Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The temperature dependence of the rate constant is governed by the Arrhenius equation, which describes the relationship between the rate constant $k$, the activation energy $E_a$, and the temperature $T$. This relationship is expressed logarithmically as $\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$. With the given rate constants $k_1 = 1.5 \times 10^3 \ \mathrm{s}^{-1}$ and $k_2 = 4.5 \times 10^3 \ \mathrm{s}^{-1}$, the ratio $\frac{k_2}{k_1}$ is 3. Since $\ln x = 2.3 \log x$, we can adapt the equation to use common logarithms: $\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.3 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.

Substituting the known values $E_a = 60 \times 10^3 \ \mathrm{J} \ \mathrm{mol}^{-1}$, $R = 8.3 \ \mathrm{J} \ \mathrm{K}^{-1} \ \mathrm{mol}^{-1}$, $T_1 = 300 \ \mathrm{K}$, and $\log 3 = 0.48$ into this formula gives $0.48 = \frac{60 \times 10^3}{2.3 \times 8.3} \left( \frac{1}{300} - \frac{1}{T_2} \right)$. Rearranging the expression to isolate the temperature term results in $\frac{1}{T_2} = \frac{1}{300} - \frac{0.48 \times 2.3 \times 8.3}{60 \times 10^3}$. Evaluating these terms leads to $\frac{1}{T_2} = 0.00333 - 0.00015$, which simplifies to $\frac{1}{T_2} \approx 0.00318 \ \mathrm{K}^{-1}$. Solving for $T_2$ yields approximately $314.4 \ \mathrm{K}$. Converting this to degrees Celsius by subtracting 273 results in $41.4^\circ \mathrm{C}$, which rounds to 41.

The radioactive decay follows the law $N_t = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$, where $N_t$ is the remaining activity after time $t$ and $t_{1/2}$ is the half-life. Given that 75% of the activity remains, we set $\frac{N_t}{N_0} = 0.75 = \frac{3}{4}$ and substitute the known half-life of 245 days into the equation to get $\frac{3}{4} = \left(\frac{1}{2}\right)^{x/245}$. Taking the logarithm on both sides results in $\log\left(\frac{3}{4}\right) = \frac{x}{245} \log\left(\frac{1}{2}\right)$, which expands to $\log 3 - 2\log 2 = -\frac{x}{245} \log 2$.

Substituting the given values $\log 3 = 0.4771$ and $\log 2 = 0.3010$ into the equation gives $0.4771 - 2(0.3010) = -\frac{x}{245}(0.3010)$. This simplifies to $0.4771 - 0.6020 = -\frac{x}{245}(0.3010)$, resulting in $-0.1249 = -\frac{x}{245}(0.3010)$. Solving for $x$ provides $x = \frac{0.1249 \times 245}{0.3010} \approx 101.66$. Rounding to the nearest integer, the value of $x$ is 102.

The Arrhenius equation describes the dependence of a reaction rate constant on temperature through the expression $k = A e^{-\frac{E_a}{RT}}$, where $E_a$ is the activation energy and $R$ is the universal gas constant. By comparing this standard form with the provided equation $k = (5.5 \times 10^{11} \mathrm{s}^{-1}) e^{-\frac{28000\,K}{T}}$, it becomes clear that the term in the exponent $-\frac{E_a}{RT}$ is equivalent to $-\frac{28000\,K}{T}$.

This direct correspondence allows us to determine the activation energy by equating $\frac{E_a}{R} = 28000\,K$. Multiplying $28000$ by the given gas constant value of $8.3 \, \mathrm{JK}^{-1} \, \mathrm{mol}^{-1}$ results in an activation energy of $232400 \, \mathrm{J} \, \mathrm{mol}^{-1}$. Converting this value into kilojoules by dividing by $1000$ yields $232.4 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$, which rounds to the nearest integer of $232$.

Setting the rate constants equal requires solving the equation $10^6 e^{-\frac{30000}{T}} = 10^4 e^{-\frac{24000}{T}}$. Rearranging the terms by dividing both sides by $10^4$ and $e^{-\frac{30000}{T}}$ gives $10^2 = e^{\frac{30000-24000}{T}}$, which simplifies to $100 = e^{\frac{6000}{T}}$. Taking the natural logarithm on both sides results in $\ln 100 = \frac{6000}{T}$, or $2 \ln 10 = \frac{6000}{T}$. Substituting the given value $\ln 10 = 2.303$ leads to $2 \times 2.303 = \frac{6000}{T}$, which simplifies to $4.606 = \frac{6000}{T}$. Solving for $T$ yields $T = \frac{6000}{4.606} \approx 1302.65$, which rounds to the nearest integer 1303 K.

