Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The overall order of a chemical reaction is determined by summing the exponents of the concentration terms in the experimental rate law, expressed as $r = k[\mathrm{NO}]^x [\mathrm{Cl}_2]^y$. By observing the rate changes as reactant concentrations are systematically varied, the specific dependency on each reactant can be isolated and quantified.

When comparing experimental runs where the concentration of $\mathrm{NO}$ doubles from $0.1$ to $0.2$ while the concentration of $\mathrm{Cl}_2$ remains constant at $0.2$, the reaction rate increases from $0.35$ to $1.40$. This fourfold increase in rate upon doubling the concentration of $\mathrm{NO}$ indicates that the order with respect to $\mathrm{NO}$ is $x = 2$, because $2^2 = 4$. Next, comparing the runs where the concentration of $\mathrm{Cl}_2$ doubles from $0.1$ to $0.2$ while keeping the concentration of $\mathrm{NO}$ constant at $0.1$, the reaction rate increases from $0.18$ to $0.35$. This twofold increase in rate confirms that the order with respect to $\mathrm{Cl}_2$ is $y = 1$, as $2^1 = 2$. Adding these exponents together, $x + y = 2 + 1 = 3$, provides the overall order of the reaction.

First-order kinetics follow an integrated rate law that correlates the rate constant with the reduction of reactant concentration over time. The fundamental expression for this calculation is:

$k = \frac{2.303}{t} \log_{10} \frac{A_0}{A_t}$

In this reaction, the initial concentration $A_0$ is 100, the remaining concentration $A_t$ is 32, and the elapsed time $t$ is 570 s. Substituting these values into the formula gives $k = \frac{2.303}{570} \log_{10} \frac{100}{32}$. Recognizing that the fraction $\frac{100}{32}$ can be written as $\frac{10^2}{2^5}$, the logarithmic term simplifies to $\log_{10} 100 - \log_{10} 2^5$, which is $2 - 5 \log_{10} 2$. Applying the approximation $\log_{10} 2 = 0.3$, the expression in the parentheses becomes $2 - 5(0.3)$, resulting in 0.5. Multiplying 2.303 by 0.5 and dividing by 570 yields a rate constant of $2 \times 10^{-3} \mathrm{s}^{-1}$.

The kinetics of a first-order reaction are governed by the integrated rate law, which defines the rate constant as $k = \frac{2.303}{t} \log \frac{[A_0]}{[A_t]}$. Given an initial concentration of 50 mol L⁻¹ and a final concentration of 10 mol L⁻¹ after 120 minutes, we substitute these values into the equation:

$k = \frac{2.303}{120} \log \frac{50}{10}$

Since the problem provides the value $\log 5 = 0.6989$, the expression simplifies to $k = \frac{2.303 \times 0.6989}{120}$. Performing the calculation yields approximately 0.013413 min⁻¹. This result is equivalent to $1.3413 \times 10^{-2} \text{ min}^{-1}$, where the value of $x$ is 1.34. Rounding to the nearest integer provides 1.

The rate constant $k$ is determined by rearranging the reaction rate law equation $k = \frac{\text{Rate}}{[A]^n}$ to solve for $k$. By substituting the standard units for reaction rate, mol L^-1 s^-1, and concentration, mol L^-1, into this equation, the expression simplifies mathematically to the final dimension of mol^{1-}n L^n-1 s^-1.

First-order kinetics dictate that the time needed for a reaction to reach a specific extent of completion depends on the rate constant through the relationship $k t = \ln \left( \frac{[A]_0}{[A]_t} \right)$, where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$. Given that 40% of the gaseous cyclobutene has isomerized, 60% of the reactant remains, establishing an initial-to-final concentration ratio of $\frac{100}{60}$.

Inserting the rate constant $k = 3.3 \times 10^{-4} \text{ s}^{-1}$ into the rate equation results in $t = \frac{\ln(100/60)}{3.3 \times 10^{-4}}$. Calculating the natural logarithm of the concentration ratio yields approximately $0.5108$, which, when divided by the rate constant, results in a duration of roughly $1547.95 \text{ seconds}$. Converting this value to minutes by dividing by 60 yields approximately $25.79 \text{ minutes}$, which rounds to 26 minutes.

