Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

A positive catalyst accelerates a chemical reaction by providing an alternative mechanism with a lower activation energy ($E_a$), which manifests as a reduced peak height on a potential energy diagram. Because the catalyst only lowers this energy barrier without altering the initial energy of the reactants or the final energy of the products, the thermodynamic stability and enthalpy change ($\Delta H$) of the system remain identical to the uncatalyzed process. The correct diagram depicts this specific lowering of the energy barrier compared to the original, uncatalyzed pathway.

In a zero-order reaction, the rate of consumption of the reactant is constant and independent of its concentration, which results in a linear relationship between the amount of reactant used and time. The half-life for such a reaction is expressed as $t_{1/2} = \frac{[A]_0}{2k}$, where $[A]_0$ is the initial concentration and $k$ is the rate constant. Given that the half-life is 50 minutes, we can determine the ratio $\frac{[A]_0}{k} = 100\,\text{min}$.

To find the time required for the concentration to reduce to one-fourth of its initial value, we calculate the amount of reactant consumed, which is $1 - \frac{1}{4} = \frac{3}{4}$ of the initial amount. Applying the zero-order integrated rate law $[A]_0 - [A]_t = kt$, where $[A]_0 - [A]_t$ represents the amount consumed, we get the equation $\frac{3}{4}[A]_0 = kt$. Rearranging this for time gives $t = \frac{3[A]_0}{4k}$. Substituting the known ratio of $\frac{[A]_0}{k} = 100\,\text{min}$ into this equation, we find $t = \frac{3}{4} \times 100$, which equals 75 minutes.

The rate of a chemical reaction is defined by the rate law equation $Rate = K[A]^1[B]^1$, where the reaction orders are explicitly provided as one for both reactants. By substituting the values from the first set of data, we find $0.1 = K(20)(0.5)$, which allows us to calculate the rate constant $K$ as 0.01. Applying this constant to the second data point leads to the equation $0.40 = 0.01(x)(0.5)$, which simplifies to $0.40 = 0.005x$, yielding $x = 80$. Finally, using the third set of data in the equation $0.80 = 0.01(40)(y)$ gives $0.80 = 0.4y$, which identifies the value of $y$ as 2.

The integrated rate law for a first-order reaction is expressed as $t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]_t} \right)$. The time $t_{50}$ represents the half-life where $[A]_t = \frac{1}{2}[A]_0$, which simplifies to $t_{50} = \frac{\ln 2}{k}$. For 87.5% completion, the remaining concentration of the reactant is $[A]_t = (100 - 87.5)\% [A]_0 = 12.5\% [A]_0 = \frac{1}{8}[A]_0$. Substituting this remaining concentration into the rate equation yields $t_{87.5} = \frac{1}{k} \ln \left( \frac{[A]_0}{\frac{1}{8}[A]_0} \right) = \frac{\ln 8}{k}$. Since $8 = 2^3$, the expression becomes $t_{87.5} = \frac{3 \ln 2}{k}$. Comparing this result with the expression for $t_{50}$ gives $t_{87.5} = 3 \times \left( \frac{\ln 2}{k} \right) = 3 \times t_{50}$, which identifies $x = 3$.

The rate law of a multi-step chemical reaction is determined by its slowest step, which acts as the rate-determining bottleneck. Based on the slow step provided, the rate is initially expressed as:
$r = k [\mathrm{NOBr}_2] [\mathrm{NO}]$
Because $[\mathrm{NOBr}_2]$ is an unstable intermediate, it must be expressed in terms of the reactants using the preceding fast equilibrium step. The equilibrium constant for this step is defined as:
$K_{eq} = \frac{[\mathrm{NOBr}_2]}{[\mathrm{NO}] [\mathrm{Br}_2]}$
Rearranging this expression to solve for the intermediate concentration gives $[\mathrm{NOBr}_2] = K_{eq} [\mathrm{NO}] [\mathrm{Br}_2]$. Substituting this relationship into the initial rate equation results in:
$r = k K_{eq} [\mathrm{NO}] [\mathrm{Br}_2] [\mathrm{NO}]$
By grouping the terms, we arrive at the final rate law $r = k' [\mathrm{NO}]^2 [\mathrm{Br}_2]$, where $k' = k K_{eq}$. The overall order of the reaction is the sum of the exponents of the concentration terms in the rate law, which is $2 + 1 = 3$.

