Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The rate of a chemical reaction is defined by the expression $r = k[A]^2[B]$, which indicates that the rate is proportional to the square of concentration $A$ and the concentration of $B$. When the concentrations of $A$ and $B$ are both doubled, the new rate becomes $r' = k(2[A])^2(2[B])$, which simplifies to $r' = k(4[A]^2)(2[B]) = 8k[A]^2[B]$, confirming that the factor $x$ is 8. A linear decrease of concentration over time characterizes a zero order reaction, which corresponds to $y = 0$ in this context. Consequently, the sum $x + y = 8 + 0 = 8$.

For a single-step elementary gas-phase reaction, the rate law is derived directly from the stoichiometric coefficients of the reactants. Given the reaction $2\mathrm{A}_{(\mathrm{g})} + \mathrm{B}_{(\mathrm{g})} \longrightarrow \mathrm{C}_{(\mathrm{g})}$, the rate law is defined by the expression $r = k P_A^2 P_B$. When the reaction begins with 1.5 atm of A and 0.7 atm of B, the initial rate is expressed as $r_1 = k(1.5)^2(0.7)$.

Determining the rate $r_2$ when the pressure of C reaches 0.5 atm requires calculating the change in reactant pressures based on stoichiometry. The formation of 0.5 atm of C consumes 1.0 atm of A and 0.5 atm of B, which leaves the system with 0.5 atm of A and 0.2 atm of B. The rate at this second state is $r_2 = k(0.5)^2(0.2)$.

Dividing the first rate by the second provides the ratio $\frac{r_1}{r_2} = \frac{k(1.5)^2(0.7)}{k(0.5)^2(0.2)}$. Simplifying this equation results in $9 \times 3.5 = 31.5$. Expressing this value as $315 \times 10^{-1}$, the final result is 315.

In a first-order gas-phase reaction where $A(g) \rightarrow 2 B(g) + C(g)$, the total pressure at any time relates to the partial pressure of the reactant. At the start of the reaction, the total pressure is equal to the initial partial pressure of $A$, denoted as $P_0$. Since the reaction produces three moles of gas for every mole of $A$ consumed, the total pressure after complete decomposition is $P_{\infty } = 3 P_0$. Given that $P_{\infty } = 300 \text{ torr}$, we determine the initial pressure $P_0 = 100 \text{ torr}$.

At $t = 23 \text{ sec}$, the total pressure $P_t$ of $200 \text{ torr}$ is the sum of the partial pressures of $A$, $B$, and $C$. If $x$ is the decrease in the partial pressure of $A$, then $P_t = (P_0 - x) + 2x + x = P_0 + 2x$. Substituting the known values $P_t = 200 \text{ torr}$ and $P_0 = 100 \text{ torr}$ gives $200 = 100 + 2x$, which simplifies to $x = 50 \text{ torr}$. The partial pressure of reactant $A$ remaining at this time is $P_A = P_0 - x = 100 - 50 = 50 \text{ torr}$.

Using the first-order rate constant formula $k = \frac{1}{t} \ln \frac{P_0}{P_A}$, we calculate $k = \frac{1}{23} \ln \frac{100}{50}$, which simplifies to $\frac{1}{23} \ln 2$. Applying the conversion $\ln 2 \approx 2.3 \times \log_{10}(2)$ and using the provided value $\log_{10}(2) = 0.301$, we find $k = \frac{2.3 \times 0.301}{23} = 3.01 \times 10^{-2} \text{s}^{-1}$. Rounding to the nearest integer as requested, the value is $3$.

The rate at which a reaction proceeds is determined by the stoichiometry of the balanced chemical equation, which relates the rate of disappearance of reactants to the rate of formation of products. For the decomposition reaction
$2\ \mathrm{N}_2\mathrm{O}_5 \rightarrow 4\ \mathrm{NO}_2 + \mathrm{O}_2$
it is evident that for every 2 moles of $\mathrm{N}_2\mathrm{O}_5$ consumed, 4 moles of $\mathrm{NO}_2$ are produced. This implies that the rate of formation of $\mathrm{NO}_2$ is double the rate at which $\mathrm{N}_2\mathrm{O}_5$ disappears.

To determine the rate of disappearance for $\mathrm{N}_2\mathrm{O}_5$, divide the change in concentration by the elapsed time. The concentration decreases by 0.25 over 30 minutes, resulting in a rate of $\frac{0.25}{30}$, which simplifies to $\frac{1}{120}$ moles per liter per minute. Applying the stoichiometric ratio, the rate of formation for $\mathrm{NO}_2$ is calculated as $2 \times \frac{1}{120}$, which simplifies to $\frac{1}{60}$. Expressing this as a decimal yields approximately 0.01666, which can be written as $16.66 \times 10^{-3}$ moles per liter per minute. Given the format provided in the question, the value of x is 17.

