Chapter 14: Problem 78
A certain first-order reaction is 35.5 percent complete in 4.90 min at \(25^{\circ} \mathrm{C}\). What is its rate constant?
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Chapter 14: Problem 78
A certain first-order reaction is 35.5 percent complete in 4.90 min at \(25^{\circ} \mathrm{C}\). What is its rate constant?
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What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of the term "rate determining."
The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)
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The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q) &+\mathrm{H}_{2} \mathrm{O}(l) \\ \longrightarrow & \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ shows first-order characteristics-that is, rate \(=\) \(k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right]\) - even though this is a second- order reaction (first order in \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) and first order in \(\mathrm{H}_{2} \mathrm{O}\) ). Explain.
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