Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The rate law establishes how the concentration of reactants influences the speed of a chemical reaction, mathematically expressed as $\text{Rate} = k[\mathrm{A}]^a[\mathrm{B}]^b$. We determine the unknown orders $a$ and $b$ by comparing how the initial rate changes when we vary one concentration while keeping the other constant. Comparing the first and second experiments, we see that doubling the concentration of A from 0.05 to 0.10 while holding B constant at 0.05 causes the rate to increase from 0.045 to 0.090. Since doubling the concentration of A results in an exactly proportional doubling of the rate, the reaction order with respect to A is 1.

Moving to the third experiment, both concentrations are doubled from their levels in the second experiment: A goes from 0.10 to 0.20, and B goes from 0.05 to 0.10. The overall rate increases by a factor of $\frac{0.72}{0.090} = 8$. Given that A contributes a factor of 2 to this increase, the remaining factor of $\frac{8}{2} = 4$ must be attributed to B. Because $2^b = 4$, the reaction order with respect to B is 2, leading to the overall rate law $\text{Rate} = k[\mathrm{A}][\mathrm{B}]^2$.

The temperature dependence of a reaction rate constant is described by the Arrhenius equation, which establishes a linear relationship between the natural logarithm of the rate constant and the reciprocal of absolute temperature. This relationship is expressed as:

$\ln \frac{K_2}{K_1} = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$

Using the known rate constant of $10^{-5} \text{ s}^{-1}$ at 400 K and the slope factor from the provided plot, we substitute the variables into the equation $\ln \frac{K_2}{10^{-5}} = 46.6 \left[\frac{1}{400} - \frac{1}{500}\right]$. Solving this expression for the rate constant results in $K_2 = 10^{-4} \text{ s}^{-1}$.

The kinetics of a zeroth-order reaction follow the integrated rate law equation $[A]_0 - [A]_t = kt$, where the rate constant $k$ remains independent of the reactant concentration over time. Using the provided half-life of 6 h, we know that the initial concentration of 0.2 M reduces to 0.1 M within that period. By substituting these values into the equation $0.2 - 0.1 = k \times 6$, we determine the rate constant $k = \frac{0.1}{6}$, which simplifies to $k = \frac{1}{60}$ M/h. To calculate the time required for the concentration to decrease from 0.5 M to 0.2 M, we apply the same rate law with our calculated value of $k$. Rearranging the equation to $0.5 - 0.2 = \frac{1}{60} \times t$ results in $0.3 = \frac{t}{60}$, which solves to a final duration of 18.0 h.

Energy profile diagrams map the reaction coordinate, allowing us to evaluate the specific activation energy barriers associated with competing product pathways. According to the plot, the activation enthalpy barrier for the formation of product C is exactly $10\,\mathrm{kJ\,mol}^{-1}$ lower than the barrier required to form product D. Because the statement claims a difference of only $5\,\mathrm{kJ\,mol}^{-1}$ instead of the actual $10\,\mathrm{kJ\,mol}^{-1}$, it inaccurately describes the reaction characteristics presented in the diagram.

The steady-state approximation simplifies kinetic analysis by assuming the net rate of formation for a short-lived intermediate is zero because it is consumed as quickly as it is produced. In the reaction sequence $A \xrightarrow{k_1} B \xrightarrow{k_2} C$, the rate of change for $B$ is determined by the difference between its production rate, $k_1[A]$, and its consumption rate, $k_2[B]$. By setting the net rate expression $\frac{d[B]}{dt} = k_1[A] - k_2[B] = 0$, we find that $k_1[A] = k_2[B]$, which directly yields $[B] = \left(\frac{k_1}{k_2}\right)[A]$.

The reaction rate is defined by the expression $r = k [A]^x [B]^y$. Comparing the initial data for the first two experiments shows that variations in the concentration of B do not alter the rate when A is held constant, meaning the reaction is zero-order with respect to B. Comparing the first and third experiments demonstrates that when the concentration of A is doubled, the reaction rate also doubles, confirming the reaction is first-order with respect to A. The rate law is therefore $r = k [A]$.

