Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The temperature dependence of the reaction rate constant is described by the Arrhenius equation in the form $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. With the rate constant quadrupling, the ratio $\frac{k_2}{k_1} = 4$, allowing for the substitution $\ln(4) = \frac{E_a}{8.314} \left(\frac{310 - 300}{300 \times 310}\right)$. Given that $\ln(4) = 2 \ln(2) = 1.386$, the expression simplifies to $1.386 = \frac{E_a}{8.314} \left(\frac{1}{9300}\right)$. Solving for the activation energy yields $E_a = 1.386 \times 8.314 \times 9300\,\mathrm{J\,mol^{-1}}$, which results in $E_a = 107.2\,\mathrm{kJ\,mol^{-1}}$.

The Arrhenius equation describes the rate constant as $k = A e^{-E_a/RT}$, where $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature. When comparing two reactions with identical pre-exponential factors, the ratio $k_2/k_1$ simplifies to $e^{(E_{a_1} - E_{a_2})/RT}$ because the $A$ terms cancel out.

Taking the natural logarithm of this ratio yields $\ln(k_2/k_1) = (E_{a_1} - E_{a_2}) / (RT)$. Substituting the provided values, where the difference in activation energy is 10,000 J mol⁻¹, the temperature is 300 K, and the gas constant is 8.314 J mol⁻¹ K⁻¹, we arrive at the calculation $\frac{10000}{8.314 \times 300} = 4$.

The temperature dependence of a reaction rate is governed by the Arrhenius equation:

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

For reaction A, where the rate doubles when the temperature rises from 300 K to 310 K, we express this as $\ln(2) = \frac{E_A}{R} \left(\frac{1}{300} - \frac{1}{310}\right)$. Since the rate for reaction B also doubles and its activation energy is twice that of reaction A, where $E_B = 2E_A$, we set up a similar equation for reaction B starting at 300 K: $\ln(2) = \frac{2E_A}{R} \left(\frac{1}{300} - \frac{1}{T}\right)$. Because the left-hand sides of both equations are identical, we equate the right-hand sides to find that $\frac{E_A}{R} \left(\frac{1}{300} - \frac{1}{310}\right) = \frac{2E_A}{R} \left(\frac{1}{300} - \frac{1}{T}\right)$.

By canceling the common term $\frac{E_A}{R}$, we simplify the relationship to $\frac{1}{300} - \frac{1}{310} = 2 \left(\frac{1}{300} - \frac{1}{T}\right)$, which further rearranges to $\frac{2}{T} = \frac{1}{300} + \frac{1}{310}$. Performing the arithmetic yields $\frac{2}{T} = \frac{610}{93000}$, resulting in $T = \frac{186000}{610}$, which equals approximately 304.92 K. The required temperature increase is 304.92 K - 300 K, which is 4.92 K.

The rate constant for an overall chemical reaction constructed from a series of elementary steps is determined by calculating the product of the rate constants of those individual steps. The net reaction $\mathrm{O}_3(g) + \mathrm{O}^{\bullet}(g) \rightarrow 2\mathrm{O}_2(g)$ is formed by summing the two provided elementary processes, so its rate constant, $k_{\text{overall}}$, is found using the relation:
$k_{\text{overall}} = k_i \times k_{ii}$
To arrive at the final value, multiply the rate constants for the two steps, $5.2 \times 10^9 \mathrm{L\,mol^{-1}\,s^{-1}}$ and $2.6 \times 10^{10} \mathrm{L\,mol^{-1}\,s^{-1}}$. The product of these values is $1.352 \times 10^{20} \mathrm{L\,mol^{-1}\,s^{-1}}$, which rounds to $1.4 \times 10^{20} \mathrm{L\,mol^{-1}\,s^{-1}}$.

The rate constant $k$ is a proportionality constant that depends exclusively on temperature and the nature of the reactants, remaining invariant with respect to changes in reactant concentrations. Because it is not a function of the concentration of species $A$ or $B$, increasing the concentration of $B$ will change the overall reaction rate but will have no effect on the value of $k$, which remains $k$.

