Chapter 16: Problem 15
Why are reaction orders not always equal to the coefficients in a chemical equation?
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These are the key concepts you need to understand to accurately answer the question.
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Chapter 16: Problem 15
Why are reaction orders not always equal to the coefficients in a chemical equation?
These are the key concepts you need to understand to accurately answer the question.
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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 role of an intermediate in a reaction mechanism.
Using chemical terminology, explain the purpose of food refrigeration.
Explain how to calculate a reaction rate from concentration-versus-time data.
Why is it necessary to divide by the coefficient in the balanced chemical equation when calculating a reaction rate? When can that step be omitted?
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