Chemical Kinetics and Rate Laws (5E)
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MCAT Chemical and Physical Foundations of Biological Systems › Chemical Kinetics and Rate Laws (5E)
A reaction relevant to food spoilage is monitored: $A \rightarrow P$. Two experiments are run with identical initial A but different temperatures. The measured initial rate at 40°C is higher than at 20°C.
Which scenario best illustrates the concept of activation energy as the cause of this difference?
At higher temperature, $E_a$ increases, which increases the rate by destabilizing reactants
At higher temperature, a larger fraction of molecules has kinetic energy exceeding $E_a$, increasing the rate constant
At higher temperature, the equilibrium constant increases, forcing the reaction to proceed faster
At higher temperature, reactant concentration is higher due to thermal expansion, increasing rate
Explanation
This question assesses understanding of chemical kinetics and activation energy. Higher temperature increases rate by providing more molecules with energy ≥ Ea, per Maxwell-Boltzmann distribution. The higher rate at 40°C vs. 20°C with same [A] reflects this. Option A is correct because it links temperature to fraction exceeding Ea. Option D is incorrect as Ea does not increase with T. When comparing temperatures, use Arrhenius equation. This explains why spoilage accelerates with warmth.
A clinical lab evaluated decomposition of a disinfectant (D) used on medical devices: $D \rightarrow$ products. The reaction was run at constant temperature and pH. The initial rate was measured at two initial concentrations.
Data:
D = 0.30 M, rate = $9.0\times10^{-5}$ M/s
D = 0.60 M, rate = $9.0\times10^{-5}$ M/s
Based on the data, what is the order of the reaction with respect to D?
Zero order, because the initial rate is unchanged when [D] doubles
Second order, because doubling [D] should double the rate but does not due to error
First order, because decomposition is typically unimolecular
Cannot be determined without time-course data
Explanation
This question assesses understanding of chemical kinetics and rate laws. Zero-order reactions have rates independent of reactant concentration, often due to saturation or external limitations. In the data, doubling [D] from 0.30 M to 0.60 M keeps the rate at $9.0×10^{-5}$ M/s. Option B is correct because this constancy indicates zero order in D. Option A is incorrect as first order would double the rate. When evaluating rate laws, look for rate invariance with concentration changes. Remember zero-order half-life depends linearly on initial concentration, unlike other orders.
A pharmaceutical lab monitored hydrolysis of a prodrug (D) to the active drug (X) in buffered plasma simulant at 37°C: $D + H_2O \rightarrow X$. Water is in large excess. Initial rates were measured at different D.
Data (37°C):
D = 0.10 mM, rate = $5.0\times10^{-8}$ M/s
D = 0.20 mM, rate = $1.0\times10^{-7}$ M/s
D = 0.40 mM, rate = $2.0\times10^{-7}$ M/s
Based on the data, what is the order of the reaction with respect to D?
Zero order in D, because water is in excess
Cannot be determined without measuring [H$_2$O]
First order in D, because rate is proportional to [D]
Second order in D, because hydrolysis is bimolecular
Explanation
This question assesses understanding of chemical kinetics and rate laws under pseudo-conditions. When one reactant is in large excess, like water here, the reaction can exhibit pseudo-order behavior with respect to the limiting reactant. In the data, doubling [D] from 0.10 mM to 0.20 mM doubles the rate from $5.0×10^{-8}$ to $1.0×10^{-7}$ M/s, and again to 0.40 mM yields $2.0×10^{-7}$ M/s. Option B is correct because this proportionality indicates first order in D. Option A is incorrect as zero order would show no rate change with [D]. When evaluating rate laws, consider excess reactants as constant to simplify to pseudo-order. A useful check is to plot rate vs. concentration; a straight line through origin suggests first order.
A biotech team examines an enzyme-catalyzed step where substrate S is converted to P. Under low S, the initial rate is proportional to S; under high S, the rate approaches a maximum and becomes insensitive to S.
Which factor most influences the reaction rate in the high-S regime?
