Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

The kinetics of a first-order reaction dictate that the natural logarithm of the reactant partial pressure decreases linearly with time, following the integrated rate equation $\ln P = \ln P_0 - kt$. Within this relationship, the slope of the plot representing the natural logarithm of pressure versus time is equivalent to the negative of the rate constant, $-k$. Given the slope of $-3.465 \times 10^4$ from the experimental data, the rate constant $k$ is $3.465 \times 10^4$.

The half-life of a first-order reaction is determined by the relationship $t_{1/2} = \frac{0.693}{k}$. By substituting the value of the rate constant into this equation, the calculation becomes $\frac{0.693}{3.465 \times 10^4}$, which simplifies to $0.2 \times 10^{-4}$ seconds. Adjusting this to the required scientific notation, the half-life is $2 \times 10^{-5}$ seconds.

First-order kinetics are governed by the relationship between the half-life and the rate constant, $t_{1/2} = \frac{0.693}{K}$, and the integrated rate law, $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$. The rate constant calculated from the given half-life of 0.3010 min is:
$K = \frac{0.693}{0.3010} \approx 2.30 \text{ min}^{-1}$
Substituting this rate constant and the time interval of 2.0 min into the integrated rate law yields:
$2.30 = \frac{2.303}{2.0} \log \frac{[A]_0}{[A]_t}$
Simplifying this expression results in the logarithm being equal to 2:
$2 = \log \frac{[A]_0}{[A]_t}$
The ratio of the initial concentration to the concentration at time 2.0 min is defined by $\frac{[A]_0}{[A]_t} = 10^2 = 100$.

The Arrhenius equation describes the dependence of a reaction's rate constant on temperature through the relationship $\ln k = \ln A - \frac{E_a}{RT}$. By expressing the reciprocal temperature as $\frac{10^3}{T}$, the equation can be rewritten as $\ln k = \ln A - \left(\frac{E_a}{10^3 R}\right) \left(\frac{10^3}{T}\right)$. In a graph plotting $\ln k$ against $\frac{10^3}{T}$, the slope of the line corresponds to the coefficient $-\frac{E_a}{10^3 R}$. Given that the slope is -18.5, we can equate this to the known parameters as follows:

$-\frac{E_a}{10^3 \times 8.31} = -18.5$

Multiplying both sides by $-10^3 \times 8.31$ results in an activation energy of $153735\ \mathrm{J\ mol^{-1}}$. Converting this value to $\mathrm{kJ\ mol^{-1}}$ yields $153.735\ \mathrm{kJ\ mol^{-1}}$, which rounds to the nearest integer of 154.

The Arrhenius equation describes the relationship between the rate constant of a chemical reaction and its temperature, which is essential for determining the activation energy $E_a$. By taking the logarithm of the Arrhenius expression for two different temperatures, we arrive at the following relationship:

$\log \frac{k_2}{k_1} = \frac{E_a}{2.3 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

In this reaction, the rate constant doubles, meaning the ratio $\frac{k_2}{k_1} = 2$. With an initial temperature $T_1$ of 300 K and a temperature increase of 9 K, the final temperature $T_2$ is 309 K. Substituting these values along with $R = 8.3$ and $\log 2 = 0.30$ into the equation gives $0.30 = \frac{E_a}{2.3 \times 8.3} \left[ \frac{9}{309 \times 300} \right]$. Solving for the activation energy leads to the expression $E_a = \frac{0.30 \times 309 \times 300 \times 2.3 \times 8.3}{9}$, which calculates to 58988.1 J/mol. Converting this value to the requested units results in approximately 59 kJ/mol.

A reaction following first-order kinetics reaching half of its initial concentration in 70 minutes implies that its half-life, $t_{1/2}$, is 70 minutes. The rate constant $k$ for such a reaction is determined by the equation $k = \frac{0.693}{t_{1/2}}$. To ensure the result is \in the required unit, we convert the half-life time into 4200 seconds. Plugging this into the formula gives $k = \frac{0.693}{4200}$, which equals 0.000165 per second. By expressing this result \in the scientific notation format specified, we find the value of $x$ to be 165.