The rate of a balanced chemical reaction is determined by its stoichiometry, where the rate of change of any species divided by its stoichiometric coefficient is constant. Given the stoichiometric ratio of $\frac{3}{6}$, the rate of production of $\mathrm{Fe}_2(\mathrm{SO}_4)_3$ is related to the rate of disappearance of $\mathrm{FeSO}_4$, which is $\frac{1.2}{1800} \mathrm{M}\,\mathrm{s}^{-1}$ calculated from a 1.2 M drop over 1800 s. Multiplying this by the stoichiometric factor yields a production rate of $\frac{1}{2} \times \frac{1.2}{1800} \mathrm{M}\,\mathrm{s}^{-1}$, which simplifies to $\frac{1}{3} \times 10^{-3} \mathrm{M}\,\mathrm{s}^{-1}$. Converting this to $333.33 \times 10^{-6} \mathrm{mol}\,\mathrm{L}^{-1}\,\mathrm{s}^{-1}$ provides the required value of 333.

First-order reactions follow kinetics where the reactant concentration approaches zero as time approaches infinity, meaning the reaction never actually completes in a finite time such as 1000 s. The half-life of this process is calculated using the formula $t_{1/2} = 0.693/k$, which results in approximately 150.65 s when applying the provided rate constant of 4.6 x 10^-3 s-1, contradicting the 500 s claim. Regarding the time required for completion percentages, the relationship is governed by the expression $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$, which shows that the time for 90% completion is 25 times larger than the time for 10% completion, effectively refuting the statement that the reverse is true.

The degree of dissociation is derived from the integrated rate equation $N_t = N_0 e^{-kt}$, which signifies the remaining reactant fraction as $e^{-kt}$, thus defining the dissociated fraction as $\alpha = 1 - e^{-kt}$, confirming this statement is accurate. Finally, the rate of reaction is measured in units of concentration per time, while the rate constant for a first-order reaction is measured in s-1, demonstrating that their units are inherently different. Since only the derivation for the degree of dissociation holds true, there is only one correct statement.

In any multi-step chemical reaction, the number of activated complexes is equal to the number of individual elementary steps, as each step necessitates passing through a distinct high-energy transition state. Since this specific sequence consists of three elementary steps, there are three activated complexes, while the two distinct chemical species that form and are subsequently consumed between the reactants and products are classified as the reaction intermediates. The overall speed of the reaction is restricted by the slowest part of the pathway, which is identified by the step possessing the highest activation energy; in this reaction, that is the second step, confirming it as the rate-determining step.

The Arrhenius equation describes the relationship between the rate constant, temperature, and activation energy using the formula $\log \frac{k_2}{k_1} = \frac{E_a}{2.3 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$. By substituting the rate constants $k_1 = 0.03$ and $k_2 = 0.05$ at their respective temperatures $T_1 = 200$ and $T_2 = 300$, we can solve for the activation energy $E_a$. The logarithmic term $\log \frac{0.05}{0.03}$ simplifies to $\log 5 - \log 3$, which is $0.70 - 0.48 = 0.22$. The temperature term $\frac{1}{200} - \frac{1}{300}$ equals $\frac{1}{600}$. Substituting these values into the rearranged equation $E_a = 0.22 \times 2.3 \times 8.3 \times 600$ results in $E_a = 2519.88$ J, which rounds to the nearest integer of 2520 J.

Radioactive decay follows an exponential law where the concentration at time $t$ is $N(t) = N_0 e^{-kt}$, with the decay constant $k$ defined by $\frac{\ln 2}{t_{1/2}}$. Equating the decay functions of substances A and B provides the basis for the calculation:

$[A]_0 e^{-\left(\frac{\ln 2}{15}\right)t} = [B]_0 e^{-\left(\frac{\ln 2}{5}\right)t}$

Substituting $[B]_0 = 4[A]_0$ simplifies the expression to $e^{-\left(\frac{\ln 2}{15}\right)t} = 4 e^{-\left(\frac{\ln 2}{5}\right)t}$. Rearranging the terms to isolate the exponential component leads to $e^{\left(\frac{\ln 2}{5} - \frac{\ln 2}{15}\right)t} = 4$. Since $\frac{\ln 2}{5} - \frac{\ln 2}{15} = \frac{2 \ln 2}{15}$, the equation becomes $e^{\frac{2 \ln 2}{15}t} = 4$, which is equivalent to $2^{\frac{2t}{15}} = 2^2$. Matching the exponents yields $\frac{2t}{15} = 2$, confirming that the concentrations equalize after $15$ min.

