Consider a complex reaction taking place in three steps with rate constants $k1$ $k{1}$ , $k2$ $k{2}$ and $k3$ $k{3}$ respectively. The overall rate constant k is given by the expression $k = sqrtfrack1k_3k_2$ $k = \sqrt{\frac{k{1}k_{3}}{k_{2}}}$ . If the activation energies of the three steps are 60, 30 and 10 kJ mol $^-1$ $^{-1}$ respectively, then the overall energy of activation in kJ mol $^-1$ $^{-1}$ is . (Nearest integer)

Numerical Answer Type:

Enter a numerical value Answer: 20 to 20 +4 marks

Solution & Explanation

### Related Formula From the Arrhenius equation, rate constants vary exponentially with temperature:

k = A cdot e^-E_a / RT

When rate constants combine multiplicatively or via roots, their corresponding activation energies combine linearly. ### Core Logic Given the overall rate constant expression:

k = left(frack_1 cdot k_3k_2 ight)^1/2

Substitute the Arrhenius expression (

k_i = A_i cdot e^-Eai/RT

) for each rate constant:

A cdot e^-E_a/RT = left[frac(A_1 cdot e^-Ea1/RT) cdot (A_3 cdot e^-Ea3/RT)A_2 cdot e^-Ea2/RT ight]^1/2

Equating the exponential terms yields the linear relationship for the overall activation energy (

E_a

fracE_aRT = frac12 left(fracE_a1RT + fracE_a3RT - fracE_a2RT ight)

E_a = fracE_a1 + E_a3 - E_a22

Substitute the given activation energy values (

E_a1 = 60, E_a2 = 30, E_a3 = 10text kJ/mol

E_a = frac60 + 10 - 302 = frac402 = 20text kJ mol^-1

The overall activation energy is

20text kJ/mol

. ### Pattern Recognition Shortcut: Convert the rate constant algebraic expression directly into an activation energy formula by swapping

k

for

E_a

, turning multiplications into additions, divisions into subtractions, and powers into multipliers. Here,

k = (k_1 k_3 / k_2)^1/2

translates directly to

E_a = frac12(E_a1 + E_a3 - E_a2)

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

Chemical Kinetics

More Chemical Kinetics Previous-Year Questions — Page 2

Q28 2025 First Order Reactions

Reaction

mathrmA(g) rightarrow 2mathrmB(g) + mathrmC(g)

is a first order reaction. It was started with pure A.

$t / textmin$ $t / \text{min}$	Pressure of system at time $t / textmm Hg$ $t / \text{mm Hg}$
10	160
$infty$ $\infty$	240

Which of the following options is incorrect?

A. $textInitial pressure of A is 80 mm Hg$
B. $textThe reaction never goes to completion$
C. $textRate constant of the reaction is 1.693 min^-1$
D. $textPartial pressure of A after 10 minute is 40 mm Hg$

Solution

### Related Formula

k = frac2.303t logleft(fracP_0P_tright)

### Core Logic For the reaction:

mathrmA(g) rightarrow 2mathrmB(g) + mathrmC(g)

- At

t=0

, pressure of

A = P_0

, while

B = 0

and

C = 0

. - At

t=infty

A

is completely consumed, leaving

2P_0

B

and

P_0

C

P_infty = 3P_0 = 240 text mm Hg implies P_0 = 80 text mm Hg

This confirms option (1) is correct. At any time

t

, pressure of

A = P_0 - x

B = 2x

C = x

P_t = P_0 + 2x = 80 + 2x

t=10 text min

P_10 = 160 text mm Hg

80 + 2x = 160 implies x = 40 text mm Hg

Thus, partial pressure of

A

after

10 text min

is:

P_A = P_0 - x = 80 - 40 = 40 text mm Hg

This confirms option (4) is correct. Now, calculate the rate constant

k

k = frac110 lnleft(frac8040 ight) = fracln 210 = 0.0693 text min^-1

Therefore, option (3) which states

k = 1.693 text min^-1

is incorrect. ### Pattern Recognition At

t=infty

, the total pressure is

3

times the initial pressure of

A

. So,

P_0 = P_infty / 3 = 80 text mm Hg

. Half-life

t_1/2 = 10 text min

since

P_A

drops from

80

40

10 text min

. Thus,

k = 0.693 / 10 = 0.0693 text min^-1

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q26 2025 First Order Reactions

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are

t_1

and

t_2

(s), respectively. The ratio

t_1 / t_2

will: Choose the correct answer from the options given below:

A. $frac43$ $\frac{4}{3}$
B. $frac32$ $\frac{3}{2}$
C. $frac34$ $\frac{3}{4}$
D. $frac23$ $\frac{2}{3}$

Solution

### Related Formula For a first-order reaction:

t = frac2.303k logleft(fracC_0C_tright)

Alternatively, using half-life

t_50\%

C_t = fracC_02^n

where

n

is the number of half-lives (

n = fractt_50\%

). ### Core Logic Step 1: Calculate

t_1

for decomposition to

frac14

of initial concentration:

C_t = fracC_04 = fracC_02^2 implies n = 2

t_1 = 2 cdot t_50\%

Step 2: Calculate

t_2

for decomposition to

frac18

of initial concentration:

