An organic compound dissociates into n parallel first-order reactions simultaneously and produces n different products P 1 , P 2 , P 3 , … P n having rate constants k , 2 k , 3 k , … nk and activation energies E , 2 E , 3 E , … nE , respectively. Calculate the overall energy of activation of the compound P .

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 n parallel reactions of dissociation of the compound P are taking place simultaneously as, P → k P 1 ; Activation energy = E ; P → 2 k P 2 ; Activation energy = 2 E ; P → 3 k P 3 ; Activation energy = 3 E ; P → nk 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 = 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, E a 1 , E a 2 . . . E a n 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 = E . k + 2 E . 2 k + 3 E . 3 k + . . . . . . . + nE . nk k + 2 k + 3 k + . . . . . . . + nk E a = E . k 1 2 + 2 2 + 3 2 + . . . . . . + n 2 k 1 + 2 + 3 + . . . . . . + n Sum of square of the first n natural numbers = n n + 1 2 n + 1 6 Sum of first n natural numbers = n n + 1 2 Therefore, E a = E . k n ( n + 1 ) ( 2 n + 1 ) 6 k n ( n + 1 ) 2 Solving the above equation, E a = ( 2 n + 1 ) 3 E Therefore, overall activation energy ( E a ) will be, 2 n + 1 3 E .

The thermal decomposition of N 2 O 5 occurs in the following steps. Step- I N 2 O 5 ⟶ slow NO 2 + NO 3 Step- II N 2 O 5 + NO 3 ⟶ fast 3 NO 2 + O 2 2 N 2 O 5 ⟶ 4 NO 2 + O 2 Suggest the rate expression.

The thermal decomposition of N 2 O 5 occurs in the following steps. Step- I N 2 O 5 ⟶ slow NO 2 + NO 3 Step- II N 2 O 5 + NO 3 ⟶ fast 3 NO 2 + O 2 2 N 2 O 5 ⟶ 4 NO 2 + O 2 But the rate of reaction is always determined by the slowest step. The slowest step in this reaction is the decomposition of N 2 O 5 . Reaction: N 2 O 5 ⟶ slow NO 2 + NO 3 Rate of reaction = k N 2 O 5 So, it is a first order reaction. Hence, the rate expression is k N 2 O 5 .

The reaction 2 NO + Br 2 ⟶ 2 NOBr , obeys the following mechanism. (i) NO + Br 2 = Fast NOBr 2 (ii) NOBr 2 + NO ⟶ Slow 2 NOBr Suggest the rate expression.

The rate expression is derived by ii step of mechanism, a slower step, thus, rate = K NOBr 2 NO . . . . . . i However, NOBr 2 is an intermediate and thus its concentration should be replaced from Eq. 1 For step 1 , equilibrium constant K C = NOBr 2 [ NO ] Br 2 ∴ NOBx 2 = K C NO Br 2 . . . . . . 2 Thus, by Eqs. 1 and 2 rate = K . K C NO 2 Br 2 rate = K ' NO 2 Br 2

The activity of a radioactive isotope falls to 12 . 5 % in 90 day. Compute the half-life and decay constant of isotope.

Given, if r 0 = 100 ; r = 12 . 5 ; t = 90 day; r 0 r = N 0 N = 100 12 . 5 ∵ λ = 2 . 303 t log 10 N 0 N = 2 . 303 90 log 10 100 12 . 5 = 2 . 31 × 10 - 2 day - 1 ∴ t 1 / 2 = 0 . 693 λ = 0 . 693 2 . 31 × 10 - 2 = 30 day Alternate solution N 0 N = 100 12 . 5 = 8 ∴ N = N 0 8 = N 0 2 3 = N 0 2 n Now, No. of halves ( n ) = 3 ∵ T = t 1 / 2 × n 90 = t 1 / 2 × 3 ∴ t 1 / 2 = 30 day

A sample of U 238 (half-life = 4 . 5 × 10 9 year) ore is found to contain 23 . 8 g of U 238 and 20 . 6 g of Pb 206 . Calculate the age of the ore.

We know, the reaction is: 92 U 238 ⟶ 82 P 206 + 8 He 2 4 + 6 - 1 e 0 A mole of Pb present = 20 . 6 206 = 0 . 1 g atom = U decay A mole of U the present = 23 . 8 238 = 0 . 1 g atom Thus, N = 0 . 1 g atom N 0 ∪ present + U decayed = 0 . 1 + 0 . 1 = 0 . 2 g atom According to the integrated law for the first-order reaction is, K = 2 . 303 t log N 0 N Now t = 2 . 303 × 4 . 5 × 10 9 0 . 693 log 10 0 . 2 0 . 1 t = 4 . 5 × 10 9 year. So, the age of the ore is t = 4 . 5 × 10 9 the year.

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The decomposition of hydrogen peroxide in an aqueous solution is a first order reaction. It can be studied by titrating quickly $10 mL$ portions of reactions mixture at various times from the $t = 0$ of reaction against a standard solution of ${KMnO}_{4}$ . Volume of ${KMnO}_{4}$ solution used in each case is proportional to the remaining concentration of $H_{2} O_{2} .$

From the following data calculate the rate constant of the reaction.

Time (seconds) $0$ $600$ $1200$

${KMnO}_{4}$ solution used $(mL)$ $22.8$ $13.8$ $8.2$

Important Questions on Chemical Kinetics

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 20

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

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 22

Find out the percentage of the reactant molecules crossing over the activation energy barrier at

325 K,

given that,

∆ H_{325} = 0.12 kcal, E_{a (b)} = + 0.02 kcal .

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 25

The thermal decomposition of $N_{2} O_{5}$ occurs in the following steps.

Step- $I$ $N_{2} O_{5} \overset{slow}{⟶} {NO}_{2} + {NO}_{3}$

Step- $II$ $N_{2} O_{5} + {NO}_{3} \overset{fast}{⟶} 3 {NO}_{2} + O_{2}$

$2 N_{2} O_{5} ⟶ 4 {NO}_{2} + O_{2}$

Suggest the rate expression.

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 26

The reaction $2 NO + {Br}_{2} ⟶ 2 NOBr$ , obeys the following mechanism.

(i) $NO + {Br}_{2} \overset{Fast}{=} {NOBr}_{2}$
(ii) ${NOBr}_{2} + NO \overset{Slow}{⟶} 2 NOBr$

Suggest the rate expression.

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 27

Given the following steps in the mechanism for a chemical reaction:

$A + B ⟶ C$ (fast)

$B + C ⟶ D + E$ (Slow)

$D + F ⟶ A + E$ (fast)

At any time $[C]$ is directly proportional to $[A] .$
(a) What is the stoichiometric equation for the reaction?
(b) Which species, if any, are catalysts in this reaction ?
(c) Which species, if any, are intermediates in this reaction?
(d) Write the rate law for the rate-determining step.
(e) Write the rate law for this reaction.
(f) What is overall order of the reaction ?

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 28

The activity of a radioactive isotope falls to

12.5 %

90

day. Compute the half-life and decay constant of isotope.

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 30

A sample of

U^{238}

(half-life

= 4.5 \times 10^{9}

year) ore is found to contain

23.8 g

U^{238}

and

20.6 g

{Pb}^{206} .

Calculate the age of the ore.

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Alpha Question Bank for Engineering: Chemistry>Chapter 10 - Chemical Kinetics>Exercise-1>Q 31

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Rate of formation of product at $20$ seconds is:

Time (seconds)	$0$	$600$	$1200$
${KMnO}_{4}$ solution used $(mL)$	$22.8$	$13.8$	$8.2$

Important Questions on Chemical Kinetics

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