Chemical Kinetics – NEET UG ChemistryPractice Questions & PYQs

Determining the rate law requires evaluating how changes in individual reactant amounts affect the overall speed of the transformation. Comparing the first and second experiments reveals that increasing the amount of $\mathrm{Br}_2$ from 0.05 to 0.10 has zero impact on the initial rate of $5.7\times 10^{-5}$, proving that the reaction is zeroth order with respect to bromine. Observing the shift between the second and third trials, where doubling the acid concentration from 0.05 to 0.10 scales the rate proportionally from $5.7\times 10^{-5}$ to $1.2\times 10^{-4}$, demonstrates a first-order dependence on hydrogen ions. Incorporating these findings with the remaining data for acetone yields a first-order rate dependence on $\mathrm{CH}_3\mathrm{COCH}_3$ as well, culminating in the expression Rate $= k[\mathrm{CH}_3\mathrm{COCH}_3][\mathrm{H}^{+}]$.

The half-life of a first-order reaction is independent of the initial concentration of the reactant and is given by the expression $t_{1/2} = \frac{\ln 2}{k}$. When the concentration of reactant A drops to half of its initial value of 0.5 M, the concentration terms completely cancel out from the integrated rate equation. Therefore, the resulting time expression depends solely on the rate constant and simplifies directly to $\frac{\ln 2}{k}$, making the initial concentration value extraneous.

For a first-order reaction, the rate constant is determined using the integrated rate equation $k = \frac{2.303}{t} \log\left(\frac{a}{a-x}\right)$. Substituting the initial percentage and the completion data after 60 minutes yields $k = \frac{2.303}{60} \log\left(\frac{100}{40}\right)$, which establishes the specific rate constant. The time required for 50 percent completion is then calculated using the relation $t_{1/2} = \frac{2.303}{k} \log\left(\frac{100}{50}\right)$, resulting in an approximate duration of 45 minutes.

The overall rate law of a multi-step reaction is determined by its slowest step, which acts as the rate-determining step. For Mechanism B, the slow step involves one molecule of hydrogen and one molecule of iodine monochloride, yielding a rate law that is first order with respect to each reactant. This matches the experimentally observed rate kinetics perfectly while leaving Mechanism A inconsistent with multi-step molecularity expectations.

Reaction rates for consumed reactants are represented with negative signs, while produced substances are represented with positive signs divided by their respective stoichiometric coefficients. For the given reaction, the rate of formation of product C must include a positive sign because its concentration increases over time. Consequently, expressing this rate with a negative sign violates the fundamental sign convention of reaction rates, making that particular expression invalid.

The rate of a chemical reaction is expressed by dividing the rate of change of concentration of any reactant or product by its stoichiometric coefficient, with reactants taking a negative sign and products taking a positive sign. Equating the rate expressions for ammonia and hydrogen based on the balanced equation gives $-\frac{1}{3}\frac{d[\mathrm{H}_2]}{dt} = \frac{1}{2}\frac{d[\mathrm{NH}_3]}{dt}$. Rearranging this equality to isolate the rate of formation of ammonia yields the relationship where positive $\frac{d[\mathrm{NH}_3]}{dt}$ equals negative two-thirds times $\frac{d[\mathrm{H}_2]}{dt}$.

The rate constant for a first-order reaction is obtained from the rate law equation $\text{rate} = k[A]$, where $k$ is calculated by dividing the rate by the concentration. Substituting the given values yields a rate constant of $2.0 \times 10^{-3} \text{ s}^{-1}$. The half-life is then evaluated using the relation $t_{1/2} = 0.693 / k$, where $k$ is $2.0 \times 10^{-3} \text{ s}^{-1}$, leading to a half-life of $346.5 \text{ s}$ that rounds to $347 \text{ s}$.

The rate law expresses the dependence of reaction rate on reactant concentration through the equation $\text{Rate} = k[\text{A}]^\alpha[\text{B}]^\beta$, where the exponents represent the reaction orders. Setting up a ratio of the initial rate to the new rate after doubling the concentration of reactant B gives the exponential relation $1 / 4 = (1 / 2)^\beta$. Solving this equation for the exponent reveals that $\beta$ must equal negative two, indicating an inverse second-order dependence.

First-order reaction kinetics relate the reaction rate and reactant concentration through the expression $\text{rate} = k[A]$, allowing the rate constant to be isolated as $k = \text{rate} / [A]$. Substituting the given values yields a rate constant of $3.0 \times 10^{-2} \text{ min}^{-1}$. Combining this rate constant with the first-order half-life formula $t_{1/2} = 0.693 / k$ provides the calculation for $t_{1/2}$, resulting in a value of $23.1 \text{ min}$.

A general rate law expresses the reaction rate as $\text{rate} = k[A]^n$, where $n$ represents the reaction order. When the reaction rate equals the rate constant itself, the concentration term must evaluate to unity regardless of the reactant concentration. This condition is uniquely satisfied when the concentration is raised to the power of zero, since any non-zero number to the zeroth power equals one. Consequently, the reaction order must be zero for the rate to remain independent of concentration and identical to $k$.