Substrate concentration, because more substrate always increases collision frequency
Reaction enthalpy, because exothermic reactions are faster at high [S]
Enzyme concentration (available active sites), because the enzyme becomes saturated
Equilibrium constant, because saturation implies the reaction is at equilibrium
Explanation
This question assesses understanding of chemical kinetics in enzyme systems. In Michaelis-Menten kinetics, at high [S], rate approaches V_max, limited by [E] as sites saturate, becoming zero order in [S]. The vignette describes rate proportional to [S] at low [S] but independent at high [S]. Option B is correct because enzyme saturation limits rate in high-[S] regime. Option A is incorrect as [S] no longer affects rate at saturation. When analyzing, plot rate vs. [S] for hyperbolic curve. V_max = k_cat [E] is key for saturation.
In an industrial sterilization process, a reactive species (R) decomposes: $R \rightarrow$ products. The process engineer considers raising the temperature to increase throughput without changing initial R.
How does changing temperature affect the rate constant $k$ for this decomposition reaction?
$k$ decreases with increasing temperature because decomposition reduces reactant concentration more quickly
$k$ increases only if the reaction is zero order
$k$ increases with increasing temperature because more molecules can overcome the activation barrier
$k$ is independent of temperature for unimolecular reactions
Explanation
This question assesses understanding of chemical kinetics and rate laws, specifically the effect of temperature on the rate constant. The rate constant k for a reaction is temperature-dependent, as described by the Arrhenius equation, k = A e^(-Ea/RT), where increasing temperature reduces the exponential term's suppression, leading to a higher k. In the industrial sterilization process involving the unimolecular decomposition of R to products, the engineer aims to raise temperature to increase throughput while keeping initial [R] constant, which would accelerate the reaction via a higher k. Option B is correct because higher temperatures enable more molecules to surpass the activation energy barrier, directly increasing the rate constant. Option A is incorrect due to the misconception that faster decomposition decreases k, when in fact k itself rises with temperature regardless of concentration changes. When evaluating temperature effects on rate constants, apply the Arrhenius equation to confirm that k typically increases with temperature for reactions with positive activation energy. This principle holds for various reaction orders, making it a reliable check across kinetic scenarios.
Researchers studied oxidative damage in a lipid vesicle system where radical initiator R decomposes to radicals that consume an antioxidant (AH): $R \rightarrow 2,\cdot!R$ (slow), followed by fast radical trapping. Initial rate of antioxidant loss was measured at 25°C.
Data (25°C):
Trial 1: R = 0.20 mM, rate = $3.0\times10^{-7}$ M/s
Trial 2: R = 0.40 mM, rate = $6.0\times10^{-7}$ M/s
Trial 3: R = 0.80 mM, rate = $1.2\times10^{-6}$ M/s
Based on the data, what is the order of the reaction with respect to R for the measured initial rate?
Zero order in R, because radical trapping is fast
Second order in R, because radicals are produced in pairs
First order in R, because doubling [R] doubles the measured rate
Half order in R, because decomposition is unimolecular
Explanation
This question assesses understanding of chemical kinetics and rate laws in radical reactions. The rate law for a reaction is empirical and based on how rate changes with reactant concentrations, often reflecting the slow step. In this vignette, the initial rate of antioxidant loss doubles when [R] doubles, as seen from 0.20 mM to 0.40 mM $(3.0×10^{-7}$ to $6.0×10^{-7}$ M/s) and to 0.80 mM $(1.2×10^{-6}$ M/s), indicating proportionality to [R]. Option A is correct because this behavior shows first order in R, consistent with the slow unimolecular decomposition. Option D is incorrect as half-order would show a sqrt(2) ≈1.4 factor increase upon doubling, not 2. When evaluating rate laws, compare rate ratios to concentration ratios for order determination. Remember that for mechanisms with a slow step, the rate law often matches that step's molecularity.
In a formulation study, a drug (D) degrades by $D \rightarrow$ products in solution. The measured half-life of D is 6 hours at 25°C and remains 6 hours when the initial D is doubled, with all other conditions constant.
Based on this observation, what is the most likely reaction order with respect to D?
