Chapter 10: Problem 145
The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3}
\mathrm{CSNH}_{2}\)
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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.
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A gaseous compound decomposes on heating as per the following equation: \(\mathrm{A}(\mathrm{g}) \longrightarrow B(\mathrm{~g})+2 \mathrm{C}(\mathrm{g}) .\) After 5 minutes and 20 seconds, the pressure increases by \(96 \mathrm{~mm} \mathrm{Hg}\). If the rate constant for this first order reaction is \(5.2 \times 10^{-4} \mathrm{~s}^{-1}\), the initial pressure of \(\mathrm{A}\) is (a) \(226 \mathrm{~mm} \mathrm{Hg}\) (b) \(37.6 \mathrm{~mm} \mathrm{Hg}\) (c) \(616 \mathrm{~mm} \mathrm{Hg}\) (d) \(313 \mathrm{~mm} \mathrm{Hg}\)
A certain reaction proceeds in a sequence of three elementary steps with the rate constants \(\mathrm{k}_{1}, \mathrm{k}_{2}\) and \(\mathrm{k}_{3} .\) If the observed rate constant of the expressed as \(\mathrm{k}\) (obs) \(=\mathrm{k}(\mathrm{obs})=\left[\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\right]^{1 / 2} \mathrm{k}_{3}\), the observed energy of activa- tion of the reaction is (a) \(\frac{\mathrm{E}_{3}+\mathrm{E}_{1}}{2}\) (b) \(\frac{1}{2}\left[\frac{E_{1}}{E_{2}}\right]+E_{3}\) (c) \(\mathrm{E}_{3}+\frac{1}{2}\left[\mathrm{E}_{1}-\mathrm{E}_{2}\right]\) (d) \(\mathrm{E}_{3}\left[\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right]^{12}\)
For the following reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) The initial pressure was \(\mathrm{P}_{0}\) while pressure after time ' \(\mathrm{t}\) ' was \(\mathrm{P}_{1}\). The rate constant \(\mathrm{k}\) will be (a) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{t}}}\) (b) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{\mathrm{1}}}\) (c) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{1}}\) (d) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-2 \mathrm{P}_{\mathrm{t}}}\)
The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)
A graph plotted between concentration of reactant, consumed at any time \((x)\) and time ' \(\mathrm{t}\) ' is found to be a straight line passing through the origin. The reaction is of (a) first-order (b) zero-order (c) third-order (d) second-order
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