C_t = fracC_08 = fracC_02^3 implies n = 3

t_2 = 3 cdot t_50\%

Step 3: Find the ratio

fract_1t_2

fract_1t_2 = frac2 cdot t_50\%3 cdot t_50\% = frac23

### Pattern Recognition For first-order kinetics, every step of concentration halving takes exactly one half-life (

t_50\%

textInitial xrightarrowt_50\% frac12 xrightarrowt_50\% frac14 quad text(Total 2 text half-lives)

frac14 xrightarrowt_50\% frac18 quad text(Total 3 text half-lives)

Therefore, the ratio is simply the ratio of the number of half-lives:

2 : 3

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q42 2025 First Order Reaction Kinetics

Drug X becomes ineffective after

50\%

decomposition. The original concentration of drug in a bottle was

16text mg/mL

which becomes

4text mg/mL

in 12 months. The expiry time of the drug in months is (Assume that the decomposition of the drug follows first order kinetics). (1) 12 (2) 2 (3) 3 (4) 6

A. 12
B. 2
C. 3
D. 4

Solution

### Related Formula

N_t = N_0 left(frac12 ight)^n

### Core Logic Let's track concentration reductions:

16text mg/mL xrightarrowt_1/2 8text mg/mL xrightarrowt_1/2 4text mg/mL

This total progression constitutes exactly

2

half-lives (

n = 2

2 cdot t_1/2 = 12text months implies t_1/2 = 6text months

Since the drug becomes ineffective right after

50\%

decomposition, its functional expiry limit is exactly

1

half-life period.

textExpiry time = t_1/2 = 6text months

### Pattern Recognition For multi-step concentration halving, bypass complex integrated logarithmic rate expressions by directly applying integer half-life steps (

16 ightarrow 8 ightarrow 4

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q32 2025 Order of Reaction and Half-life

For a given reaction

mathrmRrightarrow mathrmP,mathrmt_1 / 2

is related to

[mathrmA]_0

as given in table :

$[A]_0 / mathrmmol\,L^-1$ $[A]_0 / \mathrm{mol\,L^{-1}}$	$t_1/2 / mathrmmin$ $t_{1/2} / \mathrm{min}$
0.100	200
0.025	100

Given:

log 2 = 0.30

Which of the following is true? A. The order of the reaction is

frac12

. B. If

[mathrmA]_0

1mathrmM

, then

mathrmt_1/2

200sqrt10

min C. The order of the reaction changes to 1 if the concentration of reactant changes from

0.100 mathrmM

0.500 mathrmM

. D.

t_1 / 2

800 mathrm~min

for

[mathrmA]_0 = 1.6 mathrmM

Choose the correct answer from the options given below:

A. $textA and C only$ $\text{A and C only}$
B. $textA and B only$ $\text{A and B only}$
C. $textA, B and D only$ $\text{A, B and D only}$
D. $textC and D only$ $\text{C and D only}$

Solution

### Related Formula The dependence of half-life on initial concentration is given by:

t_1/2 propto frac1[A]_0^n-1

### Step 1: Finding the reaction order (n) Using the values provided:

frac(t_1/2)_1(t_1/2)_2 = left( frac[A]_0,2[A]_0,1 right)^n-1

frac200100 = left( frac0.0250.100 right)^n-1 Rightarrow 2 = left( frac14 right)^n-1

2 = 2^-2(n-1) Rightarrow 1 = -2n + 2 Rightarrow n = frac12

Hence, statement A is correct. ### Step 2: Checking half-life at other concentrations Since

n = frac12

t_1/2 propto sqrt[A]_0

. - For

[A]_0 = 1\,mathrmM

frac200t_1/2 = sqrtfrac0.11 Rightarrow t_1/2 = 200sqrt10\,mathrmmin

Hence, statement B is correct. - For

[A]_0 = 1.6\,mathrmM

frac200t_1/2 = sqrtfrac0.11.6 = sqrtfrac116 = frac14 Rightarrow t_1/2 = 800\,mathrmmin

Hence, statement D is correct. ### Pattern Recognition Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of

[A]_0

by 4 causes reduction of

t_1/2

by 2

rightarrow

indicates a square root dependence (

t_1/2 propto sqrtA_0

), which implies

n = 0.5

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q30 2025 First Order Kinetics

In a reaction

A+B ightarrow C

, initial concentrations of A and B are related as

[A]_0=8[B]_0

. The half lives of A and B are 10 min and 40 min. respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

A. 60 min
B. 80 min
C. 20 min
D. 40 min

Solution

### Related Formula For a first-order integrated rate law expression:

[A]_t = [A]_0 e^-k_A t quad textwhere k = fracln 2t_1/2

### Core Logic We require the instantaneous concentration to be identical at time

t

[A]_t = [B]_t implies [A]_0 e^-k_A t = [B]_0 e^-k_B t

frac[A]_0[B]_0 = e^(k_A - k_B)t

### Step 1: Substituting Parameters Substitute

[A]_0 = 8[B]_0

and express rate constants in terms of half-lives:

8 = e^(k_A - k_B)t implies ln 8 = (k_A - k_B)t

3ln 2 = ln 2 left( frac1(t_1/2)_A - frac1(t_1/2)_B ight) times t

3 = left( frac110 - frac140 ight) times t implies 3 = frac340 times t implies t = 40text min.

### Pattern Recognition Shortcut: Express the concentration drop using half-life indices:

[A]_t = frac[A]_02^t/10 = frac8[B]_02^t/10

[B]_t = frac[B]_02^t/40

Equating both:

8 cdot 2^-t/10 = 2^-t/40 implies 2^3 = 2^fract10 - fract40 implies 3 = frac3t40 implies t = 40text min.

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

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