The kinetic analysis of the reaction between $\mathrm{H}_2\mathrm{O}_2$ and iodide ions relies on the precise detection of an iodine-starch complex endpoint, which necessitates that the starch indicator be freshly prepared to avoid potential contamination. Because KI is maintained in excess to ensure the complete consumption of $\mathrm{H}_2\mathrm{O}_2$, the concentration of the sodium thiosulphate solution must be kept strictly lower than that of the KI solution. Finally, the experimental time is marked immediately after the characteristic deep blue colour appears, signaling that the thiosulphate has been entirely consumed and free iodine is available to form the complex.

The relationship between the rate constant of a reaction and temperature is given by the Arrhenius equation, expressed in its logarithmic form as $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. Here, $k_1$ and $k_2$ represent the rate constants at temperatures $T_1$ and $T_2$, respectively. By assigning $T_1 = 300\ \mathrm{K}$ and $T_2 = 310\ \mathrm{K}$, we substitute the activation energy $E_a = 532611\ \mathrm{J}\ \mathrm{mol}^{-1}$ and the gas constant $R = 8.3\ \mathrm{J K}^{-1}\ \mathrm{mol}^{-1}$ to obtain $\ln\left(\frac{k_{310}}{k_{300}}\right) = \frac{532611}{8.3} \left(\frac{1}{300} - \frac{1}{310}\right)$. Simplifying the term in the parenthesis gives $\frac{10}{300 \times 310}$, leading to the calculation $\ln\left(\frac{k_{310}}{k_{300}}\right) = 6.9$. Using the provided fact that $\ln 10 = 2.3$, we can interpret 6.9 as $3 \times 2.3$, which is equal to $3 \ln 10$, or $\ln 10^3$. This equality implies that $\frac{k_{310}}{k_{300}} = 10^3$, which can be rearranged to $k_{300} = 10^{-3} k_{310}$. When this result is compared to the given expression $k_{300} = x \times 10^{-3} k_{310}$, it is evident that the value of $x$ is 1.

The relationship between the half-life of a reaction and the initial pressure of the reactant is given by $t_{1/2} \propto P_0^{1-n}$, where $n$ represents the reaction order. Data indicates that the half-life is 240 s at an initial pressure of 500 Torr, and it remains 4.0 min, which is equivalent to 240 s, when the initial pressure is reduced to 250 Torr. Because the half-life remains independent of the change in initial pressure, the exponent $1-n$ must equal zero, establishing that $n=1$.

Radioactive decay follows first-order kinetics, where the rate of decay is governed by the decay constant $\lambda = \frac{0.693}{t_{1/2}}$. The relationship between the initial activity and the activity remaining at a specific time is described by the integrated rate law $\log \frac{A_0}{A} = \frac{\lambda t}{2.303}$. Substituting the definition of the decay constant into this rate equation establishes a direct connection between the half-life and the activity ratio through the formula $\log \frac{A_0}{A} = \frac{0.693 \times t}{2.303 \times t_{1/2}}$.

Inserting 83 days for the elapsed time and 200 days for the half-life into the expression yields $\log \frac{A_0}{A} = \frac{0.693 \times 83}{2.303 \times 200}$, which simplifies to approximately 0.125. Applying the provided antilogarithm value, the ratio is $\frac{A_0}{A} = 1.333$. Calculating the reciprocal gives the remaining activity fraction $\frac{A}{A_0} = \frac{1}{1.333} \approx 0.75$, which corresponds to 75 percent.

The Arrhenius equation, $k = A e^{-\frac{E_a}{RT}}$, relates the reaction rate constant to the activation energy. When a catalyst provides a pathway with lower activation energy, the ratio of the catalyzed rate constant to the uncatalyzed rate constant is given by the expression $e^{\frac{E_a - E_a'}{RT}}$, provided the pre-exponential factor remains constant.

