Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

Bacterial growth follows first-order kinetics, where the population size increases at a rate proportional to the number of cells currently present. As time progresses, the ratio $\frac{N}{N_0}$ rises exponentially, creating a curve that begins at the starting population level and accelerates upward. This upward trajectory visually represents the characteristic exponential expansion inherent in biological reproduction systems.

A zero-order reaction follows the integrated rate law $[A]_{t}=[A]_0-kt$, which fundamentally mirrors the linear equation $y = mx + c$. Because the concentration decreases as a function of time based on this linear relationship, the resulting plot of concentration versus time produces a straight line descending with a constant negative slope of $-k$. This specific shape confirms that the rate of reaction remains uniform, independent of the concentration of the reactant present.

For any first-order reaction, the rate constant $k$ is dictated by the Arrhenius equation, $k = A e^{-\frac{E_a}{RT}}$, which relates the frequency factor and activation energy to the reaction rate at a specific temperature. The time required for 50% of the reactant to convert into the product is defined as the half-life, $t_{1/2}$, which relates to the rate constant by the expression $t_{1/2} = \frac{0.693}{k}$.

Substituting the provided values into the Arrhenius equation requires converting the activation energy to $191480 \ \text{J mol}^{-1}$ to ensure consistency with the units of the gas constant $R = 8.314 \ \text{J K}^{-1} \text{mol}^{-1}$. The product $RT$ at 1000 K equals 8314 $\text{J mol}^{-1}$, making the exponent $-\frac{191480}{8314} = -23.03$. Since $e^{-23.03}$ is numerically equivalent to $10^{-10}$, the rate constant simplifies to $k = 10^{20} \times 10^{-10} = 10^{10} \text{ s}^{-1}$.

The half-life is then calculated by evaluating $t_{1/2} = \frac{0.693}{10^{10}}$, which yields $0.693 \times 10^{-10}$ seconds. Converting this duration into picoseconds by multiplying by $10^{12}$ results in $69.3$ picoseconds. Rounding to the nearest integer gives 69 picoseconds.

Since the container volume and temperature remain constant, the total pressure is directly proportional to the total concentration of gaseous species present. Initially, only the reactant $\mathrm{N}_2\mathrm{O}_5(g)$ is present, exerting a pressure $P_0$. When the reaction reaches 50% completion, the degree of dissociation is 0.5. At this stage, the remaining pressure of $\mathrm{N}_2\mathrm{O}_5(g)$ is $P_0(1 - 0.5)$, resulting in $0.5 P_0$. Simultaneously, the products generate additional pressure based on their stoichiometric coefficients: the pressure from $\mathrm{NO}_2(g)$ is $2 \times 0.5 \times P_0 = 1.0 P_0$, while the pressure from $\mathrm{O}_2(g)$ is $\frac{1}{2} \times 0.5 \times P_0 = 0.25 P_0$. Adding these components together, the final pressure of the system equals $0.5 P_0 + 1.0 P_0 + 0.25 P_0$, which simplifies to $\frac{7}{4} P_0$.

For an elementary reaction, the rate law is dictated directly by the stoichiometric coefficients, resulting in the expression $r = k[\mathrm{A}][\mathrm{B}]$. Since concentration is inversely proportional to volume, decreasing the volume of the reaction mixture to one-third of its initial value increases the concentration of each reactant by a factor of three. Substituting these tripled concentrations into the rate expression yields a new rate of $r' = k(3[\mathrm{A}])(3[\mathrm{B}])$, which simplifies to $9k[\mathrm{A}][\mathrm{B}]$, meaning the reaction rate becomes 9 times the original rate.