For a first-order gaseous reaction where the total pressure changes over time, the rate constant is calculated by relating the partial pressure of the reactant to the total pressure of the system. In the reaction $A(g) \rightarrow 2B(g) + C(g)$, the total pressure at time $t$ is the sum of the partial pressures of all components: $P_t = P_A + P_B + P_C$. Initially, at $t=0$, the system contains only reactant A at a pressure of $0.1$ atm. As the reaction proceeds for $t=115$ seconds, if a quantity $x$ of A decomposes, the partial pressures of B and C increase by $2x$ and $x$ respectively, while the partial pressure of A becomes $0.1 - x$. Consequently, the total pressure is represented by the equation $P_t = (0.1 - x) + 2x + x = 0.1 + 2x$. Given that the total pressure at this time is $0.28$ atm, solving the equation $0.1 + 2x = 0.28$ yields $x = 0.09$ atm. The partial pressure of A at $t=115$ seconds is therefore $0.1 - 0.09 = 0.01$ atm. Applying the first-order integrated rate law $k = \frac{1}{t} \ln \frac{P_0}{P_A}$, we substitute the initial pressure $P_0 = 0.1$ and the final pressure $P_A = 0.01$ to find $k = \frac{1}{115} \ln \frac{0.1}{0.01} = \frac{\ln(10)}{115}$. Evaluating this expression provides a rate constant of $0.02$ $\mathrm{s}^{-1}$, which corresponds to $2 × 10^{-2} \mathrm{s}^{-1}$.

The integrated rate law for a first-order reaction describes the time $t$ required for a reactant to decay, expressed as
$t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$
where $[A]_0$ is the initial concentration and $[A]_t$ is the remaining concentration at time $t$. The half-life, which represents the time required for half the reactant to disappear, is defined by the expression
$t_{1/2} = \frac{2.303}{k} \log 2$
To find the time for 99.9% completion, the remaining reactant is 0.1% of the initial amount, making the concentration ratio equal to 1000. This yields the time calculation
$t_{99.9\%} = \frac{2.303}{k} \log(1000) = \frac{2.303 \times 3}{k}$
Comparing these two durations, the ratio is calculated by dividing $t_{99.9\%}$ by $t_{1/2}$, which simplifies to 3 divided by the value of $\log 2$. Since $\log 2$ is approximately 0.3, dividing 3 by 0.3 results in 10.

The rate law $r = k[A]$ identifies the process as a first-order reaction, where the half-life $t_{1/2}$ is 120 minutes. The rate constant is determined by the expression $k = \frac{0.693}{t_{1/2}}$.

For 90% decomposition, the remaining reactant is 10% of the initial concentration, which requires the integrated rate equation $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$ Substituting the value of $k$ from the half-life period results \in $t = \frac{2.303 \times 120}{0.693} \log \left( \frac{100}{10} \right)$ Because the base-10 logarithm of 10 equals 1, the time taken for the reaction to reach this point is 399 minutes.

The steady-state approximation assumes that the concentration of a reactive intermediate remains constant because its rate of production equals its rate of consumption. For the reaction sequence $A \xrightarrow{K_1} B \xrightarrow{K_2} C$, the net rate of change for the intermediate $B$ is given by the equation
$\frac{d[B]}{dt} = K_1[A] - K_2[B]$
Setting this expression to zero to reflect the steady-state condition implies that $K_1[A] = K_2[B]$. Rearranging this relationship to solve for the concentration of $B$ yields the final result of $[B] = \left(\frac{K_1}{K_2}\right)[A]$.

For a first-order reaction, the rate $r$ decreases exponentially over time following the expression $r = r_0 e^{-kt}$, where $r_0$ is the initial rate, $k$ is the rate constant, and $t$ is the elapsed time. By comparing the rates at two different times, the initial concentration terms cancel out, resulting in the relationship $\frac{r_1}{r_2} = e^{k(t_2 - t_1)}$. Substituting the given rates and times, we have $\frac{0.04}{0.03} = e^{k(20 - 10) \times 60}$, which simplifies to $\frac{4}{3} = e^{600k}$.

