Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

Zero-order reactions proceed at a steady rate independent of the remaining reactant concentration, resulting in a half-life of $t_{50\%} = \frac{a_0}{2k}$ and a total completion time of $t_{100\%} = \frac{a_0}{k}$. Because the total time required to consume the initial quantity is exactly double the time needed to consume half of it, we establish the relationship $t_{100\%} = 2 t_{1/2}$. In contrast, first-order reactions follow exponential decay kinetics where the reaction rate continuously decreases as the concentration drops, meaning the reactant is consumed asymptotically. While the half-life for such a process is $t_{50\%} = \frac{\ln 2}{k}$, the theoretical time required for 100% completion is infinite, which is expressed in the given mathematical context as $t_{100\%} = (t_{1/2})^{\infty }$.

The order of the reaction is identified by examining the units of the rate constant, which are given as $\mathrm{M}^{-1} \mathrm{s}^{-1}$. Since the general expression for the units of a rate constant is $\mathrm{M}^{1-n} \mathrm{s}^{-1}$, where $n$ represents the order of the reaction, setting the exponent equal to $-1$ gives the equation $1 - n = -1$, confirming that this is a second-order reaction. The rate law for this reaction is therefore expressed as $\mathrm{Rate} = k[X]^2$, which demonstrates that the rate is proportional to the square of the concentration of $X$. Consequently, if the concentration of $X$ is increased by a factor of four, the reaction rate increases by $4^2$, or $16$ times. Other statements are incorrect because the half-life of a second-order reaction is inversely proportional to the initial concentration rather than independent of it, and the decomposition of $\mathrm{N}_2 \mathrm{O}_5$ and the linear plot of $\ln([R]_0/[R]_t)$ versus time are both characteristic features of first-order reactions.

First-order chemical kinetics follow a property where the reactant concentration is halved over fixed, repeating time intervals known as the half-life. Because the half-life of this specific sucrose hydrolysis reaction is defined as $t_{\frac{1}{2}} = 3$ hours, we can determine the progression of the reaction by counting these distinct intervals. Over a total duration of 6 hours, the reaction completes exactly two full half-lives. During the first 3-hour interval, the initial amount of sucrose is reduced by 50%. Following this, the remaining 50% undergoes another half-life over the next 3 hours, effectively cutting the reactant concentration in half once more. Consequently, the fraction of sucrose left after 6 hours is 25% of the initial amount.

The rate of a chemical reaction is defined by the rate of change of concentration of reactants or products divided by their respective stoichiometric coefficients. For reaction II, the rate is given by the following expression:
$\text{Rate}_{II} = -\frac{1}{5} \frac{\Delta \$[\mathrm{Br}^{-}]\$}{\Delta t}$
Substituting the given rate of disappearance of the bromide ion, -\frac{\Delta \[\mathrm{Br}^{-}]$}{\Delta t} = 2 \times 10^{-4} \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$, the value of the reaction rate is$\text{Rate}_{II} = \frac{1}{5} \times 2 \times 10^{-4} = 4 \times 10^{-5} \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$. Since the rates of reaction I and II are identical at 10.10 am, the rate of reaction I is also$4 \times 10^{-5} \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$. As reaction I follows first-order kinetics, its rate is expressed by the equation$\text{Rate}_I = k[A]$. Using the concentration$[A] = 10^{-2} \mathrm{mol} \mathrm{L}^{-1}$, the first-order rate constant$k$ is calculated as:
$k = \frac{\text{Rate}_I}{[A]} = \frac{4 \times 10^{-5}}{10^{-2}} = 4 \times 10^{-3} \mathrm{min}^{-1}$

The Arrhenius equation $k = Ae^{-\frac{E_a}{RT}}$ establishes that the factor $e^{-\frac{E_a}{RT}}$ represents the fraction of molecules with kinetic energy greater than or equal to the activation energy $E_a$, meaning statement A is false, while the inverse dependence on $E_a$ correctly implies that lower activation energies yield faster reaction rates. The empirical rule that reaction rates typically double with a $10^{\circ} \mathrm{C}$ increase is correct, but statement D is invalid because the conversion to base-10 logarithms results in the equation $\log k = \log A - \frac{E_a}{2.303RT}$, which produces a slope of $-\frac{E_a}{2.303R}$ rather than $-\frac{E_a}{R}$.

