Chemical Kinetics – JEE Main ChemistryPractice Questions & PYQs

Radioactive decay follows first-order kinetics, where the decay constant $k$ is defined by the relationship $k = \frac{0.693}{t_{1/2}}$. For a sample with a half-life of 36 hours, the decay constant is $k = \frac{0.693}{36} = 0.01925 \, \text{hr}^{-1}$. Utilizing the integrated rate law $\log \frac{N_0}{N_t} = \frac{kt}{2.303}$ for an elapsed time of 24 hours, we calculate $\log \frac{N_0}{N_t} = \frac{0.01925 \times 24}{2.303} = 0.2006$. Taking the reciprocal of the antilog of 0.2006 provides the fraction of the substance remaining, yielding $\frac{N_t}{N_0} = \frac{1}{1.587} \approx 0.63$. Expressing this result as $63 \times 10^{-2}$ accounts for the remaining portion of the isotope.

Radioactive decay processes rely on the conservation of mass and atomic numbers throughout the transformation chain. When the parent nucleus ${}_{92}^{238}\mathrm{U}$ emits an alpha particle, its atomic number decreases by 2 and its mass number decreases by 4, creating an intermediate thorium isotope, ${}_{90}^{234}\mathrm{Th}$. To reach the final state of ${}_{92}^{234}\mathrm{U}$, the atomic number must increase by 2 while the mass number remains unchanged at 234. Because a single $\beta^{-}$ decay increases the atomic number by 1 while leaving the mass number constant, two such emissions are required to successfully raise the atomic number from 90 to 92.

The Arrhenius equation $k = A e^{-\frac{E}{RT}}$ describes reaction kinetics where $E$ represents the activation energy, acting as the precise energy barrier that molecules must surpass to initiate a chemical change. Because this value establishes the minimum energy required for a collision to yield products, it effectively defines the energy threshold that reactants must meet or exceed to successfully undergo a reaction.

The overall reaction rate is governed by the rate-determining step, which involves the collision between the intermediate $\mathrm{NOBr}_2$ and $\mathrm{NO}$, resulting in an initial rate expression of $r = k[\mathrm{NOBr}_2][\mathrm{NO}]$. Since reaction intermediates must not appear in the final rate law, we use the equilibrium constant expression $K_c = \frac{[\mathrm{NOBr}_2]}{[\mathrm{NO}][\mathrm{Br}_2]}$ from the first step to find that $[\mathrm{NOBr}_2] = K_c[\mathrm{NO}][\mathrm{Br}_2]$. Substituting this concentration into the rate equation yields the final expression:
$r = k \cdot K_c [\mathrm{NO}]^2 [\mathrm{Br}_2]$
This shows that the reaction order with respect to $\mathrm{NO}$ is 2.

The van't Hoff equation $\ln K_{eq} = -\frac{\Delta H}{RT} + \text{constant}$ establishes the fundamental relationship between the equilibrium constant and temperature, where the slope of a $\ln K_{eq}$ versus $\frac{1}{T}$ plot is determined by $-\frac{\Delta H}{R}$. For an exothermic reaction, the enthalpy change is defined as $\Delta H < 0$. This thermodynamic requirement ensures that the reaction is classified as exothermic based on the corresponding slope characteristics shown in the schematic plot.

Nuclear reactions strictly obey the conservation of both total nucleon number and total atomic charge, requiring these quantities to balance exactly on both sides of the process. Since the original ${}_{12}^{24}\mathrm{Mg}$ nucleus sheds a proton, which carries a charge of 1 and a mass of 1, the remaining nucleus must have an atomic number of 11 and a mass number of 23. This change is described by the reaction ${}_{12}^{24}\mathrm{Mg} \rightarrow {}_{11}^{23}\mathrm{Na} + {}_{1}^{1}\mathrm{H}$, identifying the resulting daughter product as ${}_{11}^{23}\mathrm{Na}$.

The hydrogen bomb functions on the principle of nuclear fusion, a process where two lighter atomic nuclei combine to create a single, heavier nucleus. To trigger this, a mixture containing deuterium oxide and tritium oxide is positioned around a conventional atomic bomb, which generates the extreme temperatures necessary to fuse ${}_1\mathrm{H}^2$ and ${}_1\mathrm{H}^3$. This resulting reaction releases an immense quantity of energy, significantly exceeding the power produced by standard fission weapons.

For any first-order reaction, the integrated rate equation relates the rate constant $K$ and time $t$ through the expression $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$. Because $t_{\frac{1}{4}}$ denotes the moment concentration drops to three-quarters of its initial value, the concentration ratio becomes $\frac{4}{3}$, leading to the expression $t = \frac{2.303}{K} (\log 4 - \log 3)$. Utilizing the standard logarithmic values where $\log 4$ is $2 \log 2 \approx 0.602$ and $\log 3 \approx 0.4771$, the equation becomes $\frac{2.303}{K} (2 \times 0.301 - 0.4771)$, which simplifies to $\frac{0.29}{K}$.

The molecularity of an elementary reaction describes the number of reactant molecules that must collide in a single step to trigger the chemical transformation. If a reaction involves two different reactants, at least two distinct particles must participate in this interaction, making a molecularity of $1$ impossible. By definition, a unimolecular reaction entails the transformation of only one reactant molecule, so it cannot describe a process that requires the presence of two separate species.

For a first-order reaction, the half-life, denoted as $t_{1/2}$, remains constant regardless of the initial concentration. Since the reactant concentration drops from 0.8 M to 0.4 M in 15 minutes, we have established that a single half-life interval is 15 minutes. Transitioning from 0.1 M down to 0.025 M requires two successive half-life periods, as the concentration first halves to 0.05 M and then halves again to 0.025 M. Because each period accounts for 15 minutes, the total duration for this change is 30 minutes.

