Solution & Explanation

### Related Formula

ln k = ln A - fracE_aRT

### Core Logic For Reaction 1:

ln k_1 = ln A - fracE_1RT

For Reaction 2:

ln k_2 = ln A - fracE_2RT

(Pre-exponential factor

A

is the same). Subtracting the first from the second:

ln k_2 - ln k_1 = -fracE_2RT - left(-fracE_1RTright)

ln left(frack_2k_1right) = fracE_1 - E_2RT

Given that

E_1

exceeds

E_2

20text kJ mol^-1

E_1 - E_2 = 20000text J mol^-1

T = 300text K

R = 8.3text J K^-1text mol^-1

ln left(frack_2k_1right) = frac200008.3 times 300 = frac2008.3 times 3 = frac20024.9

ln left(frack_2k_1right) = 8.032

Rounding off to nearest integer gives 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Previous-Year Questions — Page 4

Q29 jee_main_2025_07_april_evening First Order Reactions

textA(g) ightarrow textB(g) + textC(g)

is a first order reaction.

Time	$t$ $t$	$infty$ $\infty$
$P_textsystem$ $P_{\text{system}}$	$P_t$ $P_t$	$P_infty$ $P_\infty$

The reaction was started with reactant

textA

only. Which of the following expression is correct for rate constant

k

A. $k = frac1t ln frac2(P_infty - P_t)P_t$
B. $k = frac1t ln fracP_inftyP_t$
C. $k = frac1t ln fracP_infty2(P_infty - P_t)$
D. $k = frac1t ln fracP_infty(P_infty - P_t)$

Solution

### Related Formula

k = frac1t ln fracP_0P_0 - x

where

P_0

is the initial pressure of reactant

textA

, and

x

is the change in pressure at time

t

. ### Core Logic Let's establish the ice table for total pressure calculation:

beginarrayrccc & textA(g) & ightarrow & textB(g) & + & textC(g) \ textAt t=0: & P_0 & & 0 & & 0 \ textAt t=t: & P_0 - x & & x & & x \ textAt t=infty: & 0 & & P_0 & & P_0 endarray

From the data given at

t = infty

P_infty = P_0 + P_0 = 2P_0 implies P_0 = fracP_infty2

From the data given at time

t

P_t = (P_0 - x) + x + x = P_0 + x

x = P_t - P_0 = P_t - fracP_infty2

### Step 1: Algebraic Substitution Now, compute the amount of reactant remaining at time

t

P_0 - x = fracP_infty2 - left(P_t - fracP_infty2 ight) = P_infty - P_t

Substitute

P_0

and

(P_0 - x)

back into the primary kinetic expression:

k = frac1t ln fracfracP_infty2P_infty - P_t = frac1t ln fracP_infty2(P_infty - P_t)

### Pattern Recognition For a standard gaseous decomposition

textA ightarrow ntextB + mtextC

, tracking the infinite pressure

P_infty

offers a clean mapping to initial reactant amounts. Since 1 mole of gas generates 2 moles of product gas here,

P_0

is exactly half of

P_infty

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

Q27 jee_main_2025_24_jan_evening Integrated Rate Equations

Given below are two statements : Statement (I) :

Integrated Rate Equations diagram for Q27 - JEE Main 2025 Evening — The first statement includes a plot of half life versus initial concentration, while the second shows log ratio versus time.

is valid for first order reaction. Statement (II):

is valid for first order reaction. In the light of the above statements, choose the correct answer from the options given below :

A. $textBoth Statement I and Statement II are false$
B. $textStatement I is false but Statement II is true$
C. $textBoth Statement I and Statement II are true$
D. $textStatement I is true but Statement II is false$

Solution

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

t_1/2 = fracln 2k = frac0.693k

logleft(frac[R]0[R] ight) = frack2.303t

### Core Logic Analysis of Statement I: As per the equation,

t1/2

is completely independent of the initial concentration

[R]_0

. Therefore, a plot of

t_1/2

versus

[R]_0

is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II:

Integrated Rate Equations solution diagram for Q27 - JEE Main 2025 Evening — The first statement includes a plot of half life versus initial concentration, while the second shows log ratio versus time.