The enthalpy change of a reaction, denoted by $\Delta H$, is calculated by finding the difference between the forward activation energy, $E_{a(f)}$, and the reverse activation energy, $E_{a(b)}$. Using the expression $\Delta H = E_{a(f)} - E_{a(b)}$ we substitute the forward value of 20 kJ mol⁻¹ and the reverse value of 65 kJ mol⁻¹, which is the sum of 20 kJ mol⁻¹ and 45 kJ mol⁻¹. This arithmetic yields -45 kJ mol⁻¹ for the total enthalpy change. Taking the magnitude of this result gives a final value of 45.

In first-order kinetics, the concentration of a reactant at any time can be determined using the formula $[A]_t = \frac{[A]_0}{2^n}$, where $n$ represents the number of half-lives that have elapsed. Since both substances start with an identical concentration, we can express their remaining amounts after time $t$ as $\frac{[A]_0}{2^{n_A}}$ and $\frac{[A]_0}{2^{n_B}}$, with $n_A = \frac{t}{54.0}$ and $n_B = \frac{t}{18.0}$. We need the concentration of $A$ to be 16 times that of $B$, leading to the relationship $[A]_t = 16 \times [B]_t$. Substituting the decay formulas into this condition results in the equation $\frac{1}{2^{n_A}} = \frac{16}{2^{n_B}}$. Because 16 is equal to $2^4$, the expression simplifies to $\frac{1}{2^{n_A}} = \frac{2^4}{2^{n_B}}$, which further reduces to $2^{n_B} = 2^{n_A + 4}$, or $n_B = n_A + 4$. By replacing the number of half-lives with their expressions in terms of $t$, we obtain $\frac{t}{18.0} = \frac{t}{54.0} + 4$. Solving this for $t$ yields $t = 108$ minutes.

For first-order kinetics, the rate constant $k$ is defined by the relationship $k = \frac{\ln 2}{t_{1/2}}$, which allows us to determine the time required for a specific extent of reaction. When a reaction reaches 99.9% completion, the remaining concentration $C_t$ is 0.1% of the initial concentration $C_0$, making the ratio $\frac{C_0}{C_t} = 1000$.

Substituting these values into the integrated rate law $t = \frac{1}{k} \ln \left( \frac{C_0}{C_t} \right)$ with the half-life of $1\,\text{min}$ yields $t = \frac{\ln 1000}{\ln 2}$. Since $\ln 1000$ can be expressed as $3 \ln 10$, we substitute the known values $\ln 10 = 2.3$ and $\ln 2 = 0.69$ to obtain $t = \frac{3 \times 2.3}{0.69}$, resulting in 10.

The rate of reaction is mathematically described by $Rate = k[NO]^x[H_2]^y$, where $k$ is the rate constant and the exponents $x$ and $y$ denote the reaction order with respect to nitric oxide and hydrogen, respectively. To determine $x$, we compare two experiments where the concentration of $H_2$ is kept constant so that its influence on the rate is neutralized. Between the first two datasets, the concentration of $H_2$ remains at $8 \times 10^{-5} \, \text{mol L}^{-1}$ while the concentration of $NO$ increases from $8 \times 10^{-5} \, \text{mol L}^{-1}$ to $24 \times 10^{-5} \, \text{mol L}^{-1}$. Because the concentration of $NO$ is tripled and the rate increases from $7 \times 10^{-9} \, \text{mol L}^{-1}$ to $2.1 \times 10^{-8} \, \text{mol L}^{-1}$, which is also a three-fold increase, we can establish the relationship $\frac{7 \times 10^{-9}}{2.1 \times 10^{-8}} = \left( \frac{8 \times 10^{-5}}{24 \times 10^{-5}} \right)^x$. Simplifying these ratios leads to $\frac{1}{3} = \left(\frac{1}{3}\right)^x$, which confirms that the order with respect to $NO$ is 1.

For an elementary reaction, the rate law is derived directly from the stoichiometric coefficients of the balanced chemical equation, yielding the expression $Rate = k[A]^2[B_2]$. Because concentration is the ratio of moles to volume, decreasing the volume of the reaction vessel by a factor of 3 causes the concentration of each reactant to increase by a factor of 3. Substituting these new concentrations into the rate law expression results in $k(3[A])^2(3[B_2])$, which simplifies mathematically to $27k[A]^2[B_2]$. Therefore, the reaction rate increases by a factor of 27 relative to the initial rate.