Substituting the values from the first experiment into this rate law, $6.93 \times 10^{-3} = k \times 0.1$, yields a rate constant of $k = 6.93 \times 10^{-2} \text{ min}^{-1}$. Given the stoichiometry of the reaction $2A + B \rightarrow \text{Products}$, the rate of consumption of reactant A is given by $-\frac{d[A]}{dt} = 2k [A]$. The time required to consume half of the initial concentration of A is calculated using the formula $t_{1/2} = \frac{0.693}{2k}$. Substituting the value of $k$ gives $t_{1/2} = \frac{0.693}{2 \times 6.93 \times 10^{-2}} = \frac{0.693}{0.1386}$, which results in 5 minutes.

The relationship between the rates of disappearance of a reactant and the appearance of a product in the reaction $xA \rightarrow yB$ is defined by the stoichiometric coefficients as $-\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$, which rearranges to $-\frac{d[A]}{dt} = \frac{x}{y} \frac{d[B]}{dt}$. Taking the logarithm on both sides yields $\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + \log_{10} \left( \frac{x}{y} \right)$, and comparing this to the provided equation shows that $\log_{10} \left( \frac{x}{y} \right) = 0.3010$. Since $\log_{10} 2 = 0.3010$, the ratio $\frac{x}{y}$ must be 2, making $x=2$ and $y=1$. Therefore, the reaction is $2 \mathrm{C}_2 \mathrm{H}_4 \rightarrow \mathrm{C}_4 \mathrm{H}_8$.

The Arrhenius equation $k = A e^{-E_a/RT}$ establishes that the rate constant $k$ is dependent on both temperature and activation energy $E_a$, confirming that a plot of $\ln k$ versus $1/T$ yields a straight line with a negative slope of $-E_a/R$, which makes the first plot correct. Because the relationship involves an exponential decay, the rate constant decreases as $E_a$ increases, meaning the second plot is only valid if it accurately reflects this non-linear dependency. Any graphical representation that fails to show this precise exponential decline between $k$ and $E_a$ is conceptually inaccurate, confirming the second plot is wrong.

The rate of a chemical reaction is defined by the relationship between the concentration changes of the reactants and products over time, which is strictly governed by their stoichiometric coefficients in the balanced equation. For the decomposition reaction $2\mathrm{N_2O_5}(g) \rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g)$, the balanced equation indicates that for every two moles of $\mathrm{N_2O_5}$ that decompose, four moles of $\mathrm{NO_2}$ are produced. This implies that the rate of formation of $\mathrm{NO_2}$ is exactly double the absolute rate of disappearance of $\mathrm{N_2O_5}$.

To determine the rate of formation, we first calculate the rate of disappearance of $\mathrm{N_2O_5}$ by finding the change in concentration, which is 0.25 mol L$^{-1}$ (the difference between 3.00 and 2.75 mol L$^{-1}$), divided by the elapsed time of 30 minutes. By applying the stoichiometric multiplier of 2 to this rate of disappearance, we arrive at the rate of formation for the product. Following this calculation, the rate of formation of nitrogen dioxide is $1.667 \times 10^{-3}$ mol L$^{-1}$ min$^{-1}$.

The rate of a chemical reaction is described by the equation $r = k [A]^x [B]^y$, where $x$ and $y$ represent the reaction orders for reactants A and B, respectively. To determine these orders, we analyze how the rate responds to systematic changes in reactant concentrations. Observing the scenario where the concentration of A alone is doubled, the rate rises from 0.3 mol $L^{-1} s^{-1}$ to 0.6 mol $L^{-1} s^{-1}$, indicating a direct proportionality. Mathematically, this change represents a factor of 2 increase in rate for a factor of 2 increase in concentration, expressed as $2 = 2^x$, confirming that $x = 1$.

