Chapter 14: Problem 27
Define activation energy. What role does activation energy play in chemical kinetics?
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Chapter 14: Problem 27
Define activation energy. What role does activation energy play in chemical kinetics?
These are the key concepts you need to understand to accurately answer the question.
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The rate law for the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Which of these changes will change the value of \(k ?\) (a) The pressure of \(\mathrm{NO}_{2}\) is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled. (d) The temperature is decreased. (e) A catalyst is added to the container.
Determine the overall orders of the reactions to which these rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\); (b) rate \(=k ;\) (c) rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{\frac{1}{2}}\) (d) rate \(=\) \(k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)
The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ $$ \begin{array}{cc} {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.
The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q) &+\mathrm{H}_{2} \mathrm{O}(l) \\ \longrightarrow & \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ When methyl phosphate is heated in acid solution, it reacts with water: $$ \mathrm{CH}_{3} \mathrm{OPO}_{3} \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{3} \mathrm{PO}_{4} $$ If the reaction is carried out in water enriched with \({ }^{18} \mathrm{O}\) the oxygen- 18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the bond-breaking scheme in the reaction?
Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \underset{k_{-1}}{\stackrel{k_{1}}{\leftrightarrows}} \mathrm{ES} \quad \text { (fast equilibrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). ( Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)
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