Taking the logarithm of the Arrhenius equation transforms exponential temperature dependence into a linear equation matching the slope-intercept form $y = mx + c$. Specifically, expressing it as $\log k = \log A - \frac{E^*}{2.303 R} \left(\frac{1}{T}\right)$ maps $\log k$ to the vertical axis and the inverse temperature $1 / T$ to the horizontal axis. The slope of this straight-line graph directly incorporates the activation energy, making $\log k$ versus $1 / T$ the correct graphical representation.

The relationship between the activation energy of a forward reaction and its reverse counterpart is dictated by the overall enthalpy change of the system. For an exothermic reaction, the potential energy of the products lies lower than that of the reactants, making the reverse activation energy greater than the forward activation energy $E_{a}$. Conversely, an endothermic reaction features products at a higher energy state, meaning the reverse pathway encounters a smaller energy barrier than the forward route. Since thermochemical characteristics vary across different reactions, the reverse activation energy can end up being either less than or more than $E_{a}$.

The fraction of reactant converted in any given time interval for a first-order reaction depends solely on the rate constant and time, remaining entirely independent of the initial concentration. In the first scenario, $0.6 \text{ mol}$ of product is formed from $0.8 \text{ mol}$ of reactant, representing a $75\%$ conversion of reactant $A$. In the second scenario, producing $0.675 \text{ mol}$ of product from $0.9 \text{ mol}$ of reactant represents that exact same $75\%$ fraction of conversion. Because first-order fractional-life times are constant for identical percentage completions, the time required for both processes is identical at $1 \text{ hour}$.

The rate law for a zero-order reaction is expressed as $r = k[\mathrm{A}]^{0}$, meaning the reaction rate equals the rate constant $k$ and is entirely independent of the reactant concentration. This implies the reaction velocity remains completely unchanged regardless of fluctuations in the amount of starting material present. Consequently, the process proceeds at a constant speed at any concentration level of the reactant $\mathrm{A}$.

For any balanced chemical equation, the rate of consumption of reactants and the rate of formation of products are related through their stoichiometric coefficients. Equating the reaction rates yields $-\frac{1}{3}\frac{d[\mathrm{A}]}{dt} = \frac{1}{2}\frac{d[\mathrm{B}]}{dt}$ based on the stoichiometry of three moles of $\mathrm{A}$ producing two moles of $\mathrm{B}$. Rearranging this expression to isolate the rate of change of component $\mathrm{B}$ directly results in $\frac{d[\mathrm{B}]}{dt} = -\frac{2}{3}\frac{d[\mathrm{A}]}{dt}$.

The rate law for this first-order decomposition reaction is expressed as $\text{rate} = k[\mathrm{N}_{2}\mathrm{O}_{5}]$, establishing a direct proportionality between the reaction velocity and the reactant concentration. Rearranging this relation to isolate the concentration term gives $[\mathrm{N}_{2}\mathrm{O}_{5}] = \frac{\text{rate}}{k}$. Substituting the respective numerical values of $1.02 \times 10^{-4}$ and $3.4 \times 10^{-5}$ into this expression directly evaluates to a concentration value of $3$.

Catalysts and enzymes function by providing an alternative reaction pathway that significantly lowers the activation energy barrier compared to an uncatalyzed process. Because the laboratory setup lacks the biological enzyme, the uncatalyzed reaction proceeds through a high-energy mechanism requiring a much larger $E_{a}$. Introducing the enzyme alters this microscopic pathway, resulting in an activation energy value entirely distinct from the laboratory measurement.

The rate of a chemical reaction is expressed by dividing the rate of change of concentration of each species by its stoichiometric coefficient, applying a negative sign for reactants and a positive sign for products. For the given reaction, the instantaneous rate is equated as -\frac{d}[\mathrm{H}^+]$}{dt} = -\frac{d[\mathrm{BrO}_3^-]\}{dt} = -\frac{1}{3}\frac{d[\mathrm{Br}^-]\{}{dt} = \frac{1}{5}\frac{d[\mathrm{Br}_2]}{dt} = \frac{d[\mathrm{H}_2\mathrm{O}]}{dt}$. Isolating the relationship between the bromide ion consumption and bromine formation by setting$-\frac{1}{3}\frac{d}[\mathrm{Br}^-]\{}{dt} = \frac{1}{5}\frac{d[\mathrm{Br}_2]}{dt}$directly yields$\frac{d}[\mathrm{Br}^-]\{}{dt} = -\frac{3}{5}\frac{d[\mathrm{Br}_2]}{dt}$.

At the temperature where two reactions proceed at equal rates, their respective rate constants are set equal to each other, yielding the equation $10^{16} e^{-\frac{2000}{T}} = 10^{15} e^{-\frac{1000}{T}}$. Rearranging the terms by dividing both sides simplifies the relationship to $e^{-\frac{1000}{T}} = 10^{-1}$. Taking the natural logarithm on both sides and converting to base 10 using the factor $2.303$ gives $-\frac{1000}{T} \times \log_{10}(e) = -1 \times \log_{10}(10)$, which solves to $T = \frac{1000}{2.303}$ K.

Catalysts function by providing an alternative reaction pathway that requires less thermal energy for reactant molecules to reach the transition state. Biological catalysts, or enzymes, achieve this by stabilizing the transition state and securely binding substrates in an optimal orientation at their active sites. Consequently, a significantly larger fraction of reactant molecules possess sufficient kinetic energy to overcome this reduced energetic barrier at body temperature, leading to a dramatic acceleration in reaction velocity without altering the overall thermodynamic equilibrium.