The logarithmic form of the Arrhenius equation, $\ln \left( \frac{K_2}{K_1} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$, establishes the relationship between temperature changes and the rate constant. Using the activation energy $E_a = 12.6\,\text{kcal/mol}$, which converts to $12.6 \times 10^3 \times 4.2\,\text{J/mol}$, and the gas constant $R = 8.314\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}$, the ratio of rate constants for a temperature shift from 298 K to 308 K is determined through the calculation:
$\ln \left( \frac{K_2}{K_1} \right) = \frac{12.6 \times 10^{3} \times 4.2}{8.314} \left( \frac{10}{298 \times 308} \right) \approx 0.693$
Because $\ln 2$ is approximately 0.693, the ratio $\frac{K_2}{K_1}$ equals 2, verifying that the first statement is true.

The half-life of a first-order reaction is given by $t_{1/2} = \frac{\ln 2}{k}$, a relationship that shows the half-life is dependent only on the rate constant $k$. As this value remains constant regardless of the initial concentration $[A]_0$, any assertion that implies a dependency on the initial reactant amount for a first-order process is incorrect.

The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$. For decomposition to one-eighth of the initial concentration, $t_{\frac{1}{8}} = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_0/8} \right) = \frac{2.303}{k} \log(8) = \frac{2.303}{k} \times 3 \log 2$. For decomposition to one-tenth of the initial concentration, $t_{\frac{1}{10}} = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_0/10} \right) = \frac{2.303}{k} \log(10) = \frac{2.303}{k}$. The ratio of these two times is $\frac{t_{\frac{1}{8}}}{t_{\frac{1}{10}}} = 3 \log 2$. Substituting $\log 2 = 0.3$ into the expression gives $3 \times 0.3 = 0.9$. Multiplying this value by 10 yields 9.

The Arrhenius equation describes the temperature dependence of rate constants as $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$, and the first-order rate constant relates to the half-life by $k = \frac{\ln 2}{t_{1/2}}$. For the first reaction, given $t_{1/2} = 2$ hours at 500 K, the rate constant is $k_{1, 500} = \frac{\ln 2}{2} = 0.3465$ per hour. With the condition $k_{1, 500} = 2k_{1, 300}$, the Arrhenius equation becomes $\ln\left(\frac{k_{1, 500}}{k_{1, 300}}\right) = \ln 2 = \frac{E_{a1}}{R} \left(\frac{1}{300} - \frac{1}{500}\right)$, which simplifies to $\ln 2 = \frac{E_{a1}}{R} \left(\frac{1}{750}\right)$.

For the second reaction, since $E_{a2} = 0.5 E_{a1}$, substituting this into the Arrhenius equation gives $\ln\left(\frac{k_{2, 500}}{k_{2, 300}}\right) = \frac{E_{a2}}{R} \left(\frac{1}{750}\right) = 0.5 \left(\frac{E_{a1}}{R}\right) \left(\frac{1}{750}\right) = 0.5 \ln 2 = \ln \sqrt{2}$. This result implies $\frac{k_{2, 500}}{k_{2, 300}} = \sqrt{2}$. Given $k_{2, 500} = 2 k_{1, 500} = 2 \times 0.3465 = 0.693$ per hour, the rate constant for the second reaction at 300 K is $k_{2, 300} = \frac{0.693}{\sqrt{2}} \approx 0.49$ per hour. This value equals $4.9 \times 10^{-1}$ per hour, leading to 5 as the nearest integer.

The potential energy profile of a multi-step reaction is governed by the enthalpy change ($\Delta H$) of each elementary step, where endothermic processes ($\Delta H > 0$) cause an increase in potential energy and exothermic processes ($\Delta H < 0$) cause a decrease. Since the overall reaction $A \rightarrow D$ is endothermic, the potential energy of the final product $D$ must be higher than that of the initial reactant $A$. The correct profile must therefore feature an initial rise in energy from $A$ to the first intermediate $B$, followed by two successive decreases in energy during the formation of intermediate $C$ and final product $D$, such that the net energy of $D$ remains above that of $A$.

The thermal decomposition of the reactant $\mathrm{AB}(\mathrm{g})$ follows first-order kinetics, defined by the integrated rate law $\ln[\mathrm{AB}]_t = -kt + \ln[\mathrm{AB}]_0$. In the plot of $\ln[\mathrm{AB}]$ against time $t$, the slope of the linear fit is equal to $-k$. Extracting the slope from the provided graphical data yields a rate constant $k = 0.0693 \text{ min}^{-1}$. The half-life of a first-order reaction is determined by the formula $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$. Substituting the value of $k$ into this expression results in $t_{1/2} = \frac{0.693}{0.0693} = 10 \text{ min}$.

