Chapter 10: Problem 145
The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3}
\mathrm{CSNH}_{2}\),
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 10: Problem 145
The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3}
\mathrm{CSNH}_{2}\),
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
A substance reacts according to first-order kinetics. The rate constant for the reaction is \(1 \times\) \(10^{-2} \mathrm{sec}^{1} .\) Its initial concentration is IM. Its initial rate is (a) \(2 \times 10^{2} \mathrm{Ms}^{-1}\) (b) \(1 \times 10^{2} \mathrm{Ms}^{-1}\) (c) \(1 \times 10^{-2} \mathrm{Ms}^{-1}\) (d) \(2 \times 10^{-2} \mathrm{Ms}^{-1}\)
Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)
The activation energy of a reaction is \(9 \mathrm{kcal} /\) mole. The increase in the rate constant when its temperature is raised from 295 to 300 is (a) \(14.9 \%\) (b) \(28.9 \%\) (c) \(78.9 \%\) (d) \(82.9 \%\)
The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at 295 \begin{tabular}{lll} \multicolumn{2}{c} { Table \(10.6\)} \\ \hline [CI_] & [NO] & Initial rate \(\left(\mathrm{molL}^{-4} \mathrm{~s}^{-1}\right)\) \\ \hline \(0.05\) & \(0.05\) & \(1 \times 10^{-3}\) \\ \(0.15\) & \(0.05\) & \(3 \times 10^{-3}\) \\ \(0.05\) & \(0.15\) & \(9 \times 10^{-3}\) \\ \hline \end{tabular} What is the rate law? (a) \(\mathrm{r}=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) (b) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{\mathrm{2}}[\mathrm{NO}]^{2}\) (c) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) (d) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}\)
The half-life of a chemical reaction at a particular concentration is \(50 \mathrm{~min}\), when the concentration of reactants is doubled, the half-life becomes \(100 \mathrm{~min}\). Find the order. (a) zero (b) first (c) second (d) third
What do you think about this solution?
We value your feedback to improve our textbook solutions.