Chapter 16: Problem 2
Explain the difference between a reaction rate and a rate law.
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Chapter 16: Problem 2
Explain the difference between a reaction rate and a rate law.
These are the key concepts you need to understand to accurately answer the question.
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Sketch a diagram showing how the potential energy changes with the progress of an endothermic reaction. Label the curve "Initial state, ". Final state," and "Transition state." Then, draw a second curve to show the change brought about by a catalyst.
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 \(2=1.54 \times 10^{-4}\)
What is the rate of the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{NOBr}$$ given that the bromine concentration decreased by \(5.3 \times 10^{-5} \mathrm{M}\) during an interval of 38 \(\mathrm{s} ?\)
Explain the effect that area has on reactions that occur on surfaces.
How do enzymes differ from other catalysts?
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