The concept being tested is rate laws for elementary reactions. In elementary steps, the rate law is written directly from the reactant coefficients because molecularity dictates the order. This termolecular step involves two NO and one O2, so $\text{Rate} = k [\mathrm{NO}]^2 [\mathrm{O_2}]$. The rate depends on the frequency of three-molecule collisions. Choice B, $\text{Rate} = k [\mathrm{NO_2}]^2$, is a distractor that wrongly uses products and ignores the elementary step's reactants. A transferable approach is to derive rate laws solely from elementary steps' reactant coefficients, not from overall stoichiometry.

This question is about elementary reactions and unimolecular decompositions. For elementary reactions, the rate law corresponds directly to the molecularity, with the order equaling the number of molecules involved. This unimolecular step means one N2O5 molecule decomposes, giving $\text{Rate} = k [\mathrm{N_2O_5}]$. No other reactants are involved, so it's first-order. Choice B, $\text{Rate} = k [\mathrm{NO_2}][\mathrm{NO_3}]$, is incorrect because it includes products instead of the reactant in the rate law. When dealing with elementary steps, always base the rate law on the reactants' coefficients to ensure accuracy.

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.

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.

This question tests the skill of elementary reactions. The rate law for an elementary step mirrors the molecularity, with exponents matching the number of reactant molecules involved in the collision. Here, one Br and one H2 molecule react, indicating a bimolecular process, so the rate is first-order in each. This is because elementary reactions proceed via a single step, and the rate depends on the frequency of those specific collisions. A tempting distractor is choice E, $\text{Rate} = k[\mathrm{Br}]^2[\mathrm{H_2}]$, which is wrong because of the misconception of doubling the coefficient for Br without basis in the step. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.

This question tests comprehension of elementary reactions and rate laws. For elementary steps, the rate law follows directly from molecularity, mirroring the reactant stoichiometry. The bimolecular reaction between NH3 and HCl gives $ \text{Rate} = k [\mathrm{NH_3}][\mathrm{HCl}] $. Each has a coefficient of one, so first-order for each. Choice A, $ \text{Rate} = k [\mathrm{NH_4Cl}] $, is incorrect because it uses the product instead of reactants in the rate law. Always rely on the elementary step's reactant coefficients to define the rate law, not the products or overall balance.

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.

This question evaluates understanding of elementary reactions in mechanisms. The rate law for an elementary step directly follows from its molecularity, as it represents a single collision event. Here, the bimolecular collision between Cl and O3 gives $\text{Rate} = k [\mathrm{Cl}][\mathrm{O_3}]$, with each reactant to the first power. The exponents are identical to the coefficients in the balanced elementary equation. Choice A, $\text{Rate} = k [\mathrm{Cl}]^2 [\mathrm{O_3}]$, is incorrect because it assumes a termolecular collision instead of the given bimolecular one. Focus on the given elementary step's stoichiometry to derive rate laws, ignoring overall reactions unless specified.

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.

This problem involves applying the principle of elementary reactions. For an elementary step, the rate law reflects the actual molecular collision, with each reactant's concentration raised to the power of its stoichiometric coefficient. This reaction shows one NO molecule colliding with one NO₂ molecule, giving rate = k[NO][NO₂]. Both reactants appear to the first power because exactly one of each participates in the elementary collision. Choice A incorrectly squares the NO concentration, perhaps misreading the equation or applying a different reaction's pattern. Remember that elementary steps provide a direct path from balanced equation to rate law.