Elementary Reactions - AP Chemistry

In the gas phase, the following step is explicitly stated to be an elementary reaction: $2\,\text{HI}(g)\rightarrow \text{H}_2(g)+\text{I}_2(g)$ Which rate law is consistent with this step being elementary?
Answer: $\text{rate}=k[\text{HI}]^2$ This question tests understanding of elementary reactions. For an elementary step, the rate law comes directly from the stoichiometry of the reactants—the coefficients become the exponents. Since two HI molecules must collide for this reaction to occur, the rate law is rate = k[HI]². Choice B incorrectly uses the products [H₂][I₂] instead of the reactants, confusing what determines the forward reaction rate. The key principle is: elementary steps allow you to write the rate law by inspection using reactant coefficients.
A solution-phase reaction step is stated to be an elementary process: $\text{S}_2\text{O}_8^{2-}(aq)+\text{I}^-(aq)\rightarrow \text{SO}_4^{2-}(aq)+\text{SO}_4^{\bullet-}(aq)+\text{I}^{\bullet}(aq)$ Which rate law is consistent with an elementary step?
Answer: $\text{rate}=k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$ This question tests understanding of elementary reactions. For an elementary step, the rate law is determined by the molecularity—the number of reactant particles that must collide. The elementary reaction S₂O₈²⁻(aq) + I⁻(aq) → SO₄²⁻(aq) + SO₄•⁻(aq) + I•(aq) shows one S₂O₈²⁻ ion colliding with one I⁻ ion, giving rate = k[S₂O₈²⁻][I⁻]. Choice E incorrectly uses products (SO₄²⁻ and I•) in the rate law, but elementary step rate laws depend only on reactant concentrations. When writing rate laws for elementary reactions, use only the reactants with their stoichiometric coefficients as exponents.
A chemist claims the following single step in a mechanism is elementary: $\mathrm{2NO(g) + Br_2(g) \rightarrow 2NOBr(g)}$ Assuming the step is elementary, which rate law expression is consistent with molecularity for this collision?
Answer: $\text{Rate}=k[\mathrm{NO}]^2[\mathrm{Br_2}]$ This question assesses knowledge of elementary reactions and their rate laws. In elementary reactions, the rate law mirrors the molecularity, meaning the exponents in the rate law match the coefficients of the reactants in the balanced equation. This step is termolecular, involving two NO molecules and one Br2 molecule, so the rate law is $\text{Rate} = k [\mathrm{NO}]^2 [\mathrm{Br_2}]$ . The reaction occurs in a single collision event, making the rate dependent on the concentrations raised to the power of their stoichiometric coefficients. Choice C, $\text{Rate} = k [\mathrm{NOBr}]^2$ , is a common distractor but wrong because it uses products instead of reactants in the rate law. Always verify that the rate law for elementary steps is based solely on reactant concentrations and their coefficients.
In a gas-phase reaction vessel, the following elementary step is proposed: $2\,\text{NO}(g)+\text{O}_2(g)\rightarrow 2\,\text{NO}_2(g)$ Based on the definition of an elementary step (molecularity equals the stoichiometric coefficients of the reactants in the collision), which rate law is consistent with this step?
Answer: $\text{rate}=k[\text{NO}]^2[\text{O}_2]$ This question tests understanding of elementary reactions. For an elementary step, the rate law is determined directly by the stoichiometric coefficients of the reactants in the balanced equation—this relationship exists because elementary steps represent actual molecular collisions. In the given step $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$ , we have 2 molecules of NO and 1 molecule of O $_2$ colliding, so the rate law must be $\text{rate} = k[\text{NO}]^2[\text{O}_2]$ . Choice B incorrectly uses $[\text{NO}]^1$ instead of $[\text{NO}]^2$ , which would represent a collision of only one NO molecule rather than two. For elementary reactions, always use the coefficients from the balanced equation as the exponents in the rate law.
A researcher proposes the following elementary gas-phase step: $\text{CO}(g)+\text{Cl}_2(g)\rightarrow \text{COCl}_2(g)$ Which rate law must apply if the step is elementary?
Answer: $\text{rate}=k[\text{CO}][\text{Cl}_2]$ This question examines understanding of elementary reactions. Elementary steps represent single collision events, and the rate law exponents equal the stoichiometric coefficients of the reactants. For $\text{CO}(g) + \text{Cl}_2(g) \rightarrow \text{COCl}_2(g)$ , one CO molecule collides with one Cl $_2$ molecule, giving $\text{rate} = k[\text{CO}][\text{Cl}_2]$ . Choice A incorrectly shows $[\text{CO}]^2$ , which would mean two CO molecules collide with one Cl $_2$ molecule—this doesn't match the given elementary step. Always verify that your rate law exponents match the coefficients in the elementary step equation.