The stoichiometric relationship for the dissociation reaction $A \longrightarrow nB$ dictates that the rate of formation of $B$ is $n$ times the rate of disappearance of $A$, expressed as $\frac{d[B]}{dt} = -n \frac{d[A]}{dt}$. Over the initial time interval of 10 minutes, this implies that the change in concentrations follows the relationship $\Delta[B] = -n \Delta[A]$, which rearranges to $n = \frac{|\Delta[B]|}{|\Delta[A]|}$. By calculating the decrease in concentration of the reactant $A$, denoted as $|\Delta[A]|$, and the corresponding increase in concentration of the product $B$, denoted as $|\Delta[B]|$, from the values provided in the plot at $t = 10$ minutes, the ratio of these changes identifies the stoichiometric coefficient. Given that the magnitude of product formation is three times the magnitude of reactant consumption from the graphical data, the calculation yields $n = \frac{|\Delta[B]|}{|\Delta[A]|} = 3$.

The Arrhenius equation is $k = A e^{-E_a/RT}$, and the ratio of rate constants for catalyzed and uncatalyzed reactions is given by $\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{(E_{a(\text{uncat})} - E_{a(\text{cat})})/RT}$. Taking the common logarithm results \in the expression $\log\left(\frac{k_{\text{cat}}}{k_{\text{uncat}}}\right) = \frac{\Delta E_a}{2.303 RT}$. Using the given values where $\Delta E_a = 10000$ J mol$^{-1}$, $R = 8.314$ J K$^{-1}$ mol$^{-1}$, and $T = 300$ K, the expression becomes $\log\left(\frac{k_{\text{cat}}}{k_{\text{uncat}}}\right) = \frac{10000}{2.303 \times 8.314 \times 300} = 1.741$.

First-order kinetics dictate that the rate constant is inversely proportional to the half-life, allowing for the reaction speed to be established directly from the provided time interval. The standard relationship $K = \frac{0.693}{t_{1/2}}$ links the rate constant to a half-life of 6.93 minutes, resulting in a rate constant of 0.1 per minute.

The duration for 99% of the reaction to occur follows the integrated rate law equation $t = \frac{2.303}{K} \log \left( \frac{[A]_0}{[A]_t} \right)$ where $[A]_0$ is the initial amount and $[A]_t$ is the amount remaining at time t. Since 99% of the substance has reacted, the ratio of initial to final concentration is 100. The calculation involves multiplying 23.03 by the logarithm of 100, which equals 2, leading to a time of 46.06 minutes, effectively 46 minutes.

The slope of a concentration versus time curve signifies the rate of reactant depletion, and for a constant initial concentration, a steeper descent corresponds to a higher reaction rate. Since the rate constant directly correlates with this rate, two reactions of the same order can be compared by observing which curve decreases more sharply; the steeper curve indicates a larger rate constant. Therefore, as reaction 3 shows a more rapid decrease in reactant concentration than reaction 2, the rate constant $K_3$ is greater than $K_2$.

The rate constant $k$ for a first-order gaseous reaction $A (g) \rightarrow B (g) + C (g)$ can be determined using the initial pressure $P_0$ and the remaining pressure of the reactant $P_A$ at time $t$ with the equation
$k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$

For the reaction, the total pressure at time $t$ is given by $P_t = P_0 + x$, where $x$ represents the extent of decomposition. Given an initial pressure $P_0 = 1$ bar and a total pressure $P_t = 1.5$ bar at $t = 100$ minutes, we solve for $x$ to find $x = P_t - P_0 = 1.5 - 1 = 0.5$ bar. The partial pressure of reactant $A$ at time $t$ is $P_A = P_0 - x = 1 - 0.5 = 0.5$ bar.

Substituting these values into the rate equation gives
$k = \frac{2.303}{100} \log \left( \frac{1}{0.5} \right) = \frac{2.303}{100} \log(2)$
Applying $\log 2 = 0.3$, the calculation becomes $k = \frac{2.303 \times 0.3}{100} = 6.9 \times 10^{-3} \text{ min}^{-1}$.

The Arrhenius equation in base-10 logarithmic form is $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$. Comparing this to the given expression $\log_{10} k_1 = 14.34 - \frac{1.5 \times 10^4}{T}$, the term $\frac{E_{a1}}{2.303 R}$ is equivalent to $1.5 \times 10^4$. Using the gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, the activation energy for the first reaction is $E_{a1} = 1.5 \times 10^4 \times 2.303 \times 8.314 \text{ J mol}^{-1} \approx 287207 \text{ J mol}^{-1}$, or $287.2 \text{ kJ mol}^{-1}$. Since the activation energy of the second reaction is one-fifth of the first, $E_{a2} = \frac{287.2 \text{ kJ mol}^{-1}}{5} = 57.44 \text{ kJ mol}^{-1}$, which rounds to $57 \text{ kJ mol}^{-1}$.

