Chemical Kinetics — Practice Questions with Answers

Order is an experimentally determined quantity describing how the reaction rate depends on the concentration of each reactant — it can be a whole number, fractional, or even zero, and applies to the overall reaction (which may involve multiple steps).

Molecularity is a theoretical concept, referring to the actual number of reactant species (molecules/atoms/ions) participating in a single elementary reaction step — it is always a whole number (1, 2, or rarely 3), and only strictly applies to individual elementary steps, not necessarily to complex, multi-step overall reactions.

The rate constant ($k$) is the proportionality constant in the rate law equation, relating reaction rate to reactant concentration(s) — it is specific to a given reaction at a specific temperature.

A larger rate constant indicates a faster reaction (all else being equal) — it reflects the intrinsic "speed" of the reaction under given conditions, independent of the actual reactant concentrations used in any particular experiment, and its value characteristically increases with increasing temperature.

Arrhenius equation: $k = Ae^{-E_a/RT}$, where $k$ is the rate constant, $A$ is the pre-exponential (frequency) factor, $E_a$ is activation energy, $R$ is the gas constant, and $T$ is absolute temperature.

The activation energy term ($E_a$) in the exponential directly determines how sensitively the rate constant responds to temperature changes — a higher $E_a$ means the rate constant increases more dramatically with a given temperature rise (since a larger fraction of the reactant population must overcome this larger energy barrier), while a lower $E_a$ means a less pronounced sensitivity to temperature changes.

According to collision theory, a chemical reaction occurs only when reactant particles collide with sufficient energy (exceeding activation energy) and proper orientation.

Increasing the concentration of reactants increases the number density of reactant particles within a given volume, which correspondingly increases the frequency of collisions between reactant particles per unit time — since more frequent collisions statistically produce more successful (effective) collisions per unit time (assuming the fraction of successful collisions remains roughly constant), the overall reaction rate increases with increased reactant concentration.

Half-life ($t_{1/2}$) is the time required for the concentration of a reactant to decrease to exactly half of its initial value.

For first-order reactions specifically, half-life is remarkably useful because it is completely independent of the initial reactant concentration ($t_{1/2}=0.693/k$) — remaining constant throughout the entire reaction, regardless of how much reactant remains at any given point; this constant half-life is a distinctive, easily testable characteristic specifically diagnostic of first-order kinetics, unlike other reaction orders where half-life does depend on initial concentration.

An elementary reaction occurs in a single step, exactly as written, with its molecularity directly and reliably determining its rate law (order equals molecularity for elementary steps).

A complex (multi-step) reaction proceeds through a series of elementary steps (a reaction mechanism) — its overall order must be determined experimentally, and generally cannot be reliably predicted simply from the overall balanced equation's stoichiometric coefficients, since the rate is typically governed by the slowest ("rate-determining") step within the overall mechanism.

The rate-determining step is the slowest individual step within a multi-step reaction mechanism.

Significance: Since the overall reaction cannot proceed faster than its slowest constituent step (analogous to a chain being only as strong as its weakest link), the rate law and overall reaction rate of the entire multi-step process are effectively determined solely by the kinetics of this single rate-determining step, regardless of how fast the other (faster) steps in the mechanism might individually be.

According to the Maxwell-Boltzmann distribution of molecular energies, a modest increase in temperature disproportionately increases the fraction of molecules possessing energy exceeding the activation energy threshold (since this fraction depends exponentially on temperature, via the Arrhenius equation's exponential term).

For many common reactions, with typical activation energy values, this exponential relationship happens to result in approximately a doubling of the reaction rate for each 10°C temperature increase — though this is an empirical approximation ("rule of thumb"), not an exact universal law, since the precise multiplication factor actually depends on the specific activation energy of each individual reaction.

Average rate is the change in concentration divided by the corresponding finite time interval over which that change is measured: $\text{Average rate} = \dfrac{\Delta[\text{conc.}]}{\Delta t}$.

Instantaneous rate is the rate at one specific, precise moment in time, obtained mathematically as the limit of average rate as the time interval approaches zero (equivalent to the slope of the tangent to a concentration-time graph at that specific instant): $\text{Instantaneous rate} = -\dfrac{d[\text{conc.}]}{dt}$.

Chemical reactions involving a solid reactant can only occur at the solid's exposed surface, where it is directly accessible for collision with other reacting species (like a surrounding liquid or gas reactant) — the interior/bulk of the solid, buried beneath the surface, is not directly available for these collisions.