The rate constant for a first-order gas-phase reaction is derived from the integrated rate law, which expresses the progress of a reaction in terms of the initial reactant concentration and the concentration remaining at time $t$. For the reaction $A \longrightarrow B + C$, we start with an initial partial pressure $p_i$ of reactant A, while products B and C have an initial pressure of zero. As the reaction proceeds for a duration $t$, if the pressure of A decreases by an amount $x$, the pressures of B and C each increase by $x$. The total pressure of the system at time $t$, denoted as $p_t$, is the sum of the partial pressures of all species present: $(p_i - x) + x + x$, which simplifies to $p_i + x$. Rearranging this relationship to solve for $x$ yields $x = p_t - p_i$, allowing us to express the partial pressure of reactant A remaining at time $t$ as $p_i - (p_t - p_i)$, or $2p_i - p_t$. Substituting the initial pressure and the remaining partial pressure into the standard first-order integrated rate equation $k = \frac{1}{t} \ln \frac{p_i}{p_A}$ gives the final expression:
$k = \frac{1}{t} \ln \frac{p_i}{2p_i - p_t}$

For a first-order reaction where reactant $R$ converts to product $P$, the concentration of the reactant remaining at time $t$ is given by $A_t = A_0 e^{-k_1 t}$. The amount of reactant that has decomposed is found by subtracting the remaining concentration from the initial amount, calculated as $A_0 - A_0 e^{-k_1 t}$. Dividing this decomposed portion by the initial concentration $A_0$ determines the fraction of molecules decomposed, which simplifies to $1 - e^{-k_1 t}$.

The Arrhenius equation, represented as $k = A e^{-\frac{E_a}{RT}}$, serves as the fundamental principle for understanding the temperature dependence of chemical reaction rates. Since the exponential term $e^{-\frac{E_a}{RT}}$ is dimensionless, the pre-exponential factor $A$ must necessarily possess the same units as the rate constant $k$. Moreover, within this expression, a smaller activation energy $E_a$ results in a larger rate constant $k$, which indicates that reactions with lower energy barriers proceed at a faster pace at a given temperature. The remaining statements are invalid because the Arrhenius model treats $A$ and $E_a$ as constants independent of temperature, the equation is not restricted to elementary homogeneous reactions, and there is no inherent interdependency between $A$ and $E_a$ based on the relative magnitude of $E_a$ to $RT$.

The initial rate of reaction is given by the expression $r_1 = k[A]^n[B]^m$. Upon changing the reactant concentrations such that $[A]$ becomes $2[A]$ and $[B]$ becomes $\frac{1}{2}[B]$, the new rate of reaction is expressed as $r_2 = k(2[A])^n(\frac{1}{2}[B])^m$.

The ratio of the new rate to the initial rate is $\frac{r_2}{r_1} = \frac{k(2[A])^n(\frac{1}{2}[B])^m}{k[A]^n[B]^m}$. Canceling common terms $k$, $[A]^n$, and $[B]^m$ simplifies the expression to $\frac{r_2}{r_1} = 2^n \cdot (\frac{1}{2})^m$, which is equivalent to $2^n \cdot 2^{-m}$, ultimately resulting in $2^{(n-m)}$.

For a first-order reaction of the type $A(g) \rightarrow 2B(g) + C(g)$, the total pressure at any time $t$ is expressed as $P_t = P_0 + 2x$, where $P_0$ is the initial pressure of $A$ and $x$ is the decrease in partial pressure of $A$. As $t$ approaches infinity, the reaction goes to completion and all reactant $A$ is consumed, meaning $P_\infty = 3P_0$. Given $P_\infty = 240$ mm Hg, the initial pressure $P_0$ is $240 / 3 = 80$ mm Hg. At $t = 10$ min, the total pressure $P_t = 160$ mm Hg, so $160 = 80 + 2x$, which results in $x = 40$ mm Hg. The partial pressure of $A$ after 10 minutes is $P_0 - x = 80 - 40 = 40$ mm Hg.

The rate constant $k$ is calculated using the integrated rate equation $k = \frac{2.303}{t} \log\left(\frac{P_0}{P_A(t)}\right)$. Substituting the determined values, $k = \frac{2.303}{10} \log\left(\frac{80}{40}\right) = \frac{2.303 \times 0.301}{10} \approx 0.0693 \text{ min}^{-1}$. This calculation proves that the value $1.693 \text{ min}^{-1}$ provided in the option is incorrect. Furthermore, since a first-order reaction follows an exponential decay where reactant concentration approaches zero asymptotically, the statement that the reaction never goes to completion is physically accurate within the context of chemical kinetics.