With a reduction in activation energy of 10 kJ/mol, a temperature of 300 K, and the gas constant $R$ as 8.31 J/K/mol, the exponent is determined by the calculation $\frac{10000}{8.31 \times 300}$. This division equals approximately 4.011. Since the ratio is expressed as $e^{x}$, the nearest integer for $x$ is 4.

The rate of a chemical reaction is defined by the rate of change of the concentration of any reactant or product normalized by its stoichiometric coefficient, expressed as $\text{Rate} = -\frac{1}{\gamma_1}\frac{d[A]}{dt} = -\frac{1}{\gamma_2}\frac{d[B]}{dt} = \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{\gamma_4}\frac{d[D]}{dt}$. From the provided data, the rate of appearance of $D$ is $9 \text{ mmol dm}^{-3} \text{s}^{-1}$. Because this value is $1.5$ times the rate of disappearance of $B$, we calculate $-\frac{d[B]}{dt} = \frac{9}{1.5} = 6 \text{ mmol dm}^{-3} \text{s}^{-1}$. Furthermore, as the rate of disappearance of $B$ is twice that of $A$, we determine $-\frac{d[A]}{dt} = \frac{6}{2} = 3 \text{ mmol dm}^{-3} \text{s}^{-1}$. For product $C$, the concentration change of $10 \text{ mmol dm}^{-3}$ over $10 \text{ seconds}$ results in a rate of $\frac{d[C]}{dt} = \frac{20 - 10}{10} = 1 \text{ mmol dm}^{-3} \text{s}^{-1}$. Equating the reaction rate using these values gives $\text{Rate} = \frac{3}{\gamma_1} = \frac{6}{\gamma_2} = \frac{1}{\gamma_3} = \frac{9}{\gamma_4}$. These relations imply that $\gamma_1 = 3\gamma_3$, $\gamma_2 = 6\gamma_3$, and $\gamma_4 = 9\gamma_3$. Substituting these into the definition, the rate of reaction is $\frac{1}{\gamma_3} \times \frac{d[C]}{dt} = \frac{1}{\gamma_3} \times 1$. Given the stoichiometric consistency, the rate of reaction is $1 \text{ mmol dm}^{-3} \text{s}^{-1}$.

The temperature dependence of the reaction rate is defined by the Arrhenius equation, $k = A e^{-\frac{E_A}{RT}}$. By taking the natural logarithm, we obtain $\ln k = \ln A - \frac{E_A}{RT}$. To utilize base-10 logarithms, we apply the conversion $\ln x = 2.3 \log x$, leading to the expression $\log k = \log A - \frac{E_A}{2.3 RT}$, where the value $2.3$ is the provided approximation for $\ln 10$.

Substituting the known parameters $A = 4 \times 10^{13} \text{s}^{-1}$, $k = 2 \times 10^{-6} \text{s}^{-1}$, $E_A = 8.3 \times 10^3 \text{J mol}^{-1}$, and $R = 8.3 \text{J K}^{-1} \text{mol}^{-1}$ into the logarithmic equation yields:
$\log(2 \times 10^{-6}) = \log(4 \times 10^{13}) - \frac{8.3 \times 10^3}{2.3 \times 8.3 \times T}$
Utilizing logarithmic properties, the equation becomes $\log 2 - 6 = \log 4 + 13 - \frac{10^3}{2.3 T}$. Given $\log 2 = 0.30$, we substitute this value to get $0.30 - 6 = 2(0.30) + 13 - \frac{1000}{2.3 T}$. Simplifying these constants results in $-5.7 = 13.6 - \frac{1000}{2.3 T}$. Rearranging the terms gives $\frac{1000}{2.3 T} = 19.3$, which simplifies to $T = \frac{1000}{2.3 \times 19.3} \text{K}$. Calculating this value results in $T = 22.5 \text{K}$, which corresponds to $225 \times 10^{-1} \text{K}$.