In a gas-phase kinetics study, the following reaction step is explicitly stated to be an elementary step: $2\,\mathrm{NO}(g)+\mathrm{O_2}(g)\rightarrow 2\,\mathrm{NO_2}(g)$ Based on the definition of an elementary step (molecularity equals the stoichiometric coefficients of reactants in that step), which rate law is correct for this step?
Answer: $\text{Rate}=k[\mathrm{NO}]^2[\mathrm{O_2}]$ This question tests the skill of elementary reactions. For an elementary reaction, the rate law is determined directly from the balanced equation of that step, where the order with respect to each reactant equals its stoichiometric coefficient. In this case, the reaction involves two NO molecules and one O2 molecule colliding, making it termolecular, so the rate law incorporates [NO]^2 and [O2]^1. This direct relationship holds because elementary steps represent single collision events, and the molecularity defines the reaction order. A tempting distractor is choice D, Rate = k[NO2]^2, which is incorrect due to the misconception of basing the rate law on products instead of reactants. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
In a combustion model, the following elementary step is given: $\text{H}(g)+\text{O}_2(g)\rightarrow \text{HO}_2(g)$ Which rate law is consistent with this step being elementary?
Answer: $\text{rate}=k[\text{H}][\text{O}_2]$ This problem focuses on elementary reactions. In elementary steps, the rate law is determined by the molecularity—the number of molecules that must collide for the reaction to occur. For $\text{H}(g) + \text{O}_2(g) \rightarrow \text{HO}_2(g)$ , one H atom collides with one O₂ molecule, so $\text{rate} = k[\text{H}][\text{O}_2]$ . Choice A incorrectly uses $[\text{H}]^2$ , suggesting two H atoms collide with O₂, which contradicts the stoichiometry of the elementary step. The fundamental principle is that elementary steps allow direct translation from balanced equation coefficients to rate law exponents.
A gas-phase process includes the following step, which is explicitly stated to be an elementary termolecular reaction: $2\,\mathrm{SO_2}(g)+\mathrm{O_2}(g)\rightarrow 2\,\mathrm{SO_3}(g)$ Which rate law is consistent with this elementary step?
Answer: $\text{Rate}=k[\mathrm{SO_2}]^2[\mathrm{O_2}]$ This question tests the skill of elementary reactions. For elementary reactions, especially termolecular ones, the rate law uses coefficients as orders for all reactants involved. This step has two SO2 and one O2, so $rate = k[\mathrm{SO_2}]^2[\mathrm{O_2}]$ , reflecting the rare three-molecule collision. The molecularity directly informs the rate law because it's a single mechanistic step. A tempting distractor is choice D, $Rate = k[\mathrm{SO_3}]^2$ , which is wrong due to the misconception of basing the rate on products instead of reactants. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
In an aqueous solution, the following step is stated to be an elementary reaction: $\text{Br}^-(aq)+\text{H}_2\text{O}_2(aq)+\text{H}^+(aq)\rightarrow \text{HOBr}(aq)+\text{H}_2\text{O}(l)$ Using the definition of an elementary step, which rate law is correct?
Answer: $\text{rate}=k[\text{Br}^-][\text{H}_2\text{O}_2][\text{H}^+]$ This problem tests knowledge of elementary reactions. For an elementary step, the rate law is constructed by raising each reactant concentration to the power of its stoichiometric coefficient. This elementary reaction shows one Br⁻, one H₂O₂, and one H⁺ ion colliding simultaneously, so the rate law is rate = k[Br⁻][H₂O₂][H⁺]. All three species must be included because they all participate in the elementary collision event. Choice A omits H⁺, incorrectly treating it like a catalyst rather than a reactant in the elementary step. The key strategy is that elementary steps allow you to write the rate law directly from the balanced equation without experimental data.
A kinetics model includes the following single step, explicitly identified as elementary: $\mathrm{CH_3Br}(g)+\mathrm{OH^-}(aq)\rightarrow \mathrm{CH_3OH}(aq)+\mathrm{Br^-}(aq)$ Assuming the step is elementary as stated, which rate law is correct?