In first-order kinetics, the progress of a reaction is described by the equation $k = \frac{2.303}{t} \log \frac{a}{a-x}$. When 50% of the reaction is complete, the remaining reactant concentration is half the initial amount, simplifying the logarithmic component to $\log 2$. When the reaction progresses to 75% completion, only one-quarter of the reactant is left, which transforms the logarithmic part into $\log 4$. Because $\log 4$ is equivalent to $2 \log 2$, the time $t_{75\%}$ is twice the value of $t_{50\%}$, leading to a ratio of 2.

The average rate of a reaction is defined by the change in product concentration over a specific time interval, and for the reaction $A \rightarrow B$, it is calculated as $\frac{\Delta[B]}{\Delta t}$. Since the problem specifies a duration of 30 minutes, convert this to 0.5 hours to match the required units. Dividing the concentration increase of 0.2 mol L^-1 by this time duration yields the average rate:

$\text{Average Rate} = \frac{0.2}{0.5} = 0.4 \text{ mol L}^{-1} \text{ h}^{-1}$

Expressing 0.4 in the required format results in $4 \times 10^{-1}$, making the nearest integer 4.

A zero order reaction proceeds at a constant rate that is independent of the reactant concentration, defined by the rate law $r = k$, which typically appears as a flat, horizontal line when plotting rate against concentration. In contrast, first order reactions involve a rate proportional to reactant concentration and a concentration profile that decreases exponentially over time according to $C_{t} = C_{0} e^{-kt}$. Based on these fundamental kinetic principles, graph (b) represents a zero order reaction, whereas graphs (c) and (e) illustrate the distinct behaviors associated with first order reactions.

The proportion of molecules possessing kinetic energy equal to or greater than the activation energy is defined by the Arrhenius factor $e^{-E_a/RT}$. With the activation energy converted to $80900\,\mathrm{J}\,\mathrm{mol}^{-1}$, the exponent is determined by evaluating the ratio of energy to the product of the gas constant and temperature:
$\frac{E_a}{RT} = \frac{80900}{8.31 \times 700} = \frac{80900}{5817} \approx 13.907$
Since the problem defines this fraction as $e^{-x}$, equating the exponents identifies $x$ as approximately $13.907$. Rounding this value to the nearest integer yields 14.

The Arrhenius equation describes the dependence of reaction rate constants on temperature, providing a clear mathematical bridge between activation energy and kinetic change. To determine the rate constant at a specific temperature, we apply the logarithmic version of this equation:

$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

In this scenario, we assign $T_1 = 200\ \text{K}$ and $T_2 = 300\ \text{K}$, with the corresponding rate constant at $300\ \text{K}$ being $k_2 = 1.0 \times 10^{-3}\ \text{s}^{-1}$. The activation energy $E_a$ is $11.488\ \text{kJ}\ \text{mol}^{-1}$, which we convert to $11488\ \text{J}\ \text{mol}^{-1}$ to ensure unit consistency with the gas constant $R = 8.314\ \text{J}\ \text{mol}^{-1}\ \text{K}^{-1}$.

Substituting these values into the formula yields $\log\left(\frac{1.0 \times 10^{-3}}{k_1}\right) = \frac{11488}{2.303 \times 8.314} \left(\frac{1}{200} - \frac{1}{300}\right)$. Simplifying the temperature term, the difference $\frac{1}{200} - \frac{1}{300}$ equals $\frac{1}{600}$. Concurrently, the coefficient $\frac{11488}{2.303 \times 8.314}$ evaluates to $600$. Multiplying these results gives $\log\left(\frac{1.0 \times 10^{-3}}{k_1}\right) = 600 \times \frac{1}{600} = 1$. Since the logarithm of $10$ is $1$, it follows that $\frac{1.0 \times 10^{-3}}{k_1} = 10$, which results in $k_1 = 10^{-4}\ \text{s}^{-1}$. Expressing this in the requested format, the rate constant is $10 \times 10^{-5}\ \text{s}^{-1}$.

The kinetics of a first-order reaction dictate that the rate constant is determined by the integrated rate law, expressed as $k = \frac{2.303}{t} \log \frac{[\mathrm{N}_2\mathrm{O}_5]_0}{[\mathrm{N}_2\mathrm{O}_5]_t}$. To find the rate constant, we substitute the initial concentration of 2.40 x 10^-2 mol L^-1 and the concentration after 60 minutes, which is 1.60 x 10^-2 mol L^-1, into this formula.