For any reaction of order $n$, the half-life of a gaseous compound is related to its initial pressure by the proportionality $t_{1/2} \propto P_0^{1-n}$. By comparing two different experimental conditions, the order of the reaction can be determined using the ratio of the half-lives:

$\frac{t_{1/2(1)}}{t_{1/2(2)}} = \left( \frac{P_1}{P_2} \right)^{1-n}$

Substituting the provided data, where $t_{1/2(1)} = 340\ \mathrm{s}$ at an initial pressure $P_1 = 55.5\ \mathrm{kPa}$ and $t_{1/2(2)} = 170\ \mathrm{s}$ at $P_2 = 27.8\ \mathrm{kPa}$, results in the following calculation:

$\frac{340}{170} = \left( \frac{55.5}{27.8} \right)^{1-n}$

Simplifying these fractions gives $2 = 2^{1-n}$, which indicates that the exponent $1-n$ must be equal to $1$. Solving for $n$ leads to the conclusion that the order of the reaction is $0$.

Radioactive decay follows first-order kinetics, where the time elapsed is related to the initial amount and the quantity remaining by the expression $t = \frac{1}{\lambda} \ln\left(\frac{N_0}{N}\right)$. The decay constant $\lambda$ is defined by the half-life $T_{1/2}$ as $\lambda = \frac{\ln 2}{T_{1/2}}$.

Substituting the known values of $t = 100$ and $T_{1/2} = 30$ into the decay formula yields the following relationship:
$100 = \frac{30}{\ln 2} \ln\left(\frac{1}{N}\right)$
To solve this, we convert the natural logarithm to base 10 using the property $\ln x = 2.303 \log x$ and utilize the given value $\log 2 = 0.30$. Rearranging the equation results in:
$\log\left(\frac{1}{N}\right) = \frac{100 \times \log 2}{30}$
Evaluating the right side gives $\frac{100 \times 0.30}{30} = 1$, which simplifies to $\log\left(\frac{1}{N}\right) = 1$. This implies $\frac{1}{N} = 10^1$, so $N = 0.1 \mu\mathrm{g}$. Expressing this result as $1 \times 10^{-1} \mu\mathrm{g}$ shows that the required value is 1.

The Arrhenius equation, $k = A e^{-\frac{E_a}{RT}}$, relates the reaction rate constant to activation energy and temperature. By aligning the expression $k=(6.5 \times 10^{12} \mathrm{S}^{-1}) e^{-26000 \frac{\mathrm{K}}{T}}$ with the standard form, the exponent term $\frac{E_a}{RT}$ corresponds to $\frac{26000 \mathrm{K}}{T}$. This identifies the relationship $\frac{E_a}{R} = 26000 \mathrm{K}$. Using the gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, the calculation $E_a = 26000 \times 8.314 = 216164 \text{ J mol}^{-1}$ determines the activation energy. Dividing by 1000 gives $E_a = 216.164 \text{ kJ mol}^{-1}$, which rounds to the nearest integer of 216.

The reaction kinetics are governed by the rate law $Rate = k[X]$, as the reaction is first-order with respect to $X$ and independent of $Y$. Because the rate is directly proportional to the concentration of $X$, comparing experimental trials allows for the determination of the unknown variables by analyzing the change in rate relative to the change in concentration.

Between the first and second experiments, the reaction rate doubles from 2 x 10^-3 mol L^-1 min^-1 to 4 x 10^-3 mol L^-1 min^-1, which implies that the concentration of $X$ must also double from 0.1 mol L^-1, identifying $L = 0.2$ mol L^-1. Moving to the third experiment, the concentration of $X$ increases to 0.4 mol L^-1, which is a fourfold increase compared to the initial concentration in the first experiment. Given the direct proportionality, the rate must increase by the same factor, resulting in the relation $M \times 10^{-3} = 4 \times (2 \times 10^{-3})$, which simplifies to $M = 8$. Calculating the ratio requested involves the expression $M/L = 8 / 0.2$, which equals 40.

