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Chapter 12: Problem 59

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
a. Rate = k[\(\mathrm{CH}_{3}\mathrm{NC}\)] b. Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{NO}\)] c. Rate = k[\(\mathrm{O}_{3}\)] d. Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{O}\)]

Step by step solution

01

a. Rate law for \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) #

Since this reaction has only one reactant, \(\mathrm{CH}_{3} \mathrm{NC}(g)\), and its stoichiometric coefficient is 1, the rate law will be: Rate = k[\(\mathrm{CH}_{3}\mathrm{NC}\)]^1 Simplifying, we get: Rate = k[\(\mathrm{CH}_{3}\mathrm{NC}\)]
02

b. Rate law for \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) #

This reaction has two reactants, \(\mathrm{O}_{3}(g)\) and \(\mathrm{NO}(g)\), both with stoichiometric coefficients of 1. The rate law will be: Rate = k[\(\mathrm{O}_{3}\)]^1[\(\mathrm{NO}\)]^1 Simplifying, we get: Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{NO}\)]
03

c. Rate law for \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) #

Since this reaction has only one reactant, \(\mathrm{O}_{3}(g)\), and its stoichiometric coefficient is 1, the rate law will be: Rate = k[\(\mathrm{O}_{3}\)]^1 Simplifying, we get: Rate = k[\(\mathrm{O}_{3}\)]
04

d. Rate law for \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\) #

This reaction has two reactants, \(\mathrm{O}_{3}(g)\) and \(\mathrm{O}(g)\), both with stoichiometric coefficients of 1. The rate law will be: Rate = k[\(\mathrm{O}_{3}\)]^1[\(\mathrm{O}\)]^1 Simplifying, we get: Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{O}\)]

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Most popular questions from this chapter

The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter \(12 .\) What are these two procedures? Explain how each method is used to determine rate laws.

Would the slope of \(a \ln (k)\) versus \(1 / T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

The rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.
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