The rate of first order reaction is 0.04 mol L-1s-1 at 10 minutes and 0.03 mol L-1s-1 at 20 minutes after initiation…

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Q84 jee_main_2024_01_february_morning Kinetics of Radioactive Decay

The ratio of

¹⁴C¹²C

in a piece of wood is

(1)/(8)

part that of atmosphere. If half life of

¹⁴C

is 5730 years, the age of wood sample is .... years.

Numerical Answer. Answer: 17190 to 17190

Solution

Related Formula

N = (N₀)/(2ⁿ)

where $n = tt1/2$ $n = \frac{t}{t_{1/2}}$ (number of half-lives).

Alternatively, using the first-order decay formula:

t = (2.303)/(λ) ( (N₀)/(Nₜ) )

where $λ = 0.693t1/2$ $\lambda = \frac{0.693}{t_{1/2}}$.

Core Logic

The atmospheric ratio of $¹⁴C/¹²C$ $^{14}\mathrm{C}/^{12}\mathrm{C}$ acts as the initial activity or amount ( $N₀$ $N_0$) when the tree was alive. The current ratio in the wood represents the amount left at time $t$ $t$ ( $Nₜ$ $N_t$). Given that $Nₜ = (1)/(8) N₀$ $N_t = \frac{1}{8} N_0$.

Step 1: Calculate Half-lives

(Nₜ)/(N₀) = (1)/(8)

((1)/(2))ⁿ = (1)/(8) = ((1)/(2))³

So, the number of half-lives passed, $n = 3$ $n = 3$.

Step 2: Calculate Age

t = n \times t1/2

t = 3 \times 5730 years

t = 17190 years

Pattern Recognition

Whenever the remaining fraction is a perfect power of $1/2$ $1/2$ (like $1/2, 1/4, 1/8, 1/16$ $1/2, 1/4, 1/8, 1/16$), just find the exponent $n$ $n$ and multiply by $t1/2$ $t_{1/2}$. Here, $1/8 = (1/2)³ arrow 3$ $1/8 = (1/2)^3 \rightarrow 3$ half-lives.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q jee_main_2024_29_january_evening First Order Kinetics and Half Life

The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is ________

\times 10⁻²

. (Given antilog 0.2006 = 1.587)

Numerical Answer. Answer: 63 to 63

Solution

Related Formula

k = 0.693t1/2 and t = (2.303)/(k) ₁₀ ((a)/(a-x))

Core Logic

Given $t1/2 = 36 hours$ $t_{1/2} = 36\text{ hours}$, calculate the decay constant ( $k$ $k$):

k = (0.693)/(36) = 0.01925 hr⁻¹

We want to find the fraction remaining after $1 day = 24 hours$ $1\text{ day} = 24\text{ hours}$:

₁₀ ((a)/(a-x)) = (k \times t)/(2.303) = (0.01925 \times 24)/(2.303) = 0.2006

Step 1: Antilog Application

Taking the antilog on both sides:

(a)/(a-x) = 1.587 Fraction remaining ((a-x)/(a)) = (1)/(1.587) \approx 0.6301

Expressing the remaining fraction in the requested format:

0.6301 = 63 \times 10⁻²

Thus, the required integer value is 63.

Pattern Recognition

Ensure all time variables are in matching units (hours) before substituting values into first-order kinetic equations.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q82 jee_main_2024_27_jan_morning Determination of Order of Reaction

Consider the following data for the given reaction:

2HI(g) arrow H2(g) + I2(g)

Experiment	$[HI] (mol L⁻¹)$	Rate $(mol L⁻¹s⁻¹)$
1	0.005	$7.5 \times 10⁻⁴$ $7.5 \times 10^{-4}$
2	0.01	$3.0 \times 10⁻³$ $3.0 \times 10^{-3}$
3	0.02	$1.2 \times 10⁻²$ $1.2 \times 10^{-2}$

The order of the reaction is

.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Rate law relation expression:

R = k[HI]ⁿ

where $n$ $n$ represents the overall reaction order indicator.

Step 1: Set up ratios using data subsets

Comparing data from experiment 1 and experiment 2:

(R₂)/(R₁) = 3.0 \times 10⁻³7.5 \times 10⁻⁴ = ((0.01)/(0.005))ⁿ

$4 = (2)ⁿ$ $4 = (2)^n$

2² = 2ⁿ n = 2

Pattern Recognition

Doubling concentration ( $0.005 arrow 0.01$ $0.005 \rightarrow 0.01$) increases the reaction rate by 4 times ( $7.5 \times 10⁻⁴ arrow 3.0 \times 10⁻³$ ). Hence, it is a clear second-order ( $2² = 4$ $2^2 = 4$) dynamic pattern.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q83 jee_main_2024_29_jan_morning Arrhenius Equation and Activation Energy

For a reaction taking place in three steps at same temperature, overall rate constant

K = K₁K₂K₃

. If

Ea₁

,

Ea₂

and

Ea₃

are 40, 50 and 60 kJ/mol respectively, the overall Ea is ______ kJ/mol.

