/*! 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

When \(\mathrm{D}_{2}\) reacts with ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) in the presence of a finely divided catalyst, ethane with two deuteriums, \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\), is formed. (Deuterium, \(\mathrm{D}\), is an isotope of hydrogen of mass 2.) Very little ethane forms in which two deuteriums are bound to one carbon (for example, \(\left.\mathrm{CH}_{3}-\mathrm{CHD}_{2}\right)\). Use the sequence of steps involved in the reaction to explain why this is so.

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Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M}\), while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?
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