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.

This problem tests knowledge of elementary reactions. For elementary steps, the molecularity (number of molecules colliding) equals the sum of stoichiometric coefficients, and these coefficients become the exponents in the rate law. In H⁺(aq) + OH⁻(aq) → H₂O(l), one H⁺ ion collides with one OH⁻ ion, yielding rate = k[H⁺][OH⁻]. Choice C incorrectly uses the product concentration [H₂O], but rate laws for elementary steps depend only on reactant concentrations. The key strategy is to count the number of each reactant species in the elementary step and use those numbers as exponents.

This question requires applying the concept of elementary reactions. In an elementary step, the rate law exponents match exactly the stoichiometric coefficients of the reactants because the step represents a single molecular collision event. For $$ \text{Cl}(g) + \text{O}_3(g) \rightarrow \text{ClO}(g) + \text{O}_2(g) $$, one Cl atom collides with one $\text{O}_3$ molecule, giving $ \text{rate} = k[\text{Cl}][\text{O}_3] $. Choice E incorrectly shows $ [\text{Cl}]^2 $, suggesting two Cl atoms collide simultaneously, which contradicts the given elementary step. Remember that only for elementary steps can you directly translate coefficients to rate law exponents.

This problem requires understanding of elementary reactions. For an elementary step, the rate law follows directly from the stoichiometric coefficients of the reactants. Since two NO₂ molecules must collide in this elementary reaction, the rate law is rate = k[NO₂]². The coefficient 2 becomes the exponent 2 because this represents an actual bimolecular collision between two NO₂ molecules. Choice C incorrectly uses the products NO and O₂ instead of the reactant NO₂, confusing the direction of the reaction. Always use reactant concentrations with coefficients as exponents for elementary steps.

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.

This question involves applying knowledge of elementary reactions. In an elementary reaction, the rate law is determined by the molecularity - how many molecules must collide. This step requires two SO₂ molecules and one O₂ molecule to collide, giving rate = k[SO₂]²[O₂]. The exponents match the stoichiometric coefficients exactly because this represents the actual trimolecular collision event. Choice A incorrectly uses first-order dependence on SO₂, failing to account for the coefficient 2 in the elementary step. For elementary reactions, always use stoichiometric coefficients as exponents in the rate law.

This problem tests understanding of elementary reactions. For an elementary step, the rate law reflects the actual molecular collision event, with reactant concentrations raised to powers matching their coefficients. This reaction shows two H atoms colliding, so the rate law is rate = k[H]². The exponent 2 comes from needing two H atoms to collide simultaneously in this elementary step. Choice C incorrectly uses H₂ (the product) instead of H (the reactant), confusing products with reactants in the rate law. Remember that elementary steps allow direct translation from balanced equation to rate law using reactant coefficients.

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.

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.

This question tests understanding of elementary reactions. For an elementary step, the rate law is determined by the molecularity—the number of molecules that must collide for the reaction to occur. The elementary reaction Cl(g) + O₃(g) → ClO(g) + O₂(g) shows one Cl atom colliding with one O₃ molecule, giving a rate law of rate = k[Cl][O₃]. Choice C incorrectly uses products (ClO and O₂) in the rate law, but rate laws for elementary steps depend only on reactant concentrations. When dealing with elementary reactions, always write the rate law using only the reactants with their stoichiometric coefficients as exponents.