/*! 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} Problem 15 Consider the following hypotheti... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 15

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q) .\) A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval, in units of \(\mathrm{M} / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min}\), what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

Short Answer

Expert verified
(a) Moles of B at each time in the table: 0 min: 0, 10 min: 0.014, 20 min: 0.023, 30 min: 0.029, 40 min: 0.034. (b) Average rate of disappearance of A for each 10-minute interval in M/s: 0-10 min: \(2.33 \times 10^{-5}\) M/s, 10-20 min: \(1.50 \times 10^{-5}\) M/s, 20-30 min: \(1.00 \times 10^{-5}\) M/s, 30-40 min: \(8.33 \times 10^{-6}\) M/s. (c) Average rate of appearance of B between t=10 min and t=30 min in M/s: \(1.25 \times 10^{-5}\) M/s.

Step by step solution

01

(Calculate moles of B using stoichiometry)

Since the reaction has a 1:1 stoichiometry (1 mole of A produces 1 mole of B), we can determine the number of moles of B by subtracting the moles of A at each time point from the initial moles of A: Moles of B (0 min) = 0.065 - 0.065 = 0 Moles of B (10 min) = 0.065 - 0.051 = 0.014 Moles of B (20 min) = 0.065 - 0.042 = 0.023 Moles of B (30 min) = 0.065 - 0.036 = 0.029 Moles of B (40 min) = 0.065 - 0.031 = 0.034 #Step 2: Calculate the concentration of A and B at each time point#
02

(Convert moles to concentrations)

To compute the rate of disappearance of A and appearance of B in M/s, we first need to convert the number of moles at each time point to molar concentrations using the formula: concentration = moles/volume. The volume is given as 100.0 mL or 0.100 L. #Step 3: Calculate the average rate of disappearance of A for each 10-minute interval in M/s#
03

(Calculate average rate of A's disappearance)

To find the average rate of disappearance of A, we'll subtract the concentration of A at each time point from the concentration of A at the previous time point and divide the result by the time interval (10 minutes or 600 seconds): Rate of disappearance (0-10 min) = (0.065 - 0.051) / 600 = 2.33 x 10^(-5) M/s Rate of disappearance (10-20 min) = (0.051 - 0.042) / 600 = 1.50 x 10^(-5) M/s Rate of disappearance (20-30 min) = (0.042 - 0.036) / 600 = 1.00 x 10^(-5) M/s Rate of disappearance (30-40 min) = (0.036 - 0.031) / 600 = 8.33 x 10^(-6) M/s #Step 4: Calculate the average rate of appearance of B between t=10 min and t=30 min in M/s#
04

(Calculate average rate of B's appearance)

To find the average rate of appearance of B between t=10 min and t=30 min, we'll subtract the concentration of B at t=30 min from the concentration of B at t=10 min and divide the result by the time interval (20 minutes or 1200 seconds): Rate of appearance (10-30 min) = (0.029 - 0.014) / 1200 = 1.25 x 10^(-5) M/s The calculated values are: (a) moles of B at each time point, (b) average rate of disappearance of A for each 10-minute interval in M/s, and (c) average rate of appearance of B between t=10 min and t=30 min in M/s.

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Most popular questions from this chapter

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ?

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Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) : At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cl} \hline \text { Time }(\mathrm{min}) & \left.\mathrm{IC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M) \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?\) (b) What is the value of the rate constant?

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C}\), and the following data were obtained: $$ \begin{array}{ll} \hline \text { Time (s) } & \text { [CH }_{3} \text { NC] (M) } \\ \hline 0 & 0.0165 \\ 2,000 & 0.0110 \\ 5,000 & 0.00591 \\ 8,000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate theaverage rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Graph \(\left[\mathrm{CH}_{3} \mathrm{NC}\right]\) versus time, and determine the instantaneous rates in \(\mathrm{M} / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).
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