Taking the natural logarithm of both sides gives $\ln(4/3) = 600k$. Since the half-life is $t_{1/2} = \frac{\ln 2}{k}$, we can express this as $t_{1/2} = \frac{600 \ln 2}{\ln(4/3)}$. Using the relationship between natural and base-10 logarithms, this formula becomes:

$t_{1/2} = 600 \times \frac{\log 2}{\log 4 - \log 3}$

Substituting the provided values, where $\log 4$ is 0.6020, we get $t_{1/2} = 600 \times \frac{0.3010}{0.6020 - 0.477}$ seconds. Converting this to minutes, where 600 seconds equals 10 minutes, results in $t_{1/2} = 10 \times \frac{0.3010}{0.125}$, which evaluates to 24.08 minutes, rounding to 24 minutes.

The rate law for the reaction is defined by the expression $\text{Rate} = k[\mathrm{HI}]^n$, where $n$ represents the reaction order. To determine $n$, we examine how the reaction rate changes in response to variations in the concentration of HI. Comparing the first and second experimental trials, doubling the concentration from 0.005 mol L^-1 to 0.01 mol L^-1 results in the rate increasing from 7.5 x 10^-4 mol L^-1 s^-1 to 3.0 x 10^-3 mol L^-1 s^-1.

$\frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left( \frac{0.01}{0.005} \right)^n$

Simplifying the left side of the equation shows that the rate increases by a factor of 4, while the concentration ratio on the right side simplifies to 2. This results in the equation $4 = 2^n$, which confirms that the reaction order $n$ is 2.

The overall order of a reaction is defined by the sum of exponents in the rate law expression, represented as $r = K[A]^x[B]^y$. By isolating variables in the experimental data, we can determine these exponents individually. Comparing the first and fourth sets where the concentration of reactant B remains constant at 0.1, we see that increasing the concentration of A from 0.1 to 0.4 causes the rate to shift from $6 \times 10^{-3}$ to $2.4 \times 10^{-2}$. Since a 4-fold increase in the concentration of A leads to a 4-fold increase in the rate, we establish the relationship $4^x = 4$, which solves to $x = 1$. Similarly, by observing the second and third sets where the concentration of A is fixed at 0.3, we see that doubling the concentration of B from 0.2 to 0.4 results in the rate increasing from $7.2 \times 10^{-2}$ to $2.88 \times 10^{-1}$. This 4-fold increase in rate corresponds to a 2-fold change in concentration, meaning $2^y = 4$ and thus $y = 2$. Combining these individual orders gives the overall reaction order as $1 + 2 = 3$.

Since the reactants are provided in equal initial concentrations and react in a 1:1 stoichiometric ratio, the concentration of B remains identical to the concentration of A throughout the process. Substituting $[B] = [A]$ into the given rate expression simplifies the rate law to $\text{rate} = k[A]$, which confirms the reaction follows first-order kinetics. The integrated rate law for this first-order reaction is defined by the equation $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$ where $[A]_0$ represents the initial concentration and $[A]$ represents the final concentration. Substituting the provided values, where $k = 4.6 × 10^{-2} \text{ s}^{-1}$, $[A]_0 = 1 \text{ M}$, and $[A] = 0.1 \text{ M}$, the expression becomes $4.6 \times 10^{-2} = \frac{2.303}{t} \log \left( \frac{1}{0.1} \right)$ Given that $\log(10) = 1$, the equation simplifies to $t = \frac{2.303}{4.6 × 10^{-2}}$, which calculates to 50 seconds.

The amount of reactant remaining in a first-order process follows the relationship $[A] = [A]_0 (\frac{1}{2})^n$, where $n$ represents the number of half-lives that have occurred. Because the reactant concentration reduces to $\frac{1}{32}$ of its initial level, we set $(\frac{1}{2})^n = \frac{1}{32}$, which indicates that $n = 5$ half-lives have elapsed. The time for a single half-life is determined by the formula $t_{1/2} = \frac{0.693}{k}$. Given a rate constant of 20 min$^{-1}$, the duration of one half-life is $\frac{0.693}{20}$ min. Multiplying this by 5 for the total time yields $5 \times \frac{0.693}{20}$, which simplifies to $0.17325$ min. Expressing this result in the requested format, we obtain $17.325 \times 10^{-2}$ min, which rounds to 17 when taken to the nearest integer.

First-order kinetics dictate that the time required for a reactant to decrease from an initial quantity to a final quantity is determined by the integrated rate law, which is expressed as $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$. In this scenario, the reaction proceeds with a rate constant $k = 2.011 \times 10^{-3} \, \text{s}^{-1}$, starting with $7 \, \text{g}$ of reactant and ending with $2 \, \text{g}$.

