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2 NOBr(g) → 2 NO(g) + Br2(g) The equation above represents an elementary step in a chemical reactionKinetics Chemistry Question

Question

2 NOBr(g) → 2 NO(g) + Br2(g)

The equation above represents an elementary step in a chemical reaction. Which of the following is the correct expression for the rate law of the elementary step?

A.

Rate = k[NOBr]^(1/2)

B.

Rate = k[NOBr]

C.

Rate = k[NOBr]^2

✓ Correct
D.

Rate = k[NO]^2[Br2]

💡 Solution & Explanation

STEPS:

1. Understand the definition of an "elementary step":
Unlike overall chemical reactions (where the rate law must be determined experimentally), an elementary step is a reaction that occurs in a single molecular collision exactly as written. Because of this, the reaction's rate law is directly determined by the stoichiometry of its reactants.

2. Identify the reactants and their stoichiometric coefficients:
In the given equation:
2 NOBr(g)2 NO(g)+Br2(g)2\ \text{NOBr}(g) \rightarrow 2\ \text{NO}(g) + \text{Br}_2(g)
The only reactant is nitrosyl bromide, NOBr\text{NOBr}, and its stoichiometric coefficient in this step is 2. This means two molecules of NOBr\text{NOBr} must collide simultaneously for this elementary step to occur (a bimolecular process).

3. Formulate the rate law expression:
For any elementary step of the form aAproductsa\text{A} \rightarrow \text{products}, the rate law is written with the reactant raised to the power of its stoichiometric coefficient:
Rate=k[A]a\text{Rate} = k[\text{A}]^a
Substituting NOBr\text{NOBr} as reactant A\text{A} and 22 as the coefficient aa yields:
Rate=k[NOBr]2\mathbf{\text{Rate} = k[\text{NOBr}]^2}
This identifies Option C as the correct rate law.

*

WHY_OTHERS_WRONG:

  • Option A is incorrect: This represents a half-order rate law (Rate=k[NOBr]1/2\text{Rate} = k[\text{NOBr}]^{1/2}). Elementary steps cannot have fractional reaction orders because you cannot have a fraction of a molecule colliding in a single molecular event.
  • Option B is incorrect: This represents a first-order rate law (Rate=k[NOBr]\text{Rate} = k[\text{NOBr}]). This would be the rate law only if the elementary step were unimolecular, meaning it involved the decomposition of a single NOBr\text{NOBr} molecule (coefficient of 1) rather than a collision between two molecules.
  • Option D is incorrect: This expression (Rate=k[NO]2[Br2]\text{Rate} = k[\text{NO}]^2[\text{Br}_2]) is written using the concentrations of the *products* rather than the reactants. A reaction rate law describes how the forward rate depends on the concentrations of the starting reactants that must collide, not the products that have already been formed.
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