Second order, because doubling [D] does not change half-life when two molecules react
Cannot be inferred without knowing the mechanism
First order, because half-life is independent of initial concentration for first-order reactions
Zero order, because half-life is independent of initial concentration only for zero-order reactions
Explanation
This question assesses understanding of chemical kinetics and half-life dependence on order. For first-order reactions, half-life $t_{1/2}$ = ln(2)/k is independent of initial concentration. The half-life remaining 6 hours when [D] doubles indicates this independence. Option B is correct because only first order shows concentration-independent $t_{1/2}$. Option A is incorrect as zero-order $t_{1/2}$ halves when [D] doubles. When inferring order from half-life, compare across initial concentrations. This is transferable to radioactive decay, which is first order.
A researcher proposes the elementary step $2A + B \rightarrow P$ for a key oxidative reaction in mitochondria. Initial-rate data at constant temperature show that doubling A increases the rate by a factor of 4, while doubling B increases the rate by a factor of 2.
Based on the data, which rate law is most consistent with the observed kinetics?
$\text{rate}=k[A]^2[B]^2$
$\text{rate}=k[A][B]$
$\text{rate}=k[A]^4[B]$
$\text{rate}=k[A]^2[B]$
Explanation
This question assesses understanding of chemical kinetics and rate laws. For elementary steps, the rate law matches molecularity, but must be verified empirically. Data showing doubling [A] quadruples rate (order 2 in A) and doubling [B] doubles rate (order 1 in B) supports rate = $k[A]^2$[B]. Option B is correct as it matches the observed scalings. Option C is incorrect as it implies order 2 in B, which would quadruple rate on doubling [B]. When evaluating, match exponents to observed rate factors. Always confirm if the step is elementary before assuming rate law from stoichiometry.
A kinetics study of a signaling molecule degradation reports the rate law $\text{rate}=kS^0$. The experiment is repeated with S increased from 1.0 mM to 3.0 mM at the same temperature.
Which factor most influences the reaction rate when S is increased under these conditions?
The rate is unchanged because the reaction is zero order in S
The rate increases 9-fold because the reaction is second order in S
The rate triples because [S] triples
The rate decreases because higher [S] lowers the rate constant
Explanation
This question assesses understanding of chemical kinetics and zero-order reactions. For rate = $k[S]^0$, rate is constant and independent of [S]. Tripling [S] from 1.0 mM to 3.0 mM does not change the rate. Option B is correct because zero order means rate unchanged with [S]. Option C is incorrect as second order would increase 9-fold. When evaluating, note zero-order rate is k alone. This is common in saturated enzyme kinetics.
An environmental chemist examined decay of a pesticide (Pest) in river water, modeled as $\text{Pest} \rightarrow$ products. Initial rate constants were measured at two temperatures; Pest was kept the same in both trials.
Data:
At 15°C, initial rate = $1.0\times10^{-7}$ M/s
At 25°C, initial rate = $2.0\times10^{-7}$ M/s
Assuming the reaction order and initial concentration are unchanged, which statement best describes how changing temperature affects the rate constant $k$?
$k$ doubles only if the reaction is second order
$k$ increases with increasing temperature because a larger fraction of molecules exceeds $E_a$
$k$ decreases with increasing temperature because reactants are consumed faster
$k$ is unchanged with temperature; only concentrations affect rate
Explanation
This question assesses understanding of chemical kinetics and the temperature dependence of rate constants. The rate constant k increases with temperature according to the Arrhenius equation, as higher temperatures provide more molecules with energy surpassing the activation barrier Ea. In the data, the initial rate doubles from $1.0×10^{-7}$ M/s at 15°C to $2.0×10^{-7}$ M/s at 25°C with constant [Pest], indicating k doubles. Option B is correct because it explains this via increased fraction exceeding Ea. Option A is incorrect as k actually increases, not decreases, with temperature for most reactions. When evaluating temperature effects, use the Arrhenius plot of ln(k) vs. 1/T for Ea. Always confirm order is unchanged before attributing rate changes to k.