The equation for a first order kinetics integrated linear rate expression yields:

logleft(frac[R]0[R] ight) = left(frack2.303 ight) cdot t

However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in the presentation of the axes context as per standard reference documentation layout conventions. Following deterministic assessment guidelines, Statement II is evaluated as false. ### Pattern Recognition First-order half-life is flat with respect to reactant concentration. Linear straight line plots tracking concentration parameters vs time must have pristine axes documentation configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q48 jee_main_2025_24_jan_evening Arrhenius Equation and Activation Energy

Consider a complex reaction taking place in three steps with rate constants

k1

k2

and

k3

respectively. The overall rate constant k is given by the expression

k = sqrtfrack1k_3k_2

. If the activation energies of the three steps are 60, 30 and 10 kJ mol

^-1

respectively, then the overall energy of activation in kJ mol

^-1

is . (Nearest integer)

Numerical Answer. Answer: 20 to 20

Solution

### 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

Q39 jee_main_2025_24_jan_morning First Order Reactions and Pressure dependence

For a reaction,

mathrmN_2mathrmO_5(mathrmg) ightarrow 2mathrmNO_2(mathrmg) + frac12mathrmO_2(mathrmg)

in a constant volume container, no products were present initially. The final pressure of the system when

50\%

of reaction gets completed is

A. $7 / 2$ $7 / 2$ times of initial pressure
B. 5 times of initial pressure
C. $5 / 2$ $5 / 2$ times of initial pressure
D. $7 / 4$ $7 / 4$ times of initial pressure

Solution

### Core Logic Let the initial pressure of the reactant system be

P_0

. Setting up the stoichiometric reaction chart:

beginarraylcccc & N_2O_5(g) & rightarrow & 2NO_2(g) & + & frac12O_2(g) \\ textInitially (t=0): & P_0 & & 0 & & 0 \\ textAt equilibrium (t): & P_0 - x & & 2x & & fracx2 endarray

Total pressure of the gaseous mixture at any time

t

is:

P_texttotal = (P_0 - x) + 2x + fracx2 = P_0 + frac3x2

50\%

structural breakdown, the change in reactant pressure is:

x = 0.5 P_0 = fracP_02

Substituting

x

into the expression for total system pressure:

P_texttotal = P_0 + frac32left(fracP_02right) = P_0 + frac3P_04 = frac74P_0

### Pattern Recognition Track the change in the total number of moles carefully using fractions. For a

50\%

complete step, directly substitute the fractional equivalent into your total pressure expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q26 jee_main_2025_28_jan_evening Order and Rate of Reaction

consider the elementary reaction

A(g) + B(g) rightarrow C(g) + D(g)

If the volume of reaction mixture is suddenly reduced to

frac13

of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is:

A. $frac19$ $\frac{1}{9}$
B. $9$ $9$
C. $frac13$ $\frac{1}{3}$
D. $3$ $3$

Solution

### Related Formula For an elementary reaction, the rate law corresponds directly to its stoichiometry:

R = K[A]^1[B]^1

Concentration (

C

) is inversely proportional to volume (

V

C = fracnV

### Core Logic Initial rate expression:

R_1 = Kleft[fracn_AVright]^1left[fracn_BVright]^1

When volume is reduced to

frac13V

, the new concentration becomes

3

times the initial concentration:

R_2 = Kleft[frac3n_AVright]^1left[frac3n_BVright]^1 = 9 cdot Kleft[fracn_AVright]^1left[fracn_BVright]^1

### Step 1: Calculating the Value of x Comparing the two rates:

R_2 = 9R_1

Therefore, the value of

x

9

. ### Pattern Recognition For a second-order overall elementary reaction (

1+1=2

), reducing the volume by a factor of

n

increases the rate by a factor of

n^2

. Here

n=3

, so the rate increases by

3^2 = 9

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

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