In a first-order gas-phase reaction $A(g) \rightarrow 2 B(g) + C(g)$, the total pressure of the system reflects the stoichiometric decomposition of the reactant. Initially, the pressure is entirely due to $A$, starting at 800 mm Hg. As the reaction proceeds, for every amount $x$ of $A$ that decomposes, the partial pressure of $A$ decreases by $x$, while the pressures of $B$ and $C$ increase by $2x$ and $x$ respectively. The total pressure at any time is defined by the equation $P_T = (800 - x) + 2x + x = 800 + 2x$. Given that the total pressure reached 1600 mm Hg after 10 minutes, solving $800 + 2x = 1600$ yields $x = 400$ mm Hg. This confirms that the partial pressure of $A$ becomes 400 mm Hg after 10 minutes, establishing a half-life of 10 minutes.

Because 30 minutes represents three half-lives, the remaining partial pressure of $A$ reduces to 100 mm Hg. With 100 mm Hg of $A$ remaining, the amount of $A$ consumed is 700 mm Hg. Substituting this value into the total pressure relationship $P_T = 800 + 2(700)$, the final total pressure is 2200 mm Hg.

The temperature sensitivity of the rate constant $K$ is governed by the activation energy $E_a$ of a reaction. Specifically, if $E_a = 0$, the rate constant $K$ is independent of temperature, confirming that the reaction rate remains unchanged despite temperature fluctuations. Conversely, when $E_a > 0$, the rate constant $K$ increases as temperature rises, and when $E_a < 0$, the rate constant $K$ decreases as temperature rises, thereby invalidating any claims regarding inverse relationships or the presence of negative activation energy in the absence of correlation.

When a molecule undergoes two independent first-order reactions simultaneously, the overall rate of consumption corresponds to the sum of the individual rate constants. Because each rate constant is inversely proportional to its respective half-life, the effective half-life for the combined process follows the reciprocal addition rule $\frac{1}{t_{\text{eff}}} = \frac{1}{t_1} + \frac{1}{t_2}$.

Substituting the given values of $12\,\mathrm{min}$ and $3\,\mathrm{min}$ into this equation results in $\frac{1}{t_{\text{eff}}} = \frac{1}{12} + \frac{1}{3}$, which simplifies to $\frac{5}{12}$. Dividing $12$ by $5$ results in an effective half-life of $2.4\,\mathrm{min}$, which rounds to a nearest integer value of $2$.

To determine how a catalyst affects activation energy, we apply the Arrhenius principle where the rate of reaction depends on the ratio of activation energy to temperature. When the rates of the uncatalyzed and catalyzed reactions are identical, this ratio must be equivalent for both processes, provided the pre-exponential factor remains constant.

First, convert the temperatures to Kelvin to find that the uncatalyzed reaction at 327$^{\circ}$C corresponds to 600 K, while the catalyzed reaction at 27$^{\circ}$C corresponds to 300 K. By substituting these values into the relationship

$\frac{E_{a}}{T} = \frac{E_{a,f}'}{T'}$

we set up the equation $\frac{300}{600} = \frac{E_{a,f}'}{300}$, which simplifies to $E_{a,f}' = 150\,\mathrm{kJ}\,\mathrm{mol}^{-1}$. Since the enthalpy of the reaction is defined as the difference between the forward and backward activation energies, we use the equation $\Delta H = E_{a,f}' - E_{a,b}'$. Substituting the known enthalpy of $20\,\mathrm{kJ}\,\mathrm{mol}^{-1}$ and our calculated forward activation energy into $20 = 150 - E_{a,b}'$ yields an activation energy of $130\,\mathrm{kJ}\,\mathrm{mol}^{-1}$ for the catalyzed backward reaction.

First-order kinetics are governed by the relation $t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right)$, where the rate constant $k$ remains invariant throughout the decomposition. By comparing the time $t_1 = 540\,\text{s}$ for $60\%$ decomposition to the unknown time $t_2$ for $90\%$ decomposition, we establish the ratio $\frac{540}{t_2} = \frac{\ln(10/4)}{\ln(10)}$. Since the natural logarithm and common logarithm are related by the factor $\ln 10 = 2.3$, the constant terms cancel out when converting the ratio, simplifying the expression to $\frac{\log 10 - \log 4}{\log 10}$. Substituting the values $\log 10 = 1$ and $\log 4 = 2 \log 2 = 0.6$ into this equation gives $\frac{540}{t_2} = \frac{1 - 0.6}{1} = 0.4$, which identifies the time $t_2$ as $1350\,\text{s}$.