When both A and B concentrations are doubled simultaneously, the rate increases significantly from 0.3 mol $L^{-1} s^{-1}$ to 2.4 mol $L^{-1} s^{-1}$, a total increase by a factor of 8. Inserting this into the rate law formula, we get $8 = 2^x \cdot 2^y$, or $8 = 2^{x+y}$. Because $2^3$ equals 8, the sum of the orders must satisfy $x + y = 3$. Substituting the previously determined value of $x = 1$ into this sum results in $y = 2$, proving the reaction is second-order with respect to B.

The Arrhenius equation provides the fundamental relationship between the rate constant, temperature, and activation energy. First, convert the given temperatures from Celsius to Kelvin by adding 273, which yields $T_1 = 600 \mathrm{K}$ and $T_2 = 800 \mathrm{K}$. Applying the logarithmic form of the Arrhenius equation:
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$
Substituting the known values results in the expression $\log \left( \frac{1}{2.5 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{1}{600} - \frac{1}{800} \right]$. The left side simplifies to $\log(4000) \approx 3.602$, and the difference of the inverse temperatures is approximately $4.167 \times 10^{-4} \mathrm{K^{-1}}$. Solving for $E_a$ leads to a value of approximately $166,000 \mathrm{J mol^{-1}}$, which is $166 \mathrm{kJ mol^{-1}}$.

The Arrhenius equation is expressed as $k = Ae^{-E_a/RT}$, where $E_a$ is the activation energy. Taking the natural logarithm on both sides results in the linear form $\ln k = \ln A - E_a\left(\frac{1}{RT}\right)$. Comparing this to the equation of a straight line $Y = mX + c$, where $Y = \ln k$ and $X = \frac{1}{RT}$, the gradient $m$ corresponds to $-E_a$. Equating this to the given gradient $-y$ yields $E_a = y$.

The bacterial population growth for $t \le 1$ follows $N(t) = N_0 \exp(t)$, which determines the population at the moment the antibiotic reaches the wound as $N(1) = N_0 e$. For $t > 1$, the bacterial decay is described by the differential equation $\frac{dN}{dt} = -5N(t-1)^2$, where $H = t-1$ represents the time elapsed since the antibiotic reached the wound. Separating variables and integrating from $t=1$ to $t$ gives $\int_{N(1)}^{N} \frac{dN}{N} = \int_{1}^{t} -5(t'-1)^2 dt'$, which results in $\ln\left(\frac{N}{N_0 e}\right) = -\frac{5}{3}(t-1)^3$.

Solving for $N$ yields $N = N_0 e \exp\left(-\frac{5}{3}(t-1)^3\right)$, and taking the reciprocal gives the ratio $\frac{N_0}{N} = \frac{1}{e} \exp\left(\frac{5}{3}(t-1)^3\right) = \exp\left(\frac{5}{3}(t-1)^3 - 1\right)$. This expression defines a function that starts at a value of $e^{-1}$ at $t=1$ and increases rapidly as $t$ increases, characterizing the plotted curve.

For elementary reactions, the net rate of change of a chemical species depends on both the forward and reverse stoichiometric coefficients. In the forward step $A_2 \rightarrow 2A$, the production of $A$ happens at twice the rate of $A_2$ consumption, represented by $2k_1[A_2]$, while in the reverse step $2A \rightarrow A_2$, the consumption of $A$ occurs at twice the rate of the reverse reaction, represented by $2k_{-1}[A]^2$. Combining these terms gives the overall rate expression for the change in concentration of $A$ as $\frac{d[A]}{dt} = 2k_1 [A_2]-2k_{-1}[A]^2$.

This reaction follows zero-order kinetics, where the rate of decomposition remains constant over time regardless of the reactant concentration. For any zero-order process, the time required to decompose half of the initial quantity is determined by the half-life equation $t_{1/2} = \frac{a_0}{2k}$. By substituting the initial amount of 5 and the rate constant of 0.05 into this formula, the calculation yields $\frac{5}{2 \times 0.05}$, which equals 50 years.

