.

Accepted Answer

In a parallel first-order reaction, multiple products are formed simultaneously from one reactant through multiple reactions, and each reaction follows first-order kinetics.

Given that $P_{1}, P_{2}, P_{3}, \dots P_{n}$ parallel reactions of dissociation of the compound $P$ are taking place simultaneously as,

$P \overset{k}{\to} P_{1}; Activation energy = E; P \overset{2 k}{\to} P_{2}; Activation energy = 2 E; P \overset{3 k}{\to} P_{3}; Activation energy = 3 E; P \overset{nk}{\to} P_{n}; Activation energy = nE .$

The overall activation energy $(E_{a})$ for the above type of $n$ parallel reactions can be determined by the formula,

$E_{a} = \frac{E_{a_{1}} K_{1} + E_{a_{2}} K_{2} + E_{a_{3}} K_{3} . . . . . . + E_{a_{n}} K_{n}}{K_{1} + K_{2} + K_{3} . . . . . + K_{n}}$

Where, $k, 2 k, 3 k, \dots nk$ are the activation energies of parallel reaction,

and, $K_{1}, K_{2} . . . . K_{n}$ are the rate constants.

Now, putting the values in the above formula for overall activation energy,

$E_{a} = \frac{E . k + 2 E . 2 k + 3 E . 3 k + . . . . . . . + nE . nk}{k + 2 k + 3 k + . . . . . . . + nk} E_{a} = \frac{E . k (1^{2} + 2^{2} + 3^{2} + . . . . . . + n^{2})}{k (1 + 2 + 3 + . . . . . . + n)}$

Sum of square of the first $n$ natural numbers

$= \frac{n (n + 1) (2 n + 1)}{6}$

Sum of first $n$ natural numbers

$= \frac{n (n + 1)}{2}$

Therefore,

$E_{a} = \frac{E . k (\frac{n (n + 1) (2 n + 1)}{6})}{k (\frac{n (n + 1)}{2})}$

Solving the above equation,

$E_{a} = \frac{(2 n + 1)}{3} E$

Therefore, overall activation energy $(E_{a})$ will be,

$\frac{(2 n + 1)}{3} E$ .

Important Questions on Chemical Kinetics

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An organic compound dissociates into n parallel first-order reactions simultaneously and produces n different products P1, P2, P3,…Pn having rate constants k, 2k, 3k,…nk and activation energies E, 2E, 3E,…nE, respectively. Calculate the overall energy of activation of the compound P.

Important Questions on Chemical Kinetics

Learn and Prepare for any exam you want !