The Arrhenius equation is expressed in the linear form $\ln k = \ln A - \frac{E_a}{R} \left(\frac{1}{T}\right)$, where the plot of $\ln k$ versus $\frac{1}{T}$ yields a straight line with a slope equal to $-\frac{E_a}{R}$. Based on the coordinates provided in the graph for this specific reaction, the slope is determined to be $-4000\,\mathrm{K}$. Applying the relationship $E_a = -\text{slope} \times R$ with the given gas constant $R = 2\,\mathrm{cal}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$ yields $E_a = -(-4000\,\mathrm{K}) \times 2\,\mathrm{cal}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1} = 8000\,\mathrm{cal}\,\mathrm{mol}^{-1}$. Converting this value to kilocalories per mole results in $8\,\mathrm{kcal}\,\mathrm{mol}^{-1}$, which corresponds to the nearest integer.

The rate of a reaction is defined by a rate law equation where the reaction rate is proportional to the product of the reactant pressures raised to their respective orders. To isolate the reaction order with respect to $\mathrm{NO}$, we analyze experiments where the pressure of the other reactant, $\mathrm{H}_2$, remains constant. The first and second experiments provided are ideal for this purpose because both maintain an initial pressure of $\mathrm{H}_2$ at 65.6 kPa.

By dividing the rate law expression for the first experiment by that of the second, the rate constant and the $\mathrm{H}_2$ pressure term cancel out, leaving the dependency solely on the pressure of $\mathrm{NO}$. The calculation becomes:
$\frac{0.135}{0.033} = \left(\frac{40.0}{20.1}\right)^n$
This expression simplifies to approximately $4 = 2^n$. Since $2^2 = 4$, the reaction order $n$ with respect to $\mathrm{NO}$ is 2.

Since the half-life of each reactant is independent of its initial concentration, both A and B follow first-order reaction kinetics. The relationship between the rate constant $k$ and the half-life $t_{1/2}$ is defined by the expression $k = \frac{\ln 2}{t_{1/2}}$, yielding rate constants of $k_A = \frac{\ln 2}{100}$ and $k_B = \frac{\ln 2}{50}$.

The instantaneous concentration at time $t$ for each substance is given by $[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$. With equal initial concentrations $[A]_0 = [B]_0$, the requirement that $[A]_t = 4 [B]_t$ leads to the equation $e^{-k_A t} = 4 e^{-k_B t}$. Rearranging this expression results in $e^{(k_B - k_A) t} = 4$. Taking the natural logarithm of both sides provides $(k_B - k_A) t = \ln 4$, which simplifies to $(k_B - k_A) t = 2 \ln 2$. Substituting the values for $k_A$ and $k_B$ gives $(\frac{\ln 2}{50} - \frac{\ln 2}{100}) t = 2 \ln 2$. This further simplifies to $\frac{\ln 2}{100} t = 2 \ln 2$, and solving for $t$ gives $t = 200$ s.

A first-order reaction obeys the integrated rate law $t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]_t} \right)$, where the time taken depends on the ratio of initial to remaining reactant concentration. At 67% completion, the remaining concentration $[A]_t$ is 33% of the initial concentration $[A]_0$, which simplifies to a ratio of $3$ when inverted, leading to $t_{67\%} = \frac{1}{k} \ln(3)$. Since the half-life is defined as $t_{1/2} = \frac{\ln 2}{k}$, the relationship between these two time intervals is given by the ratio $\frac{t_{67\%}}{t_{1/2}} = \frac{\ln 3}{\ln 2}$, which is equivalent to $\frac{\log 3}{\log 2}$.

Using the provided value $\log 3 = 0.4771$ and the standard value $\log 2 \approx 0.301$, this ratio calculates to approximately $1.585$. Comparing $t_{67\%} = 1.585 \times t_{1/2}$ to the expression $x \times 10^{-1} \times t_{1/2}$ implies $x \times 10^{-1} = 1.585$, which results in $x = 15.85$. Rounding this value to the nearest integer provides 16.

The half-life of a reactant is linked to its initial concentration by the relationship $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$, where $n$ denotes the reaction order. Plugging in the experimental values of 100 s for an initial concentration of 0.5 mol L^-1 and 50 s for an initial concentration of 1.0 mol L^-1, the relationship becomes $\frac{100}{50} = \left( \frac{1.0}{0.5} \right)^{n-1}$. Simplifying these terms leads to the equality $2 = 2^{n-1}$, which indicates that $n-1 = 1$, identifying the order of the reaction as 2.