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Chapter 16: Problem 23

In the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{NOBr}(g)$$ doubling the \(\mathrm{Br}_{2}\) concentration doubles the rate, but doubling the NO concentration quadruples the rate. Write the rate law.

Short Answer

Expert verified
The rate law for the given reaction is Rate = k [NO]^2 [Br_2]^1.

Step by step solution

01

Identify the Reaction

The reaction under consideration is 2NO(g) + Br_{2}(g) -> 2NOBr(g). Here, Nitrogen Oxide (NO) and Bromine (Br_2) are the reactants, and Nitrosyl Bromide (NOBr) is the product.
02

Analyze the Reaction Rates

We're told that when the concentration of Br_2 is doubled, the rate of the reaction doubles. This suggests the rate of reaction is directly proportional to the concentration of Br_2. That is, the reaction is first order with respect to Br_2. Similarly, when the concentration of NO is doubled, the rate of the reaction quadruples. This suggests the rate of reaction is proportional to the square of the concentration of NO, and hence the reaction is second order with respect to NO.
03

Write the Rate Law

Knowing the order of reaction with respect to the reactants allows us to write the rate law. The rate law for a reaction is written as: Rate = k [A]^x [B]^y, where A and B are the reactants, x and y are the orders of reaction, and k is the reaction rate constant. So, in this case, the rate law will be: Rate = k [NO]^2 [Br_2]^1.

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Most popular questions from this chapter

Explain why the names activated complex and transition state are suitable for describing the highest energy point on a reaction's route from reactant to product.

The graphing calculator can run a program that can tell you the order of a chemical reaction, provided you indicate the reactant concentrations and reaction rates for two experiments involving the same reaction. Go to Appendix C. If you are using a TI-83 Plus, you can download the program RXNORDER and run the application as directed. If you are using another calculator, your teacher will provide you with key-strokes and data sets to use. At the prompts, enter the reactant concentrations and reaction rates. Run the program as needed to find the order of the following reactions. (All rates are given in M/s.) a. \(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) \(\mathrm{N}_{2} \mathrm{O}_{5} :\) conc. \(1=0.025 \mathrm{M} ;\) conc. \(2=0.040 \mathrm{M}\) rate \(1=8.1 \times 10^{-5} ;\) rate \(2=1.3 \times 10^{-4}\) b. \(2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) \(\mathrm{NO}_{2} : \mathrm{conc.} 1=0.040 \mathrm{M} ; \mathrm{conc} .2=0.080 \mathrm{M}\) rate \(1=0.0030 ;\) rate \(2=0.012\) c. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)\) \(\mathrm{H}_{2} \mathrm{O}_{2} :\) conc. \(1=0.522 \mathrm{M} ;\) conc. \(2=0.887 \mathrm{M}\) rate \(1=1.90 \times 10^{-4} ;\) rate \(2=3.23 \times 10^{-4}\) d. \(2 \mathrm{NOBr}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) NOBr: conc. \(1=1.27 \times 10^{-4} \mathrm{M} ;\) conc. \(2=\) \(4.04 \times 10^{-4} \mathrm{M}\) rate \(1=6.26 \times 10^{-5} ;\) rate \(2=6.33 \times 10^{-4}\) e. \(2 \mathrm{HI}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) HI: conc. \(1=4.18 \times 10^{-4} \mathrm{M} ;\) conc. \(2=\) \(8.36 \times 10^{-4} \mathrm{M}\) rate \(1=3.86 \times 10^{-5} ;\) rate \(2=1.54 \times 10^{-4}\)

Explain the difference between a reaction rate and a rate law.

Explain the effect that area has on reactions that occur on surfaces.

Explain what a catalyst is and how it works.
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