Answer: $\text{Rate}=k[\mathrm{CH_3Br}][\mathrm{OH^-}]$ This question tests the skill of elementary reactions. Elementary steps have rate laws where the order is the stoichiometric coefficient for each reactant. Here, one CH3Br and one OH- react, making it bimolecular with $\text{Rate} = k[\mathrm{CH_3Br}][\mathrm{OH^-}]$ . This direct link comes from the collision theory, where rate depends on reactant concentrations in the single step. A tempting distractor is choice C, $\text{Rate} = k[\mathrm{CH_3OH}][\mathrm{Br^-}]$ , which is incorrect because of the misconception of incorporating products into the rate law. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
In a radical chain process, the following step is given and explicitly stated to be elementary: $\mathrm{Cl}(g)+\mathrm{O_3}(g)\rightarrow \mathrm{ClO}(g)+\mathrm{O_2}(g)$ What is the correct rate law expression for this elementary step?
Answer: $\text{Rate}=k[\mathrm{Cl}][\mathrm{O_3}]$ This question tests the skill of elementary reactions. The rate law of an elementary reaction is based on its molecularity, using the stoichiometric coefficients of reactants as orders. In this step, one Cl and one O3 collide, making it bimolecular with $\text{Rate} = k[\mathrm{Cl}][\mathrm{O_3}]$ . This is due to the rate being proportional to the concentration of each species involved in the elementary collision. A tempting distractor is choice A, $\text{Rate} = k[\mathrm{ClO}][\mathrm{O_2}]$ , which is incorrect because of the misconception of incorporating products into the rate law. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
A textbook lists the following as an elementary step in the gas phase: $\mathrm{H(g) + HBr(g) \rightarrow H_2(g) + Br(g)}$ Which rate law is correct for this elementary step?
Answer: $\text{Rate}=k[\mathrm{H}][\mathrm{HBr}]$ This problem examines rate laws derived from elementary reactions. Elementary steps have rate laws that match their molecularity, using coefficients as exponents for reactants. The bimolecular collision here between H and HBr yields $\text{Rate} = k [\mathrm{H}][\mathrm{HBr}]$ . Each reactant appears to the first power, reflecting the single collision. Choice E, $\text{Rate} = k [\mathrm{H_2}][\mathrm{Br}]$ , is wrong as it mistakenly uses products rather than reactants. Remember, only elementary reactions permit direct use of stoichiometric coefficients for rate law orders.
A researcher proposes the following single reaction step and specifies that it is elementary: $\mathrm{NO_2}(g)+\mathrm{CO}(g)\rightarrow \mathrm{NO}(g)+\mathrm{CO_2}(g)$ Which rate law is consistent with an elementary step?
Answer: $\text{Rate}=k[\mathrm{NO_2}][\mathrm{CO}]$ This question tests the skill of elementary reactions. Elementary reactions have rate laws that directly reflect the molecularity, with orders equal to the coefficients of the reactants. Here, one NO2 and one CO react, so the rate is $k[\mathrm{NO_2}][\mathrm{CO}]$ , indicating a bimolecular process. This stems from the rate depending on the collision frequency between those specific molecules in a single step. A tempting distractor is choice C, $\text{Rate} = k[\mathrm{NO}][\mathrm{CO_2}]$ , which is wrong due to the misconception of basing the rate on products rather than reactants. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
In a kinetics model, the following is given as a single elementary step: $\text{CH}_3\text{Br}(aq)+\text{OH}^-(aq)\rightarrow \text{CH}_3\text{OH}(aq)+\text{Br}^-(aq)$ Which rate law follows from the definition of an elementary reaction step?
Answer: $\text{rate}=k[\text{CH}_3\text{Br}][\text{OH}^-]$ This question tests understanding of elementary reactions. In an elementary reaction, the rate law is written directly from the molecularity of the step—the number of reactant molecules that must collide. The elementary step CH₃Br(aq) + OH⁻(aq) → CH₃OH(aq) + Br⁻(aq) shows one CH₃Br molecule colliding with one OH⁻ ion, so the rate law must be rate = k[CH₃Br][OH⁻]. Choice C incorrectly uses products (CH₃OH and Br⁻) in the rate law, but elementary reactions have rate laws based only on reactant concentrations. For elementary reactions, always use the stoichiometric coefficients of the reactants as the exponents in the rate law.
In a reaction chamber, the following elementary step is proposed: $\text{Br}(g)+\text{H}_2(g)\rightarrow \text{HBr}(g)+\text{H}(g)$ Which rate law correctly represents the rate of this elementary step?
Answer: $\text{rate}=k[\text{Br}][\text{H}_2]$ This question examines elementary reactions. Elementary steps represent actual molecular collisions, so the rate law exponents match the stoichiometric coefficients of the reactants. For $\text{Br}(g) + \text{H}_2(g) \rightarrow \text{HBr}(g) + \text{H}(g)$ , one Br atom collides with one H₂ molecule, yielding $\text{rate} = k[\text{Br}][\text{H}_2]$ . Choice A incorrectly shows $[\text{Br}]^2$ , suggesting two Br atoms collide with H₂, which contradicts the given elementary step. Remember that for elementary reactions, coefficients in the balanced equation become exponents in the rate law.