Numerical Answer. Answer: 30 to 30

Solution

Related Formula

K = A e-Eₐ/RT

Core Logic

Given the relationship between the rate constants:

K = (K₁ \cdot K₂)/(K₃)

Substituting the Arrhenius equation for each rate constant:

A \cdot e-Eₐ/RT = A₁ \cdot e^-Eₐ₁/RT \cdot A₂ \cdot e^-Eₐ₂/RTA₃ \cdot e^-Eₐ₃/RT

Combining the exponential terms using rules of exponents:

A \cdot e-Eₐ/RT = ((A₁ \cdot A₂)/(A₃)) \cdot e^ -(Eₐ₁ + Eₐ₂ - Eₐ₃)RT

Step 1: Equating Activation Energies

By comparing the powers of $e$ $e$ on both sides, the overall activation energy $Eₐ$ $E_a$ is related to the individual steps as follows:

Eₐ = Eₐ₁ + Eₐ₂ - Eₐ₃

Substitute the given values ( $Eₐ₁ = 40$ $E_{a1} = 40$, $Eₐ₂ = 50$ $E_{a2} = 50$, $Eₐ₃ = 60$ $E_{a3} = 60$ kJ/mol):

Eₐ = 40 + 50 - 60

$Eₐ = 90 - 60$ $E_a = 90 - 60$

Eₐ = 30 kJ/mol

Pattern Recognition

When rate constants are multiplied or divided ( $K = K₁^a K₂^b / K₃^c$ $K = K_1^a K_2^b / K_3^c$), the corresponding overall activation energy follows the linear combination of the exponents: $Eₐ = a Eₐ₁ + b Eₐ₂ - c Eₐ₃$ $E_a = a E_{a1} + b E_{a2} - c E_{a3}$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q82 jee_main_2024_30_january_evening Rate of Chemical Reaction

NO₂

required for a reaction is produced by decomposition of

N₂O₅

in

CCl₄

as by equation

2N₂O5(g) arrow 4NO2(g) + O2(g)

The initial concentration of

N₂O₅

is

3 mol L⁻¹

and it is

2.75 mol L⁻¹

after 30 minutes. The rate of formation of

NO₂

is

x \times 10⁻³ mol L⁻¹ min⁻¹

, value of

x

is

Numerical Answer. Answer: 17 to 17

Solution

Related Formula

Rate of Reaction (ROR) = -(1)/(2) Δ [N₂O₅]Δ t = (1)/(4) Δ [NO₂]Δ t

Core Logic

First, find the rate of disappearance of $N₂O₅$ $\mathrm{N}_2\mathrm{O}_5$.

- Δ [N₂O₅]Δ t = - ((2.75 - 3))/(30) = (0.25)/(30) mol L⁻¹ min⁻¹

Now, equate it to the general Rate of Reaction:

ROR = -(1)/(2) Δ [N₂O₅]Δ t = (1)/(2) ((0.25)/(30)) = (0.125)/(30) = (1)/(240) mol L⁻¹ min⁻¹

Step 1: Calculate the Rate of Formation of $NO₂$ $\mathrm{NO}_2$

Rate of formation of $NO₂ = Δ [NO₂]Δ t = 4 \times ROR$

= 4 \times (1)/(240) = (1)/(60) mol L⁻¹ min⁻¹

Convert this to scientific notation to find $x$ $x$:

(1)/(60) \approx 0.01666 = 16.66 \times 10⁻³ mol L⁻¹ min⁻¹

Rounding to the nearest integer, we get $x = 17$ $x = 17$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

The rate of first order reaction is 0.04 mol L⁻¹s⁻¹$0.04\text{ mol }L^{-1}s^{-1}$ at 10 minutes and 0.03 mol L⁻¹s⁻¹$0.03\text{ mol }L^{-1}s^{-1}$ at 20 minutes after initiation. Half life of the reaction is ________ minutes. (Given 2=0.3010$\log 2=0.3010$, 3=0.4771$\log 3=0.4771$)

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up equations

Step 2: Solving for k

Step 3: Calculating half life

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Half-lives

Step 2: Calculate Age

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Antilog Application

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Set up ratios using data subsets

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating Activation Energies

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Rate of Formation of NO₂$\mathrm{NO}_2$

Chapter Mix

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Step 1: Calculate the Rate of Formation of $NO₂$ $\mathrm{NO}_2$