Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

For a first-order reaction, the rate constant $k$ is related to time $t$ by the expression $k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)$, where $[A]_0$ and $[A]_t$ represent the initial and remaining concentrations of the reactant. In the reaction $A \rightarrow 2B$, the stoichiometry dictates that for every $2$ moles of $B$ produced, $1$ mole of $A$ is consumed. Since $1$ mole of $A$ produces $0.2$ moles of $B$, the amount of $A$ consumed is $0.1$ moles, leaving $[A]_t = 1 - 0.1 = 0.9$ moles of $A$ after $100$ minutes.

The half-life $t_{1/2}$ of a first-order reaction is defined by $t_{1/2} = \frac{\ln 2}{k}$. Substituting $k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)$ into this definition, we obtain the relationship $t_{1/2} = \frac{t \ln 2}{\ln([A]_0 / [A]_t)}$. Using the given values, this becomes $t_{1/2} = \frac{100 \times 0.69}{\ln(1 / 0.9)}$. Given that $\ln(1/0.9) \approx 0.1$, the calculation simplifies to $t_{1/2} = \frac{69}{0.1}$, which results in $690$ minutes.

The Arrhenius equation describes how the rate constant of a chemical reaction changes with temperature based on the activation energy. The relationship is expressed through the integrated logarithmic form:

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

By converting the temperatures to Kelvin, we obtain $T_1 = 300$ K and $T_2 = 325$ K, and we use the given rate constant ratio $k_2/k_1 = 5$ with the gas constant $R = 8.314$ J K$^{-1}$ mol$^{-1}$. Substituting these values into the equation gives the following relation:

$\log 5 = \frac{E_a}{2.303 \times 8.314} \left( \frac{25}{300 \times 325} \right)$

Solving this expression for $E_a$ leads to a value of 52195 J mol$^{-1}$, which is equivalent to 52.195 kJ mol$^{-1}$. Rounding this to the nearest integer provides a final result of 52 kJ mol$^{-1}$.

The Arrhenius equation, given by $k=A e^{-\frac{E_a}{RT}}$, describes how the rate constant for a reaction, including an endothermic one, depends on temperature. As the temperature rises, the magnitude of the negative exponent decreases, which causes the rate constant to grow exponentially rather than linearly. This characteristic relationship illustrates that increasing thermal energy consistently enhances the reaction rate, resulting in a curve that rises sharply as the temperature increases.

The Arrhenius equation describes the dependence of a reaction rate constant on temperature, expressed in its logarithmic form as $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$. By observing the structure of the provided equation, $\log_{10} k = 20.35 - \frac{2.47 \times 10^3}{T}$, one can directly identify the slope term. Equating the temperature-dependent components reveals that the relationship between the activation energy and the provided constant is $\frac{E_a}{2.303 R} = 2.47 \times 10^3$.

To determine the activation energy, multiply the constant 2.47 x 10^3 by 2.303 and the given gas constant $R = 8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. This calculation produces $E_a = 2.47 \times 10^3 \times 2.303 \times 8.314$, which equals approximately 47294 J/mol. Converting this value into kilojoules per mole by dividing by 1000 results in 47.29 kJ/mol, which rounds to the nearest integer of 47.

The stability of hydrogen isotopes is determined by their nuclear composition, where protium ({}^{1}{1}\mathrm{H}$) and deuterium ({}^{2}_{1}\mathrm{H}$) are stable, while tritium ({}^{3}{1}\mathrm{H}$) is the only radioactive isotope \in this set. This radioactive nature involves the emission of low-energy$\beta^{-}$ particles, and the isotope possesses a half-life value greater than 12 years. These specific characteristics distinguish tritium as the isotope described.

The rate law for a chemical reaction follows the expression $\text{rate} = k[\text{conc}]^n$, where $n$ is the order of the reaction. When this equation is transformed into the logarithmic form $\log(\text{rate}) = n \log([\text{conc}]) + \log(k)$, it corresponds to the equation of a straight line $y = mx + c$, where the slope $m$ is equivalent to the reaction order $n$. By comparing the slopes of the lines for reactions A, B, C, and D, the steepness of each line directly indicates its order, establishing that the sequence $d > b > a > c$ corresponds to the observed order of the lines from the steepest slope to the shallowest slope.