The decomposition of hydrogen peroxide is a first-order reaction, meaning the rate of disappearance depends directly on the current concentration of the reactant. Since the concentration drops from 0.5 M to 0.125 M, we observe that the concentration has halved twice over the course of 50 minutes. This indicates that two half-lives have elapsed, establishing a half-life of $t_{1/2} = 25$ minutes. Given this, the rate constant is determined by the relationship $k = \frac{0.693}{25}$ per minute.

For the reaction $H_2O_2 \rightarrow H_2O + \frac{1}{2} O_2$, the stoichiometric coefficients dictate that the rate of oxygen formation is half the rate of hydrogen peroxide decomposition. Using the first-order rate law, the rate of oxygen formation is expressed as $\frac{d[O_2]}{dt} = \frac{1}{2} k [H_2O_2]$. By substituting the rate constant and the concentration of 0.05 M into the equation, the calculation $\frac{1}{2} \times \frac{0.693}{25} \times 0.05$ yields a rate of $R = 6.93 \times 10^{-4}$ mol min$^{-1}$ for the formation of oxygen.

Chemical reactions fundamentally rely on collision theory, which posits that reactant molecules must physically strike one another to initiate a chemical transformation. When a reaction order exceeds $3$, the process requires the simultaneous collision of four or more independent particles at a specific point in space and time with the necessary orientation and kinetic energy. The geometric and statistical probability of achieving such a highly synchronized event is extremely low, making these high-order interactions significantly less frequent than simpler bimolecular or termolecular collisions.

The rate law of a chemical reaction describes how the speed of product formation depends on the concentrations of the reactants, expressed mathematically by the general form $\frac{dc}{dt} = k[\mathrm{A}]^x[\mathrm{B}]^y$. By comparing the first and second experiments, holding the concentration of $\mathrm{A}$ constant at $0.1\text{ M}$ while doubling the concentration of $\mathrm{B}$ from $0.1\text{ M}$ to $0.2\text{ M}$ results in no change to the rate of formation, which remains $1.2 \times 10^{-3}\text{ mol L}^{-1}\text{s}^{-1}$. This observation indicates that the reaction is zero-order with respect to $\mathrm{B}$, meaning the exponent $y$ is zero. In the third experiment, keeping the concentration of $\mathrm{B}$ at $0.1\text{ M}$ while doubling the concentration of $\mathrm{A}$ from $0.1\text{ M}$ to $0.2\text{ M}$ causes the rate of formation to increase to $2.4 \times 10^{-3}\text{ mol L}^{-1}\text{s}^{-1}$. Because doubling the concentration of $\mathrm{A}$ precisely doubles the reaction rate, the reaction is first-order with respect to $\mathrm{A}$, making the exponent $x$ equal to one. Combining these findings leads to the resulting rate law expression:

$\frac{dc}{dt}=k[\mathrm{A}]$

The Arrhenius equation provides the mathematical framework for understanding how temperature influences chemical reaction kinetics. Since the rate constant reflects the frequency of successful molecular collisions, and this frequency is sensitive to temperature, we can quantify the energy barrier-the activation energy-by comparing the rate constants at two different temperatures. This relationship is expressed through the integrated form of the Arrhenius equation, which allows us to determine the activation energy based on the ratio of rate constants and the corresponding temperature values.

Substituting the known values into the equation $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$, the ratio $\frac{k_2}{k_1}$ is 2 since the reaction rate doubles. With the gas constant $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$, $T_1 = 300\,\mathrm{K}$, and $T_2 = 310\,\mathrm{K}$, the expression becomes $\log 2 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{310 - 300}{300 \times 310} \right]$. Utilizing the given value $\log 2 = 0.301$, the equation becomes $0.301 = \frac{E_a}{19.147} \left( \frac{10}{93000} \right)$. Rearranging the terms to isolate $E_a$ yields $E_a = \frac{0.301 \times 2.303 \times 8.314 \times 300 \times 310}{10}$, which calculates to approximately $53603\,\mathrm{J\,mol}^{-1}$. Converting this into kilojoules results in $53.6\,\mathrm{kJ\,mol}^{-1}$.