Increasing surface area (e.g., by breaking a solid into smaller pieces or powder) dramatically increases the total surface area exposed for the same overall mass/volume of solid — this increased exposed surface area allows significantly more collisions to occur per unit time between the solid reactant and the other reacting species, directly increasing the overall observed reaction rate.

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{2\times10^{-3}}$

$t_{1/2} = \mathbf{346.5\text{ s}}$

Integrated first-order rate law: $k = \dfrac{2.303}{t}\log\left(\dfrac{[A]_0}{[A]}\right)$

$k = \dfrac{2.303}{60}\log\left(\dfrac{0.8}{0.1}\right) = \dfrac{2.303}{60}\log(8)$

$\log(8) \approx 0.903$

$k = \dfrac{2.303\times0.903}{60} \approx \dfrac{2.080}{60} \approx \mathbf{0.0347\text{ min}^{-1}}$

Rate $\propto [A]^n$. If concentration doubles and rate increases 4-fold: $4 = 2^n$

$2^n = 4 = 2^2 \Rightarrow n = 2$

The reaction is second order with respect to that reactant.

$\ln\left(\dfrac{8\times10^{-4}}{2\times10^{-4}}\right) = \ln(4) \approx 1.386$

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{300}-\dfrac{1}{320} = \dfrac{320-300}{300\times320} = \dfrac{20}{96000} \approx 2.083\times10^{-4}$

$1.386 = \dfrac{E_a}{8.314}\times2.083\times10^{-4}$

$E_a = \dfrac{1.386\times8.314}{2.083\times10^{-4}} \approx \dfrac{11.52}{2.083\times10^{-4}} \approx \mathbf{55300\text{ J/mol} \approx 55.3\text{ kJ/mol}}$

If 40% complete, 60% of reactant remains: $[A]/[A]_0 = 0.60$

$k = \dfrac{2.303}{50}\log\left(\dfrac{1}{0.60}\right) = \dfrac{2.303}{50}\log(1.667)$

$\log(1.667) \approx 0.222$

$k = \dfrac{2.303\times0.222}{50} \approx 0.01022\text{ min}^{-1}$

For 80% completion (20% remains, $[A]/[A]_0=0.20$): $t = \dfrac{2.303}{k}\log\left(\dfrac{1}{0.20}\right) = \dfrac{2.303}{0.01022}\log(5)$

$\log(5)\approx0.699$; $t = \dfrac{2.303\times0.699}{0.01022} \approx \dfrac{1.610}{0.01022} \approx \mathbf{157.5\text{ minutes}}$

New rate $= k[2A][3B]^2 = k(2)[A](9)[B]^2 = 18\times k[A][B]^2$

Factor increase $= \dfrac{\text{new rate}}{\text{original rate}} = \dfrac{18\,k[A][B]^2}{k[A][B]^2} = \mathbf{18}$

The reaction rate increases by a factor of 18.

Integrated zero-order rate law: $[A] = [A]_0 - kt$

$[A] = 0.5 - (2\times10^{-3})(100)$

$[A] = 0.5 - 0.2$

$[A] = \mathbf{0.3\text{ M}}$

Number of half-lives: $n = 40/10 = 4$

Fraction remaining $= \left(\dfrac{1}{2}\right)^n = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}$

Fraction remaining = $\mathbf{1/16}$ (6.25%)

At 90% completion, 10% of reactant remains: $[A]/[A]_0 = 0.10$

$t = \dfrac{2.303}{k}\log\left(\dfrac{1}{0.10}\right) = \dfrac{2.303}{1.5\times10^{-3}}\log(10)$

$\log(10)=1$

$t = \dfrac{2.303\times1}{1.5\times10^{-3}} \approx \mathbf{1535.3\text{ minutes}}$

Using rate $=k[A]^n$: $\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{[A]_2}{[A]_1}\right)^n$

$\dfrac{8\times10^{-3}}{2\times10^{-3}} = \left(\dfrac{0.2}{0.1}\right)^n \Rightarrow 4 = 2^n \Rightarrow n=2$

The reaction is second order.

Finding $k$ using the first data point: $k = \dfrac{\text{rate}}{[A]^2} = \dfrac{2\times10^{-3}}{(0.1)^2} = \dfrac{2\times10^{-3}}{0.01}$

$k = \mathbf{0.2\text{ mol}^{-1}\text{L s}^{-1}}$