The temperature dependence of chemical reaction rates is governed by the Arrhenius equation, which establishes a direct relationship between the rate constant, activation energy, and temperature. When the reaction rate decreases by a factor of 3.555 as the temperature is reduced from 313 K to 303 K, we apply the integrated form of the Arrhenius equation to calculate the activation energy:

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

Here, we define $T_1$ as 303 K and $T_2$ as 313 K, using the gas constant $R = 8.314$ J mol^-1 K^-1 and the provided value $\ln 3.555 = 1.268$. Substituting these into the formula results in the following expression:

$1.268 = \frac{E_a}{8.314} \left( \frac{1}{303} - \frac{1}{313} \right)$

Calculating the difference between the reciprocal temperatures and multiplying the result by the gas constant allows us to isolate $E_a$. This process yields a value of 99,980.7 J mol^-1, which, when converted to kilojoules, corresponds to an activation energy of 100.00 kJ mol^-1.

The speed of a chemical process, such as the growth of bacteria, is inversely proportional to the time taken for the population to reach a specific density. To determine the activation energy $E_a$, we utilize the Arrhenius equation in its logarithmic form, which relates the reaction rates or times at two distinct temperatures: $\ln \left( \frac{t_1}{t_2} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$. This formula allows us to quantify the energy barrier by comparing how the duration of the process changes as the temperature shifts from $T_1 = 300\ \mathrm{K}$ to $T_2 = 400\ \mathrm{K}$.

Substituting the given values into the equation, we have $\ln \left( \frac{60}{40} \right) = \frac{E_a}{8.3} \left( \frac{1}{300} - \frac{1}{400} \right)$. By calculating the temperature difference, $\left( \frac{400-300}{120000} \right) = \frac{100}{120000} = \frac{1}{1200}$, the equation becomes $\ln \left( 1.5 \right) = \frac{E_a}{8.3} \times \frac{1}{1200}$. Using the provided logarithmic ratio value of $0.4$, we find $0.4 = \frac{E_a}{8.3 \times 1200}$. Solving for $E_a$ leads to $E_a = 0.4 \times 1200 \times 8.3$, which equals $3984\ \mathrm{J\,mol^{-1}}$. Converting this result into kilojoules gives an activation energy of $3.98\ \mathrm{kJ\,mol^{-1}}$.

The Arrhenius equation, expressed as $\log K = \frac{-E_a}{2.303RT} + \log A$, dictates that a plot of $\log K$ against $\frac{1}{T}$ yields a straight line with a slope of $\frac{-E_a}{2.303R}$. Since the activation energy is directly proportional to the steepness or magnitude of this negative slope, lines with greater downward inclination represent reactions with higher activation energies. By analyzing the slopes of the provided lines, we identify their magnitudes in the order $c > a > d > b$, which directly maps to the sequence $E_c > E_a > E_d > E_b$ for the activation energies.

The temperature dependence of chemical reaction rates is governed by the Arrhenius equation, which correlates the rate constant with the activation energy. Because the reaction rate is proportional to the concentration of molecules having energy above the threshold, a fivefold increase in these molecules effectively means the rate constant increases by a factor of 5. We first convert the temperatures from Celsius to Kelvin, yielding $T_1 = 300 \, \text{K}$ and $T_2 = 315 \, \text{K}$. Utilizing the integrated form of the Arrhenius equation to relate these variables, we have:

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

Substituting the known values, where $k_2/k_1 = 5$, $\ln 5 = 1.6094$, and $R = 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}$, results in the following expression:

$1.6094 = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{315}\right)$

Solving the arithmetic within the parentheses gives a difference of $\frac{15}{94500}$. Rearranging to isolate $E_a$ leads to the calculation $E_a = \frac{1.6094 \times 8.314 \times 94500}{15}$, resulting in a final value of 84297.47 J/mol.

