Chapter 12: Problem 4
How does temperature affect k, the rate constant? Explain.
Chapter 12: Problem 4
How does temperature affect k, the rate constant? Explain.
All the tools & learning materials you need for study success - in one app.
Get started for freeEnzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.
Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} M\) It was found that after each reaction had progressed for \(3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]\) . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.
The reaction $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-} .\) In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus 1\(/ T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{K}\) and \(y\) -intercept of 33.5 . Assume \(k\) has units of \(\mathrm{s}^{-1}\) a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\) . c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\) .
The reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ exhibits the rate law $$ \text {Rate} =k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right] $$ Which of the following mechanisms is consistent with this rate law? $$ \begin{array}{l}{\text { a. } \mathrm{NO}+\mathrm{O}_{2} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}} \\ {\mathrm{O}+\mathrm{NO} \longrightarrow \mathrm{NO}_{2}} \\ {\text { b. } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3}} \\ {\mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2}}\end{array} $$ $$ \begin{array}{l}{\text { c. } 2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}} \\ {\mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}} \\ {\mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow 2 \mathrm{NO}_{2}} \\ {\text { d. } 2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2}} \\\ {\mathrm{N}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}} \\\ {\mathrm{O}+\mathrm{NO} \longrightarrow \mathrm{NO}_{2}}\end{array} $$
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.
What do you think about this solution?
We value your feedback to improve our textbook solutions.