Bacterial growth following first-order kinetics is governed by the differential equation $\frac{dN}{dt} = kN$, which results in an exponential growth curve for the population $N$ over time $t$. The subsequent application of antibacterial medicine, where the decay rate $r$ is proportional to the square of the bacteria, follows second-order kinetics defined by $-\frac{dN}{dt} = k_2 N^2$, which yields the solution $\frac{1}{N} = k_2 t + C$ upon integration. This mathematical relationship indicates that the population decreases along a hyperbolic curve, transitioning from the previous exponential growth phase to a decay phase where the rate of decrease diminishes as the population size drops.

The overall activation energy of a complex reaction can be determined from the individual steps by applying the Arrhenius equation, which relates the temperature dependence of rate constants to their respective activation energies. Given the relationship $\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$, where $E_a$ represents the activation energy, one can derive the overall activation energy by differentiating the natural logarithm of the given rate constant expression with respect to temperature.

Starting with the expression $k = \sqrt{\frac{k_1 k_3}{k_2}}$, taking the natural logarithm on both sides leads to $\ln k = \frac{1}{2} (\ln k_1 + \ln k_3 - \ln k_2)$. Differentiating both sides with respect to temperature gives the equation $\frac{d(\ln k)}{dT} = \frac{1}{2} (\frac{d(\ln k_1)}{dT} + \frac{d(\ln k_3)}{dT} - \frac{d(\ln k_2)}{dT})$. Replacing each derivative term with its equivalent activation energy expression $\frac{E_a}{RT^2}$ yields the simplified relationship $E_a = \frac{1}{2} (E_{a_1} + E_{a_3} - E_{a_2})$. Substituting the given values of 60 kJ/mol, 30 kJ/mol, and 10 kJ/mol into this formula results in $E_a = \frac{1}{2} (60 + 10 - 30)$, which gives a final value of 20 kJ/mol.

For any reaction of order $n$, the half-life $t_{1/2}$ scales with the initial concentration $[A]_0$ according to the relationship $t_{1/2} \propto [A]_0^{1-n}$. By comparing the two sets of experimental data provided, we can write the ratio as $\frac{t_{1/2(1)}}{t_{1/2(2)}} = \left(\frac{[A]_{0(1)}}{[A]_{0(2)}}\right)^{1-n}$. Substituting the specific values gives $\frac{200}{100} = \left(\frac{0.100}{0.025}\right)^{1-n}$, which simplifies to $2 = 4^{1-n}$. Recognizing that $4$ is $2^2$, the equation becomes $2^1 = 2^{2(1-n)}$, which leads to $1 = 2 - 2n$. Solving for $n$ confirms that the reaction order is $0.5$.

The order of a reaction is an intrinsic property determined by the reaction mechanism and does not fluctuate based on the initial reactant concentration, rendering the claim that the order changes with concentration invalid. To determine the half-life at other concentrations, we establish the proportionality constant $K$ using the relationship $t_{1/2} = K[A]_0^{0.5}$. Substituting $t_{1/2} = 200\,\mathrm{min}$ and $[A]_0 = 0.1\,\mathrm{M}$ allows us to find $K = \frac{200}{\sqrt{0.1}} = 200\sqrt{10}$. With this constant, the half-life for an initial concentration of $1\,\mathrm{M}$ becomes $t_{1/2} = 200\sqrt{10} \times \sqrt{1} = 200\sqrt{10}\,\mathrm{min}$. Similarly, calculating for an initial concentration of $1.6\,\mathrm{M}$ yields $t_{1/2} = 200\sqrt{10} \times \sqrt{1.6} = 200 \times \sqrt{16} = 200 \times 4 = 800\,\mathrm{min}$.

In an energy profile diagram, the activation energy for the forward reaction reflects the total energy required to reach the transition state, given here as $E_1+E_2$, while the activation energy for the backward reaction is represented by $E_1$. Because the products occupy a higher potential energy level on the diagram than the starting reactants, they are inherently at a lower thermodynamic stability. This relationship confirms that the forward reaction requires the full sum of energies to proceed and results in a less stable product.

A catalyst functions by providing an alternative reaction pathway with lower activation energy, increasing the rate of reaction without altering the system's thermodynamic state. Because Gibbs free energy change ($\Delta G$) and enthalpy change ($\Delta H$) are state functions determined solely by the initial and final states of the reactants and products, they remain invariant regardless of the pathway. The enthalpy change, calculated as $\Delta H = E_{a,f} - E_{a,b} = 180 - 200 = -20 \text{ kJ}\,\mathrm{mol}^{-1}$, is identical for both catalyzed and uncatalyzed reactions, confirming that catalysts influence the kinetics of a process rather than its spontaneity or equilibrium position.

