Chapter 16: Problem 2
Explain the difference between a reaction rate and a rate law.
Chapter 16: Problem 2
Explain the difference between a reaction rate and a rate law.
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Get started for freeCalculate the rate of a reaction, knowing that a graph of the concentration of a product versus time had a slope of \(3.6 \times 10^{-6} \mathrm{M} / \mathrm{s}\) . The product had a coefficient of 2 .
What are enzymes, and what common features do they all share?
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 \(\mathrm{s} ?\)
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 \(2=1.54 \times 10^{-4}\)
Why are reaction orders not always equal to the coefficients in a chemical equation?
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