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 .

Derive the equilibrium constant for decomposition of H 2 O 2 from the following data Time in minute 10 15 20 25 ∞ Volume of O 2 given by H 2 O 2 6 . 30 8 . 95 11 . 40 13 . 5 35 . 75

Decomposition of H 2 O 2 is first-order reaction. H 2 O 2 decomposes as, H 2 O 2 → H 2 O + 1 2 O 2 Integrated first-order rate law equation is, k = 2 . 303 t log a a - x . . . . . . . ( 1 ) Where, a = initial concentration, a - x = amount of time left after time t . Now, volume of oxygen gas collected at any time t ( V t ) ∞ amount of H 2 O 2 decomposed ( x ) x ∝ V t Volume of oxygen gas collected in infinite time ( V ∞ ) ∞ amount of H 2 O 2 initially taken a ∝ V ∞ Put values of a and x in 1 equation, we get, k = 2 . 303 t log V ∞ V ∞ - V t k = 2 . 303 10 log 35 . 75 35 . 75 - 6 . 30 k = 0 . 2303 log 1 . 214 k = 1 . 9 × 10 - 2

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For a reaction, activation energy for the forward reaction is $50 kJ$ and for the backward reaction is $30 kJ$ . Which of the following are true for the reaction?

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Important Questions on Chemical Kinetics

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.5>Q 19

During decomposition of an activated complex

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.5>Q 20

Which of the following statements is/are correct?

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 1

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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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 2

For a consecutive first-order reaction $A \overset{k_{1}}{⟶} B \overset{k_{2}}{⟶} C$ , the values of $k_{1}$ and $k_{2}$ are $45 {hr}^{- 1}$ and $15 {hr}^{- 1}$ , respectively. If the reaction is carried out with pure $A$ at a concentration of $1.0 mol {dm}^{- 3}$ . How much time will be required for the concentration of $B$ to reach a maximum?

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 2

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 2

For a consecutive first-order reaction $A \overset{k_{1}}{⟶} B \overset{k_{2}}{⟶} C$ the values of $k_{1}$ and $k_{2}$ are $45 {hr}^{- 1}$ and $15 {hr}^{- 1}$ , respectively. If the reaction is carried out with pure $A$ at a concentration of $1.0 mol {dm}^{- 3}$ , what will be the composition of the reacting system after a time interval of $10$ minutes?

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 3

The formation in water of $d$ -potassium chromo-oxalate from its $l$ -form is a reversible reaction, which is first-order in both directions, the racemate being the equilibrium product. A polarimeter experiment at $22 ° C$ showed that after $560 s, 12 mole$ $%$ of the $l$ -isomer was converted to the $d$ -form. Find the rate constant for the forward and the reverse reactions.

Given $\ln (0.76) = - 0.2744$

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.6>Q 4

Derive the equilibrium constant for decomposition of $H_{2} O_{2}$ from the following data

Time in minute	$10$	$15$	$20$	$25$	$\infty$
Volume of $O_{2}$ given by $H_{2} O_{2}$	$6.30$	$8.95$	$11.40$	$13.5$	$35.75$

For a reaction, activation energy for the forward reaction is 50 kJ and for the backward reaction is 30 kJ. Which of the following are true for the reaction?

Important Questions on Chemical Kinetics

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For a reaction, activation energy for the forward reaction is $50 kJ$ and for the backward reaction is $30 kJ$ . Which of the following are true for the reaction?