For a zero-order reaction, the half-life is defined by the equation $t_{1/2} = \frac{[A]_0}{2k}$ where $[A]_0$ is the initial concentration and $k$ is the rate constant. Substituting the given values $t_{1/2} = 1$ hour and $[A]_0 = 2.0$ mol L$^{-1}$ into the equation gives $1 = \frac{2.0}{2k}$, which simplifies to $k = 1.0$ mol L$^{-1}$ h$^{-1}$. The rate law for this reaction is expressed as $[A]_t = [A]_0 - kt$, which can be rearranged to determine the change \in concentration as $\Delta[A] = kt$. Given the required decrease from $0.50$ to $0.25$ mol L$^{-1}$, the change is $\Delta[A] = 0.25$ mol L$^{-1}$. Setting the equation as $0.25 = 1.0 × t$ yields $t = 0.25$ hours, which is equal to 15 minutes.

For a first-order gas-phase reaction where reactant $A$ decomposes into two gaseous products, the rate constant is defined by the natural logarithm of the ratio of the initial pressure of the reactant to its pressure at any time $t$, expressed as $k = \frac{1}{t} \ln \frac{p_0}{p_A}$. At infinite time, all reactant $A$ has decomposed, meaning the total pressure $p_{\infty }$ equals the sum of the partial pressures of the products, which corresponds to $2p_0$, or $p_0 = \frac{p_{\infty }}{2}$. Given that the total pressure at time $t$ is $p_t = p_0 + p$, the remaining partial pressure of reactant $A$ is $p_0 - p = 2p_0 - p_t$, which simplifies to $p_{\infty } - p_t$ when substituting the expression for $p_0$. Combining these relations leads to the final expression for the rate constant, $k = \frac{1}{t} \ln \frac{p_{\infty }}{2 (p_{\infty } - p_t)}$.

In first-order kinetics, the duration required for a reactant concentration to decrease to a specific fraction of its initial value is determined by the expression $t = \frac{1}{k} \ln \frac{A_0}{A_t}$. For the reactant to reach one fourth of its initial concentration at time $t_1$, the calculation yields $t_1 = \frac{1}{k} \ln \frac{A_0}{A_0/4}$, which simplifies to $\frac{1}{k} \ln 4$ or $\frac{2}{k} \ln 2$. When the concentration reaches one eighth of its initial value at time $t_2$, the expression becomes $t_2 = \frac{1}{k} \ln \frac{A_0}{A_0/8}$, which simplifies to $\frac{1}{k} \ln 8$ or $\frac{3}{k} \ln 2$. Comparing these two durations by calculating the ratio $\frac{t_1}{t_2}$ results in the cancellation of the rate constant $k$ and the logarithmic factor, leaving a final ratio of $\frac{2}{3}$.

For first-order reactions, the time required for a substance to reduce to 50% of its initial concentration is constant and known as the half-life. The concentration decreases from 16 mg/mL to 4 mg/mL over 12 months, which spans two half-life cycles because the drug concentration halves from 16 mg/mL to 8 mg/mL and then from 8 mg/mL to 4 mg/mL. By dividing the total duration by the two cycles, we calculate the half-life as $t_{1/2} = 12 / 2 = 6$ months. Since the drug becomes ineffective precisely at 50% decomposition, which represents a single half-life interval, the expiry time is 6 months.

The direct proportionality between half-life and initial concentration, expressed as $t_{1/2} \propto [A_0]$, identifies this as a zero-order reaction. For such a reaction, the half-life follows the equation $t_{1/2} = \frac{[A_0]}{2k}$, which implies that the constant ratio of half-life to initial concentration is $\frac{1}{2k} = 76.92$. Solving for the rate constant gives the equation $k = \frac{1}{2 \times 76.92}$, which equals $6.5 \times 10^{-3}$ mol L^-1 min^-1.

With the rate constant established, we apply the integrated rate law for a zero-order process, $[A]_t = [A]_0 - kt$. Substituting the initial concentration of 2.5 mol L^-1, the rate constant, and the time of 10 minutes into the equation, we calculate the remaining concentration as $[A]_t = 2.5 - (6.5 \times 10^{-3} \times 10)$. This simplifies to 2.5 minus 0.065, resulting in a final concentration of 2.435 mol L^-1. Expressed in the requested format, this value is 2435 times 10^-3 mol L^-1.