A researcher states that the following is a single elementary step: $\text{N}_2\text{O}_5(g)+\text{NO}(g)\rightarrow 3\text{NO}_2(g)$ Which rate law is required by the definition of an elementary reaction step?
Answer: $\text{rate}=k[\text{N}_2\text{O}_5][\text{NO}]$ This question tests understanding of elementary reactions. For an elementary step, the rate law is written directly from the stoichiometric coefficients of the reactants in the balanced equation. The elementary reaction N₂O₅(g) + NO(g) → 3NO₂(g) shows one molecule each of N₂O₅ and NO colliding, giving rate = k[N₂O₅][NO]. Choice B incorrectly cubes the NO concentration, perhaps confusing the coefficient of the product NO₂ (which is 3) with the reactant NO (which has a coefficient of 1). Remember that for elementary reactions, only the reactant coefficients determine the rate law exponents.
A single step in a proposed mechanism is explicitly stated to be an elementary reaction in aqueous solution: $\text{S}_2\text{O}_8^{2-}(aq)+2\,\text{I}^-(aq)\rightarrow 2\,\text{SO}_4^{2-}(aq)+\text{I}_2(aq)$ Which rate law follows directly from this step being elementary?
Answer: $\text{rate}=k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]^2$ This question tests understanding of elementary reactions. In an elementary step, the rate law directly reflects the molecularity—how many particles must collide simultaneously. This reaction shows one S₂O₈²⁻ ion colliding with two I⁻ ions, so the rate law must be rate = k[S₂O₈²⁻][I⁻]². Choice A incorrectly uses [I⁻]¹ instead of [I⁻]², missing that the coefficient 2 in front of I⁻ becomes the exponent in the rate law. For elementary reactions only, the stoichiometric coefficients become the reaction orders.
A chemist identifies the following as a single elementary step in the gas phase: $\text{NO}(g)+\text{NO}_3(g)\rightarrow 2\text{NO}_2(g)$ Which rate law must be true for this elementary step?
Answer: $\text{rate}=k[\text{NO}][\text{NO}_3]$ This question tests understanding of elementary reactions. In an elementary reaction, the rate law reflects exactly how many molecules must collide for the reaction to occur. The elementary step NO(g) + NO₃(g) → 2NO₂(g) shows one NO molecule colliding with one NO₃ molecule, so the rate law must be rate = k[NO][NO₃]. Choice C incorrectly uses the product NO₂ in the rate law, but rate laws for elementary steps are based only on reactant concentrations and their stoichiometric coefficients. For elementary reactions, write the rate law directly from the balanced equation using reactant coefficients as exponents.
In the atmosphere, the following step is specified to be an elementary reaction step: $\mathrm{O}(g)+\mathrm{O_3}(g)\rightarrow 2\,\mathrm{O_2}(g)$ Using the definition of an elementary step, which rate law is correct?
Answer: $\text{Rate}=k[\mathrm{O}][\mathrm{O_3}]$ This question tests the skill of elementary reactions. In elementary reactions, the rate law is written using the reactants' stoichiometric coefficients as the orders in the rate expression. This step shows one O and one O3 reacting to form products, confirming it's bimolecular with rate proportional to $[\mathrm{O}][\mathrm{O_3}]$ . The reason is that the molecularity, or number of molecules colliding, directly translates to the reaction order for elementary processes. A tempting distractor is choice B, $\text{Rate} = k[\mathrm{O_2}]^2$ , which is incorrect due to the misconception of using products instead of reactants for the rate law. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.
A kinetics model includes the following single step, explicitly labeled as an elementary reaction step: $\text{O}(g)+\text{O}_2(g)\rightarrow \text{O}_3(g)$ Which rate law matches the elementary-step definition?
Answer: $\text{rate}=k[\text{O}][\text{O}_2]$ This question tests the concept of elementary reactions. In an elementary reaction, the rate law is determined by the molecularity - how many molecules must collide for the reaction to occur. This step shows one O atom colliding with one O₂ molecule, so the rate law is rate = k[O][O₂]. Each species appears to the first power because one of each participates in the elementary collision event. Choice D incorrectly includes [O]², perhaps confusing this with a different reaction or assuming higher orders are always needed. For elementary steps, the stoichiometric coefficients directly become the exponents in the rate law.

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