The rate constant $k$ is an inherent property of a specific chemical reaction that changes exclusively with temperature. Since it is determined by the reaction mechanism and the thermal energy of the system rather than the quantity of matter present, the numerical value of $k$ remains fixed regardless of the initial amounts of reactants $A$ and $B$. Consequently, the rate constant is fundamentally independent of the starting concentration of the reacting species.

Radioactive decay is characterized by the predictable reduction of a sample by half its existing quantity for each defined half-life interval. Begin by calculating the number of half-life cycles completed within the total timeframe by dividing the total duration of 24 hours by the half-life of 4 hours, which equals $n = \frac{24}{4} = 6$ cycles. Substitute these values into the decay equation $N_t = N_0 \left( \frac{1}{2} \right)^n$ using the starting mass of 200 g to find $N_t = 200 \left( \frac{1}{2} \right)^6$. Performing this calculation effectively requires dividing 200 by 64, which determines that 3.125 g of the isotope remains undecayed.

Nuclear reactions rely on the conservation of mass number and atomic number, which dictates that the sum of these values must be identical on both sides of the reaction. Starting with the nucleus $_{92}^{238}\mathrm{M}$, the emission of two alpha particles, where each particle contributes 2 to the atomic number and 4 to the mass number, results in a decrease of 4 in the atomic number and 8 in the mass number for the daughter nucleus $_{Y}^{X}\mathrm{N}$. This calculation leaves $_{Y}^{X}\mathrm{N}$ with an atomic number of 88 and a mass number of 230.

The subsequent decay involves two positron emissions, which effectively lower the atomic number by 1 per emission while the mass number remains constant. Applying this change to the previously derived nucleus, the atomic number of the final element $_{B}^{A}\mathrm{L}$ becomes $88 - 2$, resulting in 86, while the mass number remains 230. By subtracting the atomic number from the mass number, the number of neutrons in $_{86}^{230}\mathrm{L}$ is calculated as $230 - 86$, which equals 144.

Reaction rates are fundamentally governed by reactant concentrations, and because concentration is inversely proportional to volume, halving the container volume leads to a doubling of the concentration for each gaseous reactant. Starting with the rate law $r=k[\mathrm{O}_2][\mathrm{NO}]^2$, substituting the doubled concentrations into this expression gives a new rate of $r'=k(2[\mathrm{O}_2])(2[\mathrm{NO}])^2$. Expanding this results in $r'=8k[\mathrm{O}_2][\mathrm{NO}]^2$, which mathematically confirms that the rate of reaction increases to eight times its initial value.

The Arrhenius equation, $k = Ae^{-E_a/RT}$, provides the mathematical relationship between the velocity constant $k$ and temperature. In this expression, $A$ is defined as the frequency factor and $R$ corresponds to the gas constant. Consequently, $E_a$ denotes the energy of activation, which serves as the energy threshold that reactants must bridge to successfully transform into products.

The rate of a chemical reaction is defined by the product of the rate constant and the reactant concentrations raised to their respective orders, which means any change in the concentration of a reactant directly scales the overall reaction rate according to its specific exponent. Starting with the initial rate expression $Rate_1 = k[A]^n[B]^m$, modifying the concentrations by doubling $[A]$ and halving $[B]$ results in a new rate defined as $Rate_2 = k[2A]^n[\frac{1}{2}B]^m$. To determine the ratio of the new rate to the earlier rate, we divide $Rate_2$ by $Rate_1$, which allows the rate constant $k$ and the original concentration variables to cancel out. This leaves us with the expression $\frac{[2A]^n[\frac{1}{2}B]^m}{[A]^n[B]^m}$, which simplifies to $2^n \cdot (\frac{1}{2})^m$. Applying the rule of exponents where $\frac{1}{2}$ is $2^{-1}$, we arrive at $2^n \cdot 2^{-m}$, resulting in a final ratio of $2^{n-m}$.

Radioactive decay is governed by specific conservation principles where a beta emission effectively converts a neutron into a proton, raising the atomic number by one while leaving the mass number unchanged, whereas an alpha emission releases a helium nucleus, decreasing the atomic number by two and the mass number by four. Starting with the parent nucleus ${}^{234}_{90}\mathrm{Th}$, the two consecutive beta decays increase the atomic number twice, moving it from 90 to 91 and then to 92, all while the mass number remains constant at 234. When this intermediate nucleus undergoes an alpha decay, it loses two protons and two neutrons, which shifts the atomic number down by two units to 90 and reduces the mass number by four units to 230.

Radioactive decay dictates that a substance reduces its mass by half during every distinct time interval called its half-life. To determine the remaining quantity, first identify how many half-life cycles fit into the total elapsed time of 18 hours by dividing it by the 3-hour half-life, which gives $n = \frac{18}{3} = 6$ cycles. By applying the decay formula $N = \frac{N_0}{2^n}$, where the initial mass $N_0$ is 256 g, the calculation becomes $N = \frac{256}{2^6}$. Evaluating the denominator as 64 leads to a final remaining mass of 4 g.

The overall order of a reaction corresponds to the sum of the exponents of the concentration terms present in the rate law expression. In the rate law $R=[A][B]^2$, the exponent for species $A$ is $1$, and the exponent for species $B$ is $2$. Summing these values, $1 + 2$, results in an overall reaction order of $3$.

The linear relationship required to generate a straight line is found by rearranging the given rate equation into the form $y = mx + c$. By rewriting $\log C_0 - \log C_t = R t$ as $\log C_t = -R t + \log C_0$, we treat $\log C_t$ as the dependent variable and $t$ as the independent variable. Consequently, plotting $\log C_t$ against $t$ satisfies the equation of a straight line, where the slope is $-R$ and the intercept is $\log C_0$.