The linear graph starting from the origin indicates that the reaction follows zero-order kinetics, where the half-life $t_{1/2}$ is directly proportional to the initial concentration $[A]_0$. This relationship is defined by the expression $t_{1/2} = \frac{[A]_0}{2k}$, implying that the ratio of two different half-lives must equal the ratio of their corresponding initial concentrations. By setting up the proportion $\frac{240}{x} = \frac{4 \times 10^{-3}}{1.5 \times 10^{-3}}$, where 240 s corresponds to an initial concentration of $4 \times 10^{-3}$ and $x$ corresponds to $1.5 \times 10^{-3}$, we can isolate the unknown time. Rearranging the expression gives $x = \frac{240 \times 1.5}{4}$, which simplifies to 90 s.

Chemical kinetics problems involving gas-phase reactions often rely on tracking the total pressure changes, which correlate directly with the changing moles of gaseous species according to the stoichiometry of the reaction $2\,\mathrm{A}(\mathrm{g}) \rightarrow 4\,\mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$. By setting up a pressure-based ICE table, we designate the initial pressure of A as $P_0$ and the extent of the reaction at time $t$ as $x$, where the partial pressures become $P_A = P_0 - 2x$, $P_B = 4x$, and $P_C = x$. The total pressure at any given time $t$ is the sum of these partial pressures, yielding the expression $P_t = (P_0 - 2x) + 4x + x = P_0 + 3x$.

Given that the total pressure at 30 minutes is $300\,\mathrm{mmHg}$, we have the equation $P_0 + 3x = 300$. As the reaction proceeds to completion at infinite time, all reactant A is consumed, meaning $P_0 - 2x_{\infty } = 0$ and $x_{\infty } = P_0/2$. Consequently, the total pressure at infinite time is determined solely by the products, resulting in $P_{\infty } = 4(P_0/2) + (P_0/2) = 2.5P_0$. With the total pressure at infinity established as $600\,\mathrm{mmHg}$, solving $2.5P_0 = 600$ gives $P_0 = 240\,\mathrm{mmHg}$. Substituting this value back into the expression for the pressure at 30 minutes, $240 + 3x = 300$ simplifies to $3x = 60$, which reveals that $x = 20\,\mathrm{mmHg}$. Since the partial pressure of C is defined as $x$, the final value is $20\,\mathrm{mmHg}$.

For first-order kinetics, the relationship between concentration and time follows the integrated rate law, ensuring the rate constant $k$ remains consistent throughout the reaction. Determining this constant using the known data at 20 minutes and the initial concentration allows for the calculation of the unknown time $x$ associated with the concentration 0.0650 M.

The rate expression $k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]_t} \right)$ at 20 minutes yields $k = \frac{1}{20} \ln \left( \frac{0.6500}{0.00065} \right)$, which simplifies to $\frac{3 \ln 10}{20}$ min⁻¹. Rearranging to solve for $t$ gives $t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]_t} \right)$, and substituting the constant with the target concentration of 0.0650 M results in $x = \frac{20}{3 \ln 10} \ln \left( \frac{0.6500}{0.0650} \right)$. Because the ratio inside the logarithm is 10, the expression simplifies to $\frac{20}{3}$, which equals approximately 6.67, rounding to the nearest integer of 7.

The rate of a first-order reaction is defined by the rate law:
$\text{Rate} = k[A]$
where $k$ is the rate constant and $[A]$ is the molar concentration of the reactant. Because concentration is an intensive property, the reaction rate depends solely on the molarity rather than the total volume of the solution.

For Run 1, the concentration is $10\,\text{M}$, so $\text{Rate}_1 = 10k$. For Run 2, increasing the volume of the same $10\,\text{M}$ solution does not change the concentration, so $\text{Rate}_2 = 10k$. For Run 3, the addition of water dilutes the solution, resulting in a new concentration of $[A] = \frac{10\,\text{M} \times 100\,\text{mL}}{100\,\text{mL} + 100\,\text{mL}} = 5\,\text{M}$, which yields $\text{Rate}_3 = 5k$. Comparison of these results confirms that $\text{Rate}_3 < \text{Rate}_1 = \text{Rate}_2$.

The Arrhenius equation $k = A e^{-E_a / RT}$ describes the rate constant of a reaction, where $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the universal gas constant, and $T$ is the temperature. For two reactions with identical pre-exponential factors at the same temperature, the ratio of the rate constants $k_1$ and $k_2$ simplifies to the following relationship:
$\ln \frac{k_2}{k_1} = \frac{E_{a1} - E_{a2}}{RT}$
Substituting the given values $E_{a1} - E_{a2} = 20000$ J mol$^{-1}$, $R = 8.3$ J K$^{-1}$ mol$^{-1}$, and $T = 300$ K into the equation yields:
$\ln \frac{k_2}{k_1} = \frac{20000}{8.3 \times 300} = \frac{200}{24.9} \approx 8.03$
Rounding this result to the nearest integer gives 8.