The Arrhenius equation $K=Ae^{-E_a/RT}$ dictates that the rate constant $K$ is exponentially dependent on the activation energy $E_a$, meaning an enzyme increases the reaction rate by lowering this energy barrier to a new value $E_a'$. Given that the catalyzed rate $K'$ is $10^6$ times larger than $K$, the relationship is expressed as $e^{-E_a'/RT} = 10^6 e^{-E_a/RT}$. Taking the natural logarithm of both sides leads to $-\frac{E_a'}{RT} = \ln(10^6) - \frac{E_a}{RT}$, which simplifies to $\frac{E_a - E_a'}{RT} = 6 \ln(10)$. Isolating the difference in activation energy, $E_a' - E_a$, results in $-RT \times 6 \times \ln(10)$, which evaluates to $-6(2.303)RT$ after applying the standard conversion factor.

The rate constant for any chemical reaction follows the Arrhenius equation, $k = A e^{-\frac{E_a}{RT}}$. Because the reaction rates and the pre-exponential factors are identical for both the uncatalyzed process at 700 K and the catalyzed process at 500 K, we set their rate expressions equal to each other to yield $\frac{E_a}{700R} = \frac{E_a - 30}{500R}$. Simplifying this by removing the common gas constant $R$ results in $\frac{E_a}{700} = \frac{E_a - 30}{500}$, which cross-multiplies to $5 E_a = 7 E_a - 210$. This algebraic solution reveals the initial activation energy is 105 kJ/mol, and subtracting the 30 kJ/mol reduction identifies the activation energy of the catalyzed reaction as 75 kJ/mol.

For any first-order reaction, the relationship between time and the extent of completion is governed by the rate constant $k$, defined by the formula $k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]_t} \right)$. Since the reaction conditions are fixed, the rate constant determined from the initial completion percentage will remain consistent when evaluating the second scenario.

When 75% of the reaction is complete, the amount of reactant remaining is 25%, making the ratio $\frac{[A]_0}{[A]_t} = \frac{1}{0.25}$, which is 4. Substituting this into the rate equation gives $k = \frac{\ln 4}{90} = \frac{2 \ln 2}{90}$, which simplifies to $k = \frac{\ln 2}{45}$ per minute.

For 60% completion, the remaining reactant is 40%, making the ratio $\frac{[A]_0}{[A]_t} = \frac{1}{0.4} = 2.5$. By rearranging the rate law to solve for time, $t = \frac{\ln 2.5}{k}$, and substituting our previously determined $k$, we obtain $t = \frac{\ln 2.5}{(\ln 2 / 45)} = 45 \times \frac{\ln 2.5}{\ln 2}$. Using the conversion $\frac{\ln 2.5}{\ln 2} = \frac{\log 2.5}{\log 2}$ along with the provided values $0.40$ and $0.30$, the expression becomes $t = 45 \times \frac{0.40}{0.30}$, which calculates to 60 minutes.

In any chemical reaction, the rate at which reactants are consumed is strictly governed by their stoichiometric coefficients, ensuring that the normalized rate of the reaction remains consistent regardless of the species being monitored. For the reaction $2A + 3B + \frac{3}{2}C \rightarrow 3P$, we express this by equating the rate of change of moles for each reactant divided by its respective coefficient:
$\frac{1}{2}\left(-\frac{dn_A}{dt}\right) = \frac{1}{3}\left(-\frac{dn_B}{dt}\right) = \frac{1}{3/2}\left(-\frac{dn_C}{dt}\right)$
Simplifying these expressions by multiplying to isolate the individual rates shows that $-\frac{dn_A}{dt} = \frac{2}{3}\left(-\frac{dn_B}{dt}\right) = \frac{4}{3}\left(-\frac{dn_C}{dt}\right)$, which corresponds to the relationship $\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}$.

The Arrhenius equation describes the dependency of the rate constant $k$ on temperature $T$ through the linear relation $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$. When this relationship is represented graphically, the slope of the line corresponds to the term $-\frac{E_a}{R}$. Given that the calculated slope from the plotted data is equal to $-2$, setting $-\frac{E_a}{R} = -2$ directly results in an activation energy of $E_a = 2R$.

At chemical equilibrium, the rate of the forward reaction must equal the rate of the reverse reaction, which allows us to relate the rate expressions through the equilibrium constant $K_{\mathrm{eq}}$. Because $K_{\mathrm{eq}} = \frac{k_f}{k_b}$ and is defined by the reaction stoichiometry as $K_{\mathrm{eq}} = \frac{[\mathrm{N}_2][\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{H}_2]^2[\mathrm{NO}]^2}$, we can set the forward rate $k_f[\mathrm{NO}]^2[\mathrm{H}_2]$ equal to the product of the reverse rate constant $k_b$ and the equilibrium concentration expression. Solving this equality for the reverse reaction rate shows that it must be equal to $\frac{k_b[\mathrm{N}_2][\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{H}_2]}$.