The thermal decomposition of dinitrogen pentoxide follows first-order kinetics, allowing for the application of the integrated rate law, $k = \frac{2.303}{t} \log \frac{P_i}{P_A}$, where $P_i$ is the initial pressure and $P_A$ is the reactant pressure at time $t$. Given $P_i = 0.6$ atm, the stoichiometry of $2\mathrm{N}_2\mathrm{O}_5(\mathrm{g}) \longrightarrow 2\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$ reveals that if $2p$ represents the decrease in $\mathrm{N}_2\mathrm{O}_5$ pressure, the partial pressures at time $t$ become $0.6 - 2p$ for $\mathrm{N}_2\mathrm{O}_5$, $2p$ for $\mathrm{N}_2\mathrm{O}_4$, and $p$ for $\mathrm{O}_2$. The total pressure $x$ is the sum of these, $x = 0.6 + p$, which implies $p = x - 0.6$ and $P_A = 0.6 - 2(x - 0.6) = 1.8 - 2x$. Substituting $k = 4.606 \times 10^{-2} \mathrm{s}^{-1}$ and $t = 100$ s into the rate law yields $4.606 \times 10^{-2} = \frac{2.303}{100} \log \frac{0.6}{1.8 - 2x}$, which simplifies to $2 = \log \frac{0.3}{0.9 - x}$. Solving $100 = \frac{0.3}{0.9 - x}$ results in $0.9 - x = 0.003$ and $x = 0.897$ atm. Given the required format $897 \times 10^{-3}$ atm, the missing integer value is 897.

An energy profile diagram for a multi-step reaction depicts the potential energy changes as reactants traverse transition states to become products. Since the initial transformation from $A$ to $B$ is the slow, endothermic step, the diagram must feature the highest energy barrier at the start, with the potential energy of $B$ sitting above that of $A$. The subsequent fast, exothermic steps towards $C$ and $D$ must show a continuous decline in energy, ensuring that the total energy of the final product $D$ is lower than the initial reactant $A$ to satisfy the condition of net heat evolution, denoted by $\Delta H < 0$.

Radioactive decay is fundamentally a nuclear process that operates independently of external physical conditions such as temperature, distinguishing it from chemical reactions that rely on thermal energy to overcome an activation energy barrier. Because the activation energy for radioactive decay is zero, the decay constant $\lambda$ remains unchanged regardless of any fluctuations in temperature. Consequently, claiming that the decay constant increases as temperature rises is incorrect, as nuclear stability is governed by internal forces rather than environmental thermal energy.

The Arrhenius equation is expressed as $\log k = \log A - \frac{Ea}{2.303RT}$, which represents a linear plot of $\log k$ against $\frac{1}{T}$ with a slope of $-\frac{Ea}{2.303R}$. Because the slope is negative, its absolute magnitude is directly proportional to the activation energy $Ea$, meaning a steeper negative slope corresponds to a higher activation energy. Given that the provided plot displays the steepest slope for reaction 2 and the shallowest slope for reaction 3, the activation energies follow the order $Ea_2 > Ea_1 > Ea_3$.

The reaction follows first-order kinetics, where the rate constant is governed by the equation $k = \frac{2.303}{t} \log \frac{[A_0]}{[A_t]}$. By inspecting the graph at time 25 s, where the concentration of A decreases from 50 g/L to 10 g/L, we determine the rate constant using $k = \frac{2.303}{25} \log \frac{50}{10}$. Given that $\log 5 = 0.699$, the rate constant is $k = \frac{2.303 \times 0.699}{25}$ per second.

The time required for the concentration to reduce to 2.5 g/L can be calculated using the same integrated rate law with an initial concentration of 50 g/L. Setting this equal to the derived rate constant yields $\frac{2.303 \times 0.699}{25} = \frac{2.303}{t} \log \frac{50}{2.5}$. Simplifying the equation gives $\frac{0.699}{25} = \frac{1}{t} \log 20$. With $\log 20 = 1.301$, solving for $t$ results in $t = \frac{1.301 \times 25}{0.699} \approx 46.53$ s, which rounds to 47 s.

The rate of a multi-step reaction is governed by its slowest step, which implies the rate law is initially defined as $r = k_2 [A] [B_2]$. Since $[A]$ is an intermediate, we must substitute it using the fast equilibrium expression $\frac{k_1}{k_{-1}} = \frac{[A]^2}{[A_2]}$, which yields $[A] = \left( \frac{k_1 [A_2]}{k_{-1}} \right)^{1/2}$. Substituting this term back into the rate law results in $r = k' [A_2]^{1/2} [B_2]$, where $k'$ represents the combined rate constants. Summing the reaction orders with respect to the reactants $[A_2]$ and $[B_2]$ produces an overall order of $\frac{1}{2} + 1 = 1.5$.