In a zero-order reaction, the rate of consumption is independent of the reactant concentration, resulting in a linear decrease of $[R]$ over time $t$ that appears as a straight-line plot. In contrast, a first-order reaction exhibits a rate directly proportional to the reactant concentration, leading to an exponential decay profile where $[R]$ versus $t$ forms a curve rather than a line. Since reaction (i) depicts this exponential relationship and reaction (ii) shows a constant linear decline, they correspond to first-order and zero-order reactions, respectively.

For a first-order reaction, the time required for any fractional completion is directly related to the half-life, $t_{1/2}$. When 50% of a reaction is complete, the time elapsed is equal to the half-life, meaning $t_{1/2} = 100 \text{ s}$. For 75% completion, the remaining reactant concentration is 25% of the initial value, which corresponds to the passage of two half-lives, as $25\% = (1/2)^2 \times 100\%$. The total time required for 75% completion is calculated as $t_{75\%} = 2 \times t_{1/2}$. Substituting the known half-life, the calculation $2 \times 100 \text{ s} = 200 \text{ s}$ matches the provided time data perfectly, which confirms that the reaction follows first-order kinetics.

The decomposition of a gaseous sample at 518°C follows the rate law $r = k P^n$, where $r$ is the rate, $P$ is the partial pressure of the reactant, and $n$ is the order of the reaction. Because the rate constant $k$ remains unchanged at a fixed temperature, the relationship between two different rates and their corresponding partial pressures can be expressed as the ratio $\frac{r_1}{r_2} = (\frac{P_1}{P_2})^n$.

Given the initial pressure of 363 Torr, the partial pressure at 5% reaction is $P_1 = 344.85$ Torr with a rate $r_1 = 1.00$ Torr $s^{-1}$. At 33% reaction, the partial pressure is $P_2 = 241.7$ Torr with a rate $r_2 = 0.5$ Torr $s^{-1}$. Substituting these specific values into the ratio formula yields $\frac{1.00}{0.5} = (\frac{344.85}{241.7})^n$, which simplifies to $2 = (1.42)^n$. Solving this equality confirms that the order of the reaction $n$ is 2.

The decomposition reaction follows the stoichiometry $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$. If the initial pressure of $N_2O_5$ is $P_0 = 50 mmHg$, and it decomposes by an amount $2x$, the total pressure $P_t$ at any time is given by the sum of partial pressures $P_t = (P_0 - 2x) + 4x + x = P_0 + 3x$.

Given $P_t = 87.5 mmHg$ at $t = 50$ minutes, substituting these values gives $87.5 = 50 + 3x$, resulting in $3x = 37.5$ or $x = 12.5 mmHg$. The remaining pressure of $N_2O_5$ is $P_0 - 2x = 50 - 2(12.5) = 25 mmHg$, which demonstrates that the half-life of the first-order reaction is $50$ minutes.

At $t = 100$ minutes, equivalent to two half-lives, the remaining pressure of $N_2O_5$ is $P_0/4 = 12.5 mmHg$. Setting the partial pressure of the remaining reactant equal to $P_0 - 2x' = 12.5$ yields $2x' = 37.5$ or $x' = 18.75 mmHg$. The total pressure of the gaseous mixture at this time is $P_{100} = P_0 + 3x' = 50 + 3(18.75) = 106.25 mmHg$.

The rate constant $k$ for a first-order reaction is determined by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ day}^{-1}$. Since a $1/4^{\text{th}}$ conversion implies that $1/4$ of the reactant has been consumed, the remaining concentration $[A]_t$ is $\frac{3}{4}[A]_0$. Substituting these values into the integrated first-order rate law $t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]_t} \right)$ yields $t = \frac{1}{0.0693} \ln \left( \frac{4}{3} \right)$. Utilizing the provided logarithmic values where $\ln \left( \frac{4}{3} \right) = 2 \ln 2 - \ln 3 = 2(0.693) - 1.1 = 0.286$, the time is calculated as $t = \frac{0.286}{0.0693} \approx 4.1 \text{ days}$.