The reaction rate $r$ is determined by the expression $r = k[A]^x[B]^y$, where $x$ and $y$ represent the reaction orders with respect to reactants A and B. By isolating the effect of each concentration, we determine these orders by comparing experiments where one reactant concentration remains constant. Comparing experiments I and II, where [A] is fixed at 0.1, reveals that doubling [B] from 0.1 to 0.2 increases the rate from $6.00 \times 10^{-3}$ to $2.40 \times 10^{-2}$, which means $2^y = 4$ and $y = 2$. Comparing experiments I and III, where [B] is fixed at 0.1, reveals that doubling [A] from 0.1 to 0.2 increases the rate from $6.00 \times 10^{-3}$ to $1.20 \times 10^{-2}$, which means $2^x = 2$ and $x = 1$.

Substituting the rate law $r = k[A][B]^2$ with values from experiment I yields $6.00 \times 10^{-3} = k(0.1)(0.1)^2$, confirming the rate constant $k = 6$. Applying this to experiment IV, where the rate is $7.20 \times 10^{-2}$ and [B] is 0.2, the expression $7.20 \times 10^{-2} = 6(X)(0.2)^2$ results in $X = 0.3$. Finally, for experiment V, the rate is $2.88 \times 10^{-1}$ with [A] at 0.3, so $2.88 \times 10^{-1} = 6(0.3)(Y)^2$ gives $Y^2 = 0.16$, leading to $Y = 0.4$.

Radioactive decay follows first-order kinetics, meaning the rate of disintegration is directly proportional to the amount of material remaining at any given time. Because the objective is to determine the time required for a 90% reduction in the substance, it follows that 10% of the initial sample remains, establishing an initial-to-final concentration ratio of $A_0/A = 10$. The time is calculated using the integrated rate law formula $t = \frac{2.303}{k} \log_{10} \left( \frac{A_0}{A} \right)$ where the decay constant $k$ is derived from the half-life $t_{1/2}$ as $k = 0.693 / t_{1/2}$. Substituting the half-life of 6.93 years into this expression yields $t = \frac{2.303 \times 6.93}{0.693} \log_{10}(10)$ and since the value $6.93 / 0.693$ equals 10 and the logarithm of 10 is 1, the total duration calculates to 23.03 years.

First-order decomposition follows the integrated rate law $[C]_t = [C]_0 e^{-kt}$, where the rate constant $k$ is linked to the half-life $t_{1/2}$ by the expression $k = \frac{\ln 2}{t_{1/2}}$. Because compound B has a shorter half-life of $180\text{ s}$ compared to compound A's $300\text{ s}$, B disappears more rapidly, allowing the concentration of A to eventually exceed that of B by the specified factor.

Since the initial concentrations are identical, the requirement $[A]_t = 4[B]_t$ simplifies to $e^{-k_A t} = 4 e^{-k_B t}$, which rearranges to $e^{(k_B - k_A)t} = 4$. Taking the natural logarithm of both sides results in $(k_B - k_A)t = \ln 4$, or equivalently $(k_B - k_A)t = 2 \ln 2$. Substituting the specific rate constants $k_A = \frac{\ln 2}{300}$ and $k_B = \frac{\ln 2}{180}$ gives:
$\left(\frac{\ln 2}{180} - \frac{\ln 2}{300}\right)t = 2 \ln 2$
Dividing throughout by $\ln 2$ simplifies the expression to $\left(\frac{1}{180} - \frac{1}{300}\right)t = 2$, which reduces to $\frac{2}{900}t = 2$ and yields a final time of $900\text{ s}$.

Elementary reactions that occur in a single step follow rate laws where the rate is proportional to the concentration of reactants, typically expressed as $Rate = k[A]^n$. Because the rate of a zero-order reaction remains constant regardless of reactant concentration, the reaction cannot be an elementary one-step process. Consequently, these reactions must proceed through a more complex mechanism, making a zero-order reaction a multistep reaction.