The rate of a first-order reaction depends linearly on the reactant concentration, defined by the relationship $Rate = k[A]$. The rate constant $k$ is determined using the integrated rate equation:

$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$

Substituting the known values of $t = 40\ \text{min}$, $[A]_0 = 0.1\ \text{M}$, and $[A]_t = 0.025\ \text{M}$ into the expression gives $k = \frac{2.303}{40} \log(\frac{0.1}{0.025})$. Since $\frac{0.1}{0.025}$ equals $4$ and $\log 4 \approx 0.6020$, the equation simplifies to $k = \frac{2.303 \times 0.6020}{40}$, which results in $k = 3.47 \times 10^{-2}\ \text{min}^{-1}$. Applying this rate constant to the target concentration of $0.01\ \text{M}$ yields $Rate = 3.47 \times 10^{-2} \times 0.01$, which equals $3.47 \times 10^{-4}\ \text{M/min}$.

The temperature coefficient rule states that for every 10°C rise in temperature, the reaction rate doubles. With a temperature increase of 50°C, there are 5 successive doubling intervals. The total increase in rate is determined by the calculation $2^5 = 32$, indicating that the reaction rate becomes 32 times faster.

The rate law of a reaction is fundamentally governed by the stoichiometry of its slowest elementary step, known as the rate-determining step. If the slow step comprises only reactants found in the overall balanced equation, the rate expression is simply the product of their concentrations. However, if the slow step involves chemical intermediates produced by preceding fast equilibrium processes, we must substitute the concentrations of those intermediates using the corresponding equilibrium expressions to ensure the final rate law is formulated in terms of the actual reactants.

Applying this logic to mechanism A, the slow step is an elementary reaction involving $\mathrm{Cl}_2$ and $\mathrm{H}_2\mathrm{S}$. This directly leads to the rate law Rate $=k[\mathrm{Cl}_2][\mathrm{H}_2\mathrm{S}]$, which is entirely consistent with the experimental equation provided.

For mechanism B, the slow step involves $\mathrm{Cl}_2$ and the intermediate species $\mathrm{HS}^-$. To determine the rate law, we must account for the preceding fast equilibrium $\mathrm{H}_2\mathrm{S} \Leftrightarrow \mathrm{H}^+ + \mathrm{HS}^-$. The equilibrium expression provides the relation [\mathrm{HS}^-] = \frac{K[\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}^+]\}$, where$K$is the equilibrium constant. When we substitute this intermediate concentration into the rate expression for the slow step, we obtain Rate$= k' \frac{[\mathrm{Cl}_2][\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}^+]$}$. Because this derivation results in an inverse dependence on the concentration of hydrogen ions that is absent from the given rate law, mechanism B is inconsistent with the observed kinetic data.

In a zero-order reaction, the half-life is determined by the initial concentration of the reactant $[A]_0$ and the rate constant $k$ according to the expression $t_{1/2} = \frac{[A]_0}{2k}$. Substituting the provided initial concentration of 2.0 mol L⁻¹ and the half-life of 1 hour into the rearranged equation $k = \frac{[A]_0}{2t_{1/2}}$ shows that the rate constant is 1 mol L⁻¹ h⁻¹.

Since the rate of a zero-order reaction is independent of the reactant concentration, the rate constant is defined by $k = \frac{\Delta[A]}{\Delta t}$, where $\Delta[A]$ is the change in concentration over time. Given the decrease from 0.50 mol L⁻¹ to 0.25 mol L⁻¹, the change in concentration is 0.25 mol L⁻¹. Dividing this change by the rate constant of 1 mol L⁻¹ h⁻¹ results in a total time of 0.25 hours.

For a first-order reaction, the rate constant $k$ is related to the half-life by the equation $k = \frac{0.693}{t_{1/2}}$. Substituting the given half-life of 6.93 minutes, we calculate the rate constant as $k = \frac{0.693}{6.93} = 0.1 \text{ min}^{-1}$.