Performing the substitution leads to the expression $k = \frac{2.303}{60} \log 1.5$. By applying the identity $\log 1.5 = \log 3 + \log 5 - 1$ along with the provided constants $\log 3 = 0.477$ and $\log 5 = 0.699$, we calculate the logarithm value to be 0.176. Computing the final result gives $k \approx 6.76 \times 10^{-3} \text{ min}^{-1}$, which rounds to the nearest integer of 7.

The Arrhenius equation describes the dependence of reaction rate constants on temperature, establishing that a graphical representation of $\log k$ versus $\frac{1}{T}$ produces a straight line with a gradient of $-\frac{E_a}{2.303 R}$. Since the slope of this line is 10000, we can relate the rate constants at two different temperatures using the expression:
$\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Substituting the known rate constants $10^{-4} s^{-1}$ and $10^{-5} s^{-1}$ along with the temperature 500 K into the equation yields:
$\log \left( \frac{10^{-4}}{10^{-5}} \right) = 10000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$
The logarithmic term simplifies to $\log(10)$, which is equal to 1. This reduces the equation to $1 = 10000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$. Dividing both sides by 10000 results in $0.0001 = \frac{1}{500} - \frac{1}{T_2}$, which rearranges to $\frac{1}{T_2} = 0.002 - 0.0001 = 0.0019$. Calculating the reciprocal gives 526.31 K, which rounds to 526 K.

In chemical kinetics, the rate of a reaction is linked to the stoichiometry of the balanced equation, where the rate of change of each participating species is proportional to its stoichiometric coefficient. For this specific reaction, the consumption of 3 moles of $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ corresponds to the formation of 2 moles of $\mathrm{Cr}_{2}(\mathrm{SO}_{4})_{3}$, which defines the relationship $-\frac{1}{3} \frac{d[\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}]}{dt} = \frac{1}{2} \frac{d[\mathrm{Cr}_{2}(\mathrm{SO}_{4})_{3}]}{dt}$. By rearranging this equation, the rate of disappearance of ethanol is seen to be $\frac{3}{2}$ times the rate of appearance of $\mathrm{Cr}_{2}(\mathrm{SO}_{4})_{3}$. Substituting the known value of 2.67 mol min⁻¹ into this expression gives the result $\frac{3}{2} \times 2.67 = 4.005$ mol min⁻¹. Rounding this figure to the nearest integer results in 4.

First-order chemical reactions are characterized by the rate equation $k = \frac{2.303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right)$, where $k$ is the rate constant and the term $\frac{[A]_0}{[A]_t}$ represents the reciprocal of the fraction $f$ of the substance remaining at time $t$. The rate constant is determined by the half-life through the relationship $k = \frac{\ln 2}{t_{1/2}}$. By substituting this expression for $k$ into the integrated rate law, the relationship becomes $\log_{10}\left(\frac{1}{f}\right) = \frac{\ln 2 \times t}{2.303 \times t_{1/2}}$.

Substituting the provided values, where $t = 9 \text{ h}$ and $t_{1/2} = \frac{10}{3} \text{ h}$, the calculation requires evaluating the ratio $\frac{0.693 \times 9}{2.303 \times \frac{10}{3}}$. Simplifying the numerator and denominator leads to $\frac{18.711}{23.03}$, which results in approximately 0.8124. This value corresponds to $81.24 \times 10^{-2}$. Rounding to the nearest integer, the final value is 81.

The viral inactivation process follows first-order kinetics because the rate of change is directly proportional to the amount of virus present. For such a reaction, the relationship between the rate constant $K$, the time interval $t$, and the initial and remaining concentrations $A_0$ and $A_t$ is expressed by the equation:

$K = \frac{2.303}{t} \log_{10} \frac{A_0}{A_t}$

Given that 10% of the virus is inactivated in the first minute, starting with an initial quantity of 100 units results in a remaining amount of 90 units at 1 min. Substituting these values into the formula yields $K = 2.303 \times \log_{10} \frac{100}{90}$. Using logarithmic properties, this simplifies to $2.303 \times (\log_{10} 10 - \log_{10} 3^2)$, which is $2.303 \times (1 - 2 \log_{10} 3)$. Applying the provided value $\log_{10} 3 = 0.477$, the calculation becomes $2.303 \times (1 - 0.954)$, or $2.303 \times 0.046$, which equals 0.105938 min^-1. Expressing this in the required format of 10^-3 min^-1 results in 105.938, which rounds to the nearest integer of 106.