Because the half-life of the decomposition reaction is independent of the initial concentration, the process follows first-order kinetics. For such reactions, the rate constant $k$ relates to the half-life $t_{1/2}$ through the equation $k = \frac{2.303 \log 2}{t_{1/2}}$ while the time $t$ required for a specific extent of decomposition is governed by the integrated rate law $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given that 80% of $\mathrm{AB}_2$ decomposes, the remaining amount is 20%, resulting \in a concentration ratio $\frac{[A]_0}{[A]_t} = \frac{100}{20} = 5$. Equating the two expressions for $k$ using the half-life of 200 s yields the relationship $\frac{2.303 \log 2}{200} = \frac{2.303}{t} \log 5$ Canceling the common constant 2.303 and substituting the provided value $\log 2 = 0.30$ along with $\log 5 = 1 - \log 2 = 0.70$ leads to the equation $\frac{0.30}{200} = \frac{0.70}{t}$ Solving for $t$ gives $t = \frac{0.70 × 200}{0.30}$, which results in 467 s.

The Arrhenius equation defines the relationship between the rate constant $k$ and temperature $T$ as $\ln k = \ln A - \frac{E_A}{RT}$. In the expression $\ln k = 33.24 - \frac{2.0 \times 10^4}{T}$, the slope term $\frac{E_A}{R}$ is equivalent to $2.0 \times 10^4 \mathrm{K}$.

Solving for the activation energy $E_A$ involves calculating the product of this value and the gas constant $R = 8.3 \mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. This multiplication yields $E_A = 2.0 \times 10^4 \times 8.3 \mathrm{J}\,\mathrm{mol}^{-1} = 16.6 \times 10^4 \mathrm{J}\,\mathrm{mol}^{-1}$. Converting this magnitude into kilojoules per mole results in $166 \mathrm{kJ}\,\mathrm{mol}^{-1}$.

For a first-order reaction, the time elapsed is governed by the integrated rate law $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$, while the half-life is defined as $t_{1/2} = \frac{0.693}{k}$. When 90% of the reaction is complete, the concentration of the reactant drops to 10% of its initial value, which results in the following expression:
$t_{90\%} = \frac{2.303}{k} \log(10) = \frac{2.303}{k}$
Dividing this specific time for completion by the half-life provides the ratio $\frac{2.303}{0.693}$, which simplifies to approximately 3.32.

The Arrhenius equation provides the mathematical relationship between the rate constant of a reaction and temperature, allowing us to determine how the rate constant changes when moving from one temperature to another. By utilizing the logarithmic form of this equation, we can relate the rate constants $k_1$ and $k_2$ at absolute temperatures $T_1$ and $T_2$, respectively:

$\log \frac{k_1}{k_2} = \frac{E_a}{2.303 R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$

To find the value of $x$, we substitute the given parameters into the equation: $T_1 = 700\,\mathrm{K}$, $k_1 = 6.36 \times 10^{-3}\,\mathrm{s}^{-1}$, $T_2 = 600\,\mathrm{K}$, $E_a = 209,000\,\mathrm{J\,mol}^{-1}$, and $R = 8.31\,\mathrm{J\,K^{-1}\,mol}^{-1}$. The target rate constant is $k_2 = x \times 10^{-6}\,\mathrm{s}^{-1}$. Substituting these values, we get:

$\log \left( \frac{6.36 \times 10^{-3}}{x \times 10^{-6}} \right) = \frac{209,000}{2.303 \times 8.31} \left( \frac{1}{600} - \frac{1}{700} \right)$

Simplifying the expression within the parentheses gives $\frac{1}{600} - \frac{1}{700} = \frac{100}{420,000} \approx 0.000238$. Multiplying this by the pre-factor $\frac{209,000}{19.148}$ results in a value of approximately $2.6$. The equation now simplifies to $\log \frac{6360}{x} \approx 2.6$, which means $\frac{6360}{x} = 10^{2.6}$. Since $10^{2.6}$ is approximately $398$, we calculate $x$ as $\frac{6360}{398}$, which yields $16$.