To determine the time required for 99% of the reaction to be complete, we use the integrated rate law formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$. When the reaction reaches 99% completion, starting from an initial concentration $[A]_0$ of 100, the remaining concentration $[A]_t$ is 1. Substituting these values into the rate law yields $0.1 = \frac{2.303}{t} \log \frac{100}{1}$. Since the logarithm of 100 is 2, the expression simplifies to $0.1 = \frac{2.303 \times 2}{t}$. Solving for $t$ results in $t = \frac{4.606}{0.1}$, which gives a total time of 46.06 minutes.

The rate of a chemical reaction is defined by relating the change in concentration of each species to its stoichiometric coefficient, such that the rate of change for each reactant and product, divided by its respective coefficient, is equal. For the reaction $\frac{1}{2} A \rightarrow 2 B$, the rate of disappearance of $A$ is represented by $-\frac{1}{1/2} \frac{d[A]}{dt}$ and the rate of appearance of $B$ is represented by $\frac{1}{2} \frac{d[B]}{dt}$. Equating these two expressions leads to $-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$. Dividing both sides of this equation by 2 yields the final relationship $-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.

The enthalpy change of a chemical reaction is determined by the difference between the activation energy of the forward reaction and the activation energy of the reverse reaction, expressed as $\Delta H = E_f - E_b$. A catalyst provides an alternative pathway that lowers the activation energy of both the forward and reverse reactions by the same value, which means the difference between them remains constant regardless of the catalyst's presence. Using the provided initial values, the enthalpy change is $\Delta H = 180 - 200 = -20$ kJ mol^-1. Reducing both activation energies by 100 kJ mol^-1 results in a new forward activation energy of 80 kJ mol^-1 and a new reverse activation energy of 100 kJ mol^-1, and their difference remains $\Delta H = 80 - 100 = -20$ kJ mol^-1, confirming that the magnitude of the enthalpy change stays constant at 20 kJ mol^-1.

Isotopes are defined by sharing an identical atomic number despite having different atomic masses. Because a neutron possesses an atomic number of 0 and an atomic mass of 1, any change in the number of neutrons within a nucleus alters the total mass number without changing the elemental identity, as seen in the transformation $_{92}\mathrm{U}^{238} + _{0}\mathrm{n}^{1} \rightarrow _{92}\mathrm{U}^{239}$. In contrast, radioactive processes such as alpha, beta, and positron emission involve the release of charged particles that change the proton count, resulting in the transmutation of the original atom into a completely different chemical element.

The order of a chemical reaction relative to a specific reactant is determined by observing how the reaction rate or half-life responds to concentration changes. Because the half-life remains constant when the concentration of B is doubled, the reaction is first-order with respect to B, and because the rate doubles when the concentration of A is doubled, the reaction is also first-order with respect to A, leading to an overall order of $n = 1 + 1 = 2$. For any reaction of order $n$, the rate constant unit is defined by $(L)^{n-1} (mol)^{1-n} s^{-1}$. Substituting $n = 2$ into this formula gives $(L)^{2-1} (mol)^{1-2} s^{-1}$, resulting in final units of $L mol^{-1} s^{-1}$.

Radioactive decay follows an exponential decrease over time, where the relationship between the initial activity and the remaining activity depends on the substance's half-life. To determine how long it takes for the initial activity, which is 10 times the safe limit, to reach the safe level, we use the formula $t = \frac{2.303}{\lambda} \log_{10}\left(\frac{A_0}{A}\right)$ with the decay constant defined by $\lambda = \frac{0.693}{30}$. Plugging in these values gives $t = \frac{2.303 \times 30}{0.693} \times \log_{10}(10)$. Since the logarithm of 10 to the base 10 is 1, the calculation simplifies to $\frac{2.303 \times 30}{0.693}$, which results in a time of 100 days.

The rate law for a reaction that is second order with respect to carbon monoxide is defined as $r = k[CO]^2$, where $k$ is the rate constant. Representing the initial concentration as $a$, the starting rate is $r_1 = k a^2$. When the concentration is doubled to $2a$, the new rate becomes $r_2 = k(2a)^2$, which simplifies to $r_2 = 4 k a^2$. Because the original rate was $k a^2$, this new rate is equivalent to $r_2 = 4 r_1$, meaning the reaction rate increases by a factor of 4.