Plugging these values into the equation, we get $t = \frac{2.303}{2.011 \times 10^{-3}} \log \frac{7}{2}$. Using the property of logarithms where $\log \frac{7}{2} = \log 7 - \log 2$, we calculate the value as $0.845 - 0.301$, which equals $0.544$. Multiplying $2.303$ by $0.544$ results in $1.252832$. When this value is divided by the rate constant of $2.011 \times 10^{-3} \, \text{s}^{-1}$, the calculation yields approximately $622.989$. Rounding this result to the nearest integer provides a final time of $623$ seconds.

The Arrhenius equation, $k = A e^{-\frac{E_a}{R T}}$, serves as the fundamental relationship for understanding how activation energy and temperature dictate reaction rates. Since the rate constant $k$ is inversely related to the exponential of the activation energy $E_a$, a higher energy barrier inherently leads to a smaller rate constant, confirming the first statement. The temperature coefficient, defined as the ratio $\frac{k_{T+10}}{k_T}$, reflects a reaction's sensitivity to temperature changes and is inherently higher for reactions with larger activation energies. While the rate constant changes with temperature, the magnitude of this change is not uniformly greater at lower temperatures, making the third statement inaccurate, whereas rearranging the Arrhenius equation into $\ln k = \ln A - \frac{E_a}{R} \left(\frac{1}{T}\right)$ confirms that a plot of $\ln k$ versus $\frac{1}{T}$ yields a straight line with a slope of $-\frac{E_a}{R}$.

The relationship between the half-life $t_{\frac{1}{2}}$ of a reaction and the initial pressure $P$ of the gaseous reactant is defined by the proportionality $t_{\frac{1}{2}} \propto P^{1-n}$, where $n$ represents the order of the reaction. This principle allows us to compare how changes in initial pressure affect the half-life, providing a clear path to determine the reaction order.

By utilizing the first two data points provided, where an initial pressure of $50\,\mathrm{mm\,Hg}$ results in a half-life of $4\,\mathrm{s}$ and an initial pressure of $100\,\mathrm{mm\,Hg}$ results in a half-life of $2\,\mathrm{s}$, we can set up the ratio $\frac{4}{2} = \left(\frac{50}{100}\right)^{1-n}$. This equation simplifies to $2 = \left(\frac{1}{2}\right)^{1-n}$, which can be rewritten as $2 = (2)^{n-1}$. Comparing the exponents, we find that $n - 1 = 1$, which confirms that $n = 2$.

Chemical kinetics parameters, including reaction order, molecularity, and rate constants, are defined by distinct mathematical relationships and experimental constraints. In zero-order kinetics, the half-life follows $t_{1/2} = \frac{[A]_0}{2k}$, indicating that it depends directly on the initial concentration; as this concentration drops, the half-life also decreases, making the first statement accurate. Furthermore, a substance in a chemical equation does not necessarily influence the rate if the reaction order with respect to that reactant is zero, and while reaction order can be fractional, molecularity must be an integer, identifying the third statement as the only incorrect one. Finally, the rate constant units are consistent with the reaction order, specifically mol L⁻¹ s⁻¹ for zero-order and mol⁻¹ L s⁻¹ for second-order reactions, confirming there is exactly one incorrect statement in the set.

In first-order kinetics, the concentration of the reactant decreases exponentially, which means the reaction consumes half of the remaining substance during each consecutive interval of its half-life, denoted as $t_{1/2}$. After the first half-life, 50% of the initial reactant is consumed, leaving 50% behind. During the subsequent half-life, half of that remaining amount is consumed, which corresponds to an additional 25% of the original reactant, bringing the total completion to 75%. Thus, the time required for 75% completion is equivalent to two half-lives, expressed by the relationship $T_{75} = 2 t_{1/2}$ Since the half-life is 30 min, substituting this value results in a total time of 60 min.

The reaction order $n$ is determined by the units of the rate constant, which follow the general expression (mol L^-1)^(1-n) s^-1. This formula can be rewritten as (L mol^-1)^(n-1) s^-1, allowing us to identify the order by comparing the exponent of the concentration term to the units of the given rate constant. Since the rate constant 4.6 × 10^-5 has units of L mol^-1 s^-1, the exponent n - 1 must correspond to 1. Solving the equation $n - 1 = 1$ leads to $n = 2$, confirming that the reaction is of the second order.

The rate of a chemical reaction is defined by the rate law $Rate = k[X]^n$, which depends on the concentration of reactants and the rate constant $k$ raised to the power of the reaction order $n$. Because a plot of the rate of formation against time does not provide the concentration-time data required to isolate the exponent $n$, it is impossible to uniquely determine the reaction order from the graph alone. Consequently, the only valid conclusion is that the order of the reaction cannot be determined, rendering statement (B) as the only correct claim.