For a first-order reaction, the half-life period is constant and depends only on the rate constant, given by $t_{1/2} = \frac{0.693}{k}$, which makes any statement asserting its independence from the initial concentration of the reactant valid. However, the integrated rate law dictates that the logarithmic relationship is defined by $\log \frac{[R]}{[R_0]} = - (\frac{k}{2.303}) t$, establishing that a plot of $\log \frac{[R]}{[R_0]}$ versus time results in a linear graph with a slope of $-\frac{k}{2.303}$. Consequently, any proposition in Statement II that contradicts this specific negative slope or misidentifies the mathematical relationship between the log-ratio and time is inaccurate, which explains why the second statement fails to be true.

The concentration of any reactant following first-order kinetics after time $t$ is determined by the formula $[Reactant]_t = [Reactant]_0 (1/2)^{t/t_{1/2}}$. Since both reactants follow first-order kinetics, their concentrations at any time $t$ are described by $[A]_t = 8[B]_0 (1/2)^{t/10}$ and $[B]_t = [B]_0 (1/2)^{t/40}$ based on the given initial concentration relationship $[A]_0 = 8[B]_0$ and the respective half-lives of $10 \text{ min}$ and $40 \text{ min}$.

Setting the concentration expressions for $A$ and $B$ equal to each other gives $8[B]_0 (1/2)^{t/10} = [B]_0 (1/2)^{t/40}$. Dividing both sides by $[B]_0$ results in $8(1/2)^{t/10} = (1/2)^{t/40}$, which rearranges to $8 = (1/2)^{t/40 - t/10}$. Expressing $8$ as $2^3$ and the right side as $2^{t/10 - t/40}$ allows for a direct comparison of exponents, resulting in $3 = t/10 - t/40$. Solving this equation by finding a common denominator yields $3 = 3t/40$, which simplifies to a final time of $t = 40 \text{ min}$.

The Arrhenius law characterizes the rate constant $K$ using the exponential form $K = A e^{-E_a/RT}$, where $E_a$ represents the activation energy. When the overall rate constant is expressed as $K = \frac{K_1 K_2}{K_3}$, the temperature dependence dictates that the effective activation energy is governed by the relation $E_a = E_{a1} + E_{a2} - E_{a3}$. Consequently, we find the overall value by performing the operation $E_a = 40 + 50 - 60 = 30$ kJ/mol.

The order of the reaction with respect to reactant $A$ is determined by the specific kinetic relationship provided, where the time taken to reduce the concentration to $\frac{1}{4}$ of its initial value is exactly twice the time taken to reduce it to $\frac{1}{2}$ of its initial value. This doubling of time for successive fractional reductions is the mathematical signature of a first-order reaction.

For reactant $B$, the linear graph of concentration versus time with a negative slope indicates that the rate of consumption remains constant regardless of the concentration. This behavior is the definitive characteristic of a zero-order reaction, as expressed by the linear rate equation $[B]_t = -kt + [B]_0$. Combining these findings, the overall order of the reaction is the sum of the individual orders, which results in $1 + 0 = 1$.

For a typical gas-phase decomposition reaction represented by the equation
$A \rightarrow B + C$
the integrated rate law for a first-order process is determined by relating the partial pressure of the reactant at a specific time $t$ to the total observed pressure. Initially, the system contains reactant $A$ at a partial pressure $P_i$, while products $B$ and $C$ are absent. After time $t$, if $x$ represents the decrease in the partial pressure of $A$, then $x$ also represents the corresponding increase in the partial pressure of both products $B$ and $C$.

