Chapter 13: Problem 111
The slowest step in a reaction mechanism is called the _______-_______ step.
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Chapter 13: Problem 111
The slowest step in a reaction mechanism is called the _______-_______ step.
These are the key concepts you need to understand to accurately answer the question.
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Consider the decomposition of ozone \(\left(\mathrm{O}_{3}\right)\) to oxygen \(\left(\mathrm{O}_{2}\right)\) \(2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g)\) The rate law for this reaction is: Rate \(=k\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right] .\) How is the rate of this reaction affected by the concentration of oxygen?
Consider the reaction: \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow\) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) Initial concentrations and rates for this reaction are given in the table below. $$\begin{array}{|c|c|c|c|} \hline {\text { Experiment }} & {\begin{array}{c} \text { Initial concentration } \\ \text { (mol/L) } \\ \text { [NO] } \end{array}} & \begin{array}{c} \text { Initial rate of } \\ \text { [ } \mathbf{H}_{2} \text { ] } \end{array} & \begin{array}{c} \text { formation of } \mathbf{N}_{2} \\ (\mathbf{m o l} / \mathbf{L} \text { min } \mathbf{)} \end{array} \\ \hline 1 & 0.0060 & 0.0010 & 1.8 \times 10^{-4} \\ 2 & 0.0060 & 0.0020 & 3.6 \times 10^{-4} \\ 3 & 0.0010 & 0.0060 & 0.30 \times 10^{-4} \\ 4 & 0.0020 & 0.0060 & 1.2 \times 10^{-4} \\ \hline \end{array}$$ (a) From the data given, determine the order for each of the reactants, \(\mathrm{NO}\) and \(\mathrm{H}_{2}\), show your reasoning, and write the overall rate law for the reaction. (b) Calculate the value of the rate constant, \(k\), for the reaction. Include units. (c) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of \(\mathrm{H}_{2}\) has been consumed. (d) The following sequence of elementary steps is a proposed mechanism for the reaction. I. \(\mathrm{NO}+\mathrm{NO} \rightarrow \mathrm{N}_{2} \mathrm{O}_{2}\) II. \(\quad \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2} \mathrm{O}\) III. \(\quad \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) Based on the data present, explain why the first step cannot be the rate determining step.
True or false? The orders in a rate law are equal to the balancing coefficients in the slowest elementary step in a mechanism.
Why are catalysts important to industrial chemical processes? Why are they important to biological chemical processes?
Indicate whether each reaction is endothermic or exothermic: (a) \(\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{4}+2 \mathrm{O}_{2} \Delta E_{\mathrm{rxn}}=+890 \mathrm{~kJ}\) (b) \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \Delta E_{\mathrm{rxn}}=-890 \mathrm{~kJ}\) (c) \(\mathrm{S}+\mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+\) Heat (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \Delta E_{\mathrm{rxn}}=-92 \mathrm{~kJ}\) (e) Heat \(+\mathrm{NH}_{4} \mathrm{NO}_{3} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (f) \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \Delta E_{\mathrm{rxn}}=-479 \mathrm{~kJ}\)
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