Chemical Kinetics – NEET UG ChemistryPractice Questions & PYQs

For a first-order reaction, the rate at any time is directly proportional to the reactant concentration, allowing the rate constant to be determined from rates at different times via $k = \frac{1}{t_2 - t_1} \ln \left(\frac{r_1}{r_2}\right)$. Substituting the respective time points and rates into this expression yields the rate constant. Applying this value to the half-life relation $t_{1/2} = \frac{0.693}{k}$ results in a half-life period of 24.1 s.

Heterogeneous catalysis involving gases on solid surfaces depends directly on the fraction of the surface area covered by the reactant molecules. At low pressures, the extent of this surface coverage is directly proportional to the partial pressure of the gas. Consequently, the reaction rate varies linearly with the reactant pressure, establishing first-order kinetics.

A catalyst functions by providing an alternative reaction pathway with a lower activation energy barrier, thereby increasing the reaction rate without being consumed. Thermodynamic state functions such as enthalpy, entropy, and internal energy depend exclusively on the initial and final states of the system rather than the pathway. Consequently, the only quantity altered during the addition of a catalyst is the activation energy.

The half-life of a first-order reaction is expressed by the relation $t_{1/2} = \frac{0.693}{k}$, demonstrating that it depends solely on the rate constant rather than the reactant concentration. Consequently, changing the starting amount of the reactant leaves the time required for half the reaction to complete unchanged. This independence of half-life from initial concentration is the hallmark characteristic of first-order kinetics.

The units of the rate constant reveal that the reaction follows zero-order kinetics, for which the integrated rate equation is expressed as $x = Kt$. Multiplying the reaction duration converted into seconds by the given rate constant yields the amount of reactant consumed. This calculation directly produces a final product concentration of 0.72 M.

The Arrhenius equation in logarithmic form is given by $\ln K = \ln A - \frac{E_a}{RT}$, which relates the rate constant to temperature and activation energy. By plotting $\ln K$ on the vertical axis against the inverse of temperature on the horizontal axis, the resulting straight line has a slope equal to $-\frac{E_a}{R}$. Therefore, multiplying the magnitude of this negative slope by the universal gas constant allows direct calculation of the activation energy.

The variation of reaction rates with temperature is governed by the integrated Arrhenius equation, expressed as $\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$. Substituting the absolute temperatures corresponding to 20°C and 35°C, along with a rate doubling factor and the gas constant value of 8.314 J mol^-1 K^-1, provides the necessary parameters. Evaluating this expression calculates the activation energy to be 34673 J mol^-1, or 34.7 kJ mol^-1.

Writing the Arrhenius rate expression at two distinct temperatures gives $\ln k_1 = \ln A - \frac{E_a}{RT_1}$ and $\ln k_2 = \ln A - \frac{E_a}{RT_2}$. Subtracting the first equation from the second eliminates the frequency factor and combines the temperature terms into $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. Factoring out a negative sign inside the temperature parentheses rearranges this expression into the standard form $\ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$.

The general rate law for a reaction involving reactants $A$ and $B$ is expressed as $\text{rate} = k[A]^a[B]^b$, where $a$ and $b$ are the reaction orders with respect to each reactant. Doubling the concentration of reactant $B$ alone doubles the rate, indicating that the reaction is first order with respect to $B$. When the concentrations of both reactants are doubled simultaneously, the overall rate increases by a factor of 8. Using the relationship $(2)^a(2)^1 = 8$, the calculation simplifies to $2^a = 4$, which shows that the reaction is second order with respect to $A$. Incorporating these individual orders into the initial expression gives $\text{rate} = k[A]^2[B]$.

The multiplicative increase in a reaction rate over a temperature interval is calculated using the relation $r_2 / r_1 = 2^{\Delta T / 10}$, where $\Delta T$ represents the total change in temperature. Finding the difference between 100°C and 10°C gives a temperature span of 90°C. Dividing this span by the ten-degree interval yields an exponent of nine, meaning the rate increases by a factor of two raised to the ninth power, resulting in a 512-fold enhancement.

The rate law for any reaction is given by $\text{rate} = k[A]^n$, where $n$ represents the overall order ofいは the reaction. Rearranging this expression to solve for the rate constant yields $k = \frac{\text{rate}}{[A]^n}$. For a zero order reaction, the order is zero, meaning the rate constant has the same units as the rate of the reaction itself. Therefore, substituting concentration over time gives the unit mol L^-1 s^-1 without any dependence on reactant concentrations.

The rate of a chemical reaction is defined by relating the change in concentration of each species to its stoichiometric coefficient in the balanced chemical equation. For the given reaction, equating the consumption of the reactant to the formation of the products produces the expression $\text{Rate} = -\frac{1}{2}\frac{d[\mathrm{N}_2\mathrm{O}_5]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO}_2]}{dt} = \frac{d[\mathrm{O}_2]}{dt}$

Replacing the individual derivative terms with their respective rate constants provided yields $\frac{1}{2}k[\mathrm{N}_2\mathrm{O}_5] = \frac{1}{4}k'[\mathrm{N}_2\mathrm{O}_5] = k''[\mathrm{N}_2\mathrm{O}_5]$ Eliminating the common concentration term on all sides leaves the equivalence $\frac{1}{2}k = \frac{1}{4}k' = k''$, which easily rearranges to give $k' = 2k$ and $k'' = \frac{k}{2}$.