Consequently, the total pressure $P_t$ measured at time $t$ is the sum of these individual partial pressures, given by $P_t = (P_i - x) + x + x$, which simplifies to $P_t = P_i + x$. Rearranging this expression allows us to identify the remaining partial pressure of $A$ as $P_i - x = 2 P_i - P_t$. Substituting these terms into the standard first-order kinetic equation yields the final expression
$k = \frac{2.303}{t} \log \frac{P_i}{2 P_i - P_t}$

The Arrhenius equation, $k = A e^{-E_a/RT}$, provides the fundamental link between reaction rates and activation energy, allowing us to determine the temperature dependence of a multi-step reaction mechanism. When an overall rate constant $k$ is expressed as a mathematical combination of individual elementary rate constants, we can substitute the Arrhenius expression for each constant into the formula. For the reaction $k = \frac{k_1 k_2}{k_3}$, this substitution leads to the equality $A e^{-E_a/RT} = \frac{(A_1 e^{-E_{a1}/RT})(A_2 e^{-E_{a2}/RT})}{A_3 e^{-E_{a3}/RT}}$.

By simplifying the exponential terms on the right side, we find that the overall activation energy relates to the individual energies according to the equation $E_a = E_{a1} + E_{a2} - E_{a3}$. Substituting the provided values of $E_a = 400 \text{ kJ mol}^{-1}$, $E_{a1} = 300 \text{ kJ mol}^{-1}$, and $E_{a2} = 200 \text{ kJ mol}^{-1}$ into this relationship results in the equation $400 = 300 + 200 - E_{a3}$. Solving this expression, we determine that $E_{a3} = 100 \text{ kJ mol}^{-1}$.

First-order kinetics are governed by the integrated rate equation $K = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]_t} \right)$, which links the reaction duration to the ratio of initial to remaining reactant concentration. When a reaction reaches 90% completion, 10% of the starting material remains, resulting in the rate expression $K = \frac{\ln(10)}{t_{90\%}}$, while 99.9% completion leaves 0.1% remaining, yielding $K = \frac{\ln(1000)}{t_{99.9\%}}$. Since the rate constant $K$ is identical for both cases, we can equate the expressions as $\frac{\ln(1000)}{t_{99.9\%}} = \frac{\ln(10)}{t_{90\%}}$, which simplifies to $\frac{3 \ln(10)}{t_{99.9\%}} = \frac{\ln(10)}{t_{90\%}}$. By canceling the logarithmic term from both sides, it becomes evident that the time required for 99.9% completion is 3 times the time required for 90% completion.

Radioactive decay follows an exponential pattern where the quantity of a radioisotope is halved over each successive half-life period. For ${}^{14}\mathrm{C}$ dating, the reduction in the ratio of ${}^{14}\mathrm{C}$ to ${}^{12}\mathrm{C}$ indicates the time elapsed since the organism ceased interaction with the atmosphere.

Because the current ratio is 1/8 of the atmospheric level, and this fraction represents three half-life intervals as $(1/2)^3 = 1/8$, we can determine that the sample has aged by three half-lives. Using the provided half-life value of 5730 years, the total age $t$ is calculated as $t = 3 \times 5730 = 17190$ years.

For any first-order reaction, the rate constant $k$ is inversely proportional to the half-life $t_{1/2}$ according to the relation $k = \frac{\ln 2}{t_{1/2}}$, and the time $t$ required to reach a specific completion fraction is defined by $t = \frac{1}{k} \ln \left( \frac{1}{1 - \text{fraction}} \right)$.

Since the ratio of the half-lives is given as $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{5}{2}$, the ratio of the rate constants is inverted as $\frac{k_2}{k_1} = \frac{5}{2}$. The time $t_1$ to reach two-thirds completion for the first reaction is $\frac{1}{k_1} \ln(3)$, and the time $t_2$ to reach four-fifths completion for the second reaction is $\frac{1}{k_2} \ln(5)$. Expressing the ratio of these times results in $\frac{t_1}{t_2} = \frac{k_2}{k_1} \times \frac{\ln 3}{\ln 5}$. Because the ratio of natural logarithms is equivalent to the ratio of common logarithms, substituting the provided constants leads to $\frac{t_1}{t_2} = \frac{5}{2} \times \frac{0.477}{0.699}$, which simplifies to $1.7$, or $17 \times 10^{-1}$.