The order of a reaction is defined as the sum of the powers of the concentration terms in the experimentally determined rate law expression. Because reaction mechanisms can involve complex elementary steps, the overall order is not restricted to integers and can frequently be zero or fractional. Furthermore, unlike stoichiometric coefficients which are fixed by the balanced chemical equation, reaction order must always be determined through empirical rate measurements. Consequently, stating that reaction orders are strictly whole numbers is entirely incorrect.

The rate law for the reaction is expressed generally as $\text{Rate} = k[A]^a[B]^b$, where the exponents denote the reaction orders for each species. By comparing experimental runs II and III, the concentration of reactant $A$ is held constant at 0.3 while the concentration of reactant $B$ increases from 0.2 to 0.4. Because this doubles the concentration of $B$ and causes the initial rate to jump from 7.2 x 10^-2 to 2.88 x 10^-1, which is a factor of four, the reaction order with respect to $B$ must be two. Turning next to runs I and IV, the concentration of reactant $B$ remains constant at 0.1 while the concentration of reactant $A$ changes from 0.1 to 0.4. This fourfold increase in the concentration of $A$ scales the initial rate proportionally from 6.0 x 10^-3 to 2.40 x 10^-2, proving that the reaction order with respect to $A$ is one. Combining these findings gives the complete rate expression as $\text{Rate} = k[A][B]^2$.

The rate constant of a chemical reaction is independent of reactant concentrations and is strictly a function of temperature, as mathematically described by the Arrhenius equation. Altering the concentration of reactants will change the overall reaction rate, but the proportionality constant itself remains unaffected. Consequently, the only way to increase the numerical value of the rate constant is by raising the temperature of the system.

The relationship between the rates of consumption of reactants and formation of products in a chemical reaction is established by dividing the rate of change of each species by its stoichiometric coefficient in the balanced equation. For the given decomposition, the rate expression equates the disappearance of dinitrogen pentoxide to half the rate of formation of nitrogen dioxide and twice the rate of formation of oxygen gas. Multiplying the given rate of disappearance of dinitrogen pentoxide by two yields the rate of formation of nitrogen dioxide as $1.25 \times 10^{-2}$ mol L^-1 s^-1. Similarly, dividing the rate of disappearance by two gives the rate of formation of oxygen gas as $3.125 \times 10^{-3}$ mol L^-1 s^-1.

The rate law expression relates the reaction rate to reactant concentrations in the general form $\text{rate} = k[A]^m[B]^n$, where $m$ and $n$ represent the individual reaction orders. When only the initial concentration of reactant $A$ is doubled, the reaction rate also doubles, establishing that the order with respect to $A$ is 1. When the initial concentrations of both $A$ and $B$ are doubled simultaneously, the rate increases by a factor of 8. Substituting the order of $A$ into this combined scaling gives $(2)^1(2)^n = 8$, which simplifies to $2^n = 4$, demonstrating that the order with respect to $B$ is 2. Combining these exponents yields the complete rate law expression as $\text{rate} = k[A][B]^2$.

The rate of a chemical reaction is related to the stoichiometric coefficients of its reactants and products by equating the individual rates of change divided by their respective coefficients, giving the expression $-\frac{1}{3}\frac{d[\mathrm{H}_2]}{dt} = \frac{1}{2}\frac{d[\mathrm{NH}_3]}{dt}$. Using this relationship, the rate of consumption of hydrogen is expressed as $\frac{3}{2}$ times the rate of formation of ammonia. Substituting the given rate of formation of ammonia as $2 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$ into this expression yields $3 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$.

The rate of a chemical reaction is defined by relating the rate of change in concentration of each species to its stoichiometric coefficient, using a negative sign for reactants and a positive sign for products. Applying this principle to the given reaction yields the rate expression -\frac{1}{5} \frac{d}$[\mathrm{Br}^{-}]\${}{dt} = \frac{1}{3} \frac{d[\mathrm{Br}_{2}]}{dt}. Rearranging this equation to solve for the rate of appearance of bromine directly results \in \frac{d[\mathrm{Br}_{2}]}{dt} = -\frac{3}{5} \frac{d}$[\mathrm{Br}^{-}]\${}{dt}.

For a first-order reaction, the half-life is independent of the initial concentration and is related to the rate constant through the expression $t_{1/2} = \frac{0.693}{k}$. Substituting the given half-life value of 1386 seconds into this expression yields $1386 = \frac{0.693}{k}$. Rearranging this equation results in $k = \frac{0.693}{1386}$, which simplifies to $0.5 \times 10^{-3}$ s$^{-1}$.