/*! 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 3 Consider the following hypotheti... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 3

Consider the following hypothetical reaction: $$\mathrm{X}(g) \longrightarrow \mathrm{Y}(g)$$ A 200.0-mL flask is filled with \(0.120\) moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{X} & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline\end{array}$$ (a) Make a similar table for the appearance of Y. (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two twenty-minute intervals. (c) What is the average rate of appearance of \(\mathrm{Y}\) between the 20 - and 60-minute intervals?

Short Answer

Expert verified
In this problem, we were asked to find the appearance of product Y and the average reaction rates of the disappearance of reactant X and appearance of product Y during different time intervals. We used stoichiometry and the concept of reaction rates to create a table for moles of Y at different time intervals. We then calculated the molar concentrations of X and Y and determined the average rates of disappearance and appearance for X and Y, respectively. The average disappearance rates of X were around -0.085/1200 M/s and -0.090/1200 M/s, and the average rate of appearance of Y between the 20- and 60-minute intervals was approximately 0.160/2400 M/s.

Step by step solution

01

Finding the moles of Y at different time intervals

To find the moles of Y at different intervals, we can use the stoichiometry of the reaction. At the start, there are no moles of Y, so we can subtract the moles of X at a given time from the initial moles of X to find the moles of Y at that time. Here's the table of moles of Y at different time intervals: $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{Y} & 0 & 0.017 & 0.035 & 0.049 & 0.054 \\\ \hline\end{array}$$ Now, let's move on to part (b), which is to calculate the average rate of disappearance of X in M/s for the first two twenty-minute intervals.
02

Finding the molar concentrations of X at different intervals

To find the molar concentrations of X, we divide the moles of X by the volume of the flask (200.0 mL or 0.200 L): $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { [X] } (\mathrm{M}) & 0.600 & 0.515 & 0.425 & 0.355 & 0.330 \\\ \hline\end{array}$$
03

Calculating the average rate of disappearance of X for the first two intervals

To find the average rate of disappearance of X, we'll calculate the change in concentration of X over the time interval for the first two twenty-minute intervals: Interval 1: \((0.515 - 0.600) \,\mathrm{M} / (20\,\text{min} * 60\,\text{s/min}) = -0.085\text{/}1200\,\mathrm{M/s}\) Interval 2: \((0.425 - 0.515) \,\mathrm{M} / (20\,\text{min} * 60\,\text{s/min}) = -0.090\text{/}1200\,\mathrm{M/s}\) Now, we'll move on to part (c), which is to find the average rate of appearance of Y between the 20- and 60- minute intervals.
04

Finding the molar concentrations of Y at different intervals

To find the molar concentrations of Y, we divide the moles of Y by the volume of the flask (200.0 mL or 0.200 L): $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { [Y] } (\mathrm{M}) & 0 & 0.085 & 0.175 & 0.245 & 0.270 \\\ \hline\end{array}$$
05

Calculating the average rate of appearance of Y between the 20 and 60-minute intervals

To find the average rate of appearance of Y, we'll calculate the change in concentration of Y over the time interval between 20 and 60 minutes: $$\frac{(0.245 - 0.085)\,\mathrm{M}}{(60-20)\,\text{min} * 60\,\text{s/min}} = \frac{0.160}{2400}\,\mathrm{M/s}$$ In conclusion, (a) We created a table for the appearance of Y, with the moles of Y at different time intervals. (b) The average disappearance of X in the first two twenty-minute intervals is around \(-0.085/1200\,\mathrm{M/s}\) and \(-0.090/1200\,\mathrm{M/s}\), respectively. (c) The average rate of appearance of Y between the 20- and 60-minute intervals is approximately \(\frac{0.160}{2400}\,\mathrm{M/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the principle of conservation of mass, where the number of atoms for each element is conserved in a reaction.
In the given reaction, \( \mathrm{X}(g) \rightarrow \mathrm{Y}(g) \), stoichiometry tells us how the moles of X relate to the moles of Y. Since one mole of X yields one mole of Y (1:1 ratio), for every mole of X consumed, one mole of Y is produced.
This stoichiometric relationship helps us track the changes in moles over time. By subtracting the moles of Y from the initial moles of X, we can calculate the amount of X remaining at any point. This straightforward relationship simplifies the tracking of reactants and products in this reaction.
Concentration Changes
Concentration changes over time play a crucial role in understanding reaction rates. Concentration is usually expressed in molarity (M), which is moles of solute per liter of solution.
For our exercise, the concentration of reactant X changes as it reacts to form Y. Initially, \( 0.120 \) moles of X are placed in a \( 200.0 \) mL flask. The concentration \[ \text{[X]} = \frac{\text{moles of X}}{\text{Volume of flask (L)}} \] allows us to calculate how fast X is being consumed.
The data provided shows decreasing concentrations of X as follows: \[ 0.600 \ \text{M}, \, 0.515 \ \text{M}, \, 0.425 \ \text{M}, \, 0.355 \ \text{M}, \, 0.330 \ \text{M}. \]
This change is crucial in determining the rate of disappearance of X. By calculating the change in concentration over time, we can deduce the speed at which X is disappearing, revealing how the rate of reaction might vary at different intervals.
Moles Calculation
Calculating moles accurately is critical in chemical reactions. In our exercise, moles of X are initially given, and we are asked to determine the formation of Y over time.
To find the moles of Y, we take advantage of the stoichiometry. We start with \( 0.120 \) moles of X and track how it decreases over time. At each time point, the decrease in moles of X is equal to the increase in moles of Y.
For instance, at 20 minutes, the moles of X reduced from \( 0.120 \) to \( 0.103 \). Thus, \( 0.017 \) moles of Y have formed: \[ \text{Initial moles of X} - \text{moles of X at 20 min} = \text{moles of Y at 20 min} \] Further calculations use the same approach.
  • At 40 min: \( 0.120 - 0.085 = 0.035 \) moles of Y
  • At 60 min: \( 0.120 - 0.071 = 0.049 \) moles of Y
  • At 80 min: \( 0.120 - 0.066 = 0.054 \) moles of Y
Understanding the shift in mole numbers helps us determine overall changes and reactant-product transitions throughout the reaction.

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Most popular questions from this chapter

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

Hypofluorous acid, HOF, is extremely unstable at room temperature. The following data apply to the decomposition of HOF to \(\mathrm{HF}\) and \(\mathrm{O}_{2}\) gases at a certain temperature. $$\begin{array}{cl}\hline \text { Time (min) } & \text { [HOF] } \\\\\hline 1.00 & 0.607 \\\2.00 & 0.223 \\\3.00 & 0.0821 \\\4.00 & 0.0302 \\\5.00 & 0.0111 \\\\\hline \end{array}$$ (a) By plotting the data, show that the reaction is first-order. (b) From the graph, determine \(k\). (c) Using \(k\), find the time it takes to decrease the concentration to \(0.100 \mathrm{M}\) (d) Calculate the rate of the reaction when [HOF] \(=0.0500 M\).

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), to sulfur dioxide and chlorine gases is a first-order reaction. $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At a certain temperature, the half-life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(7.5 \times 10^{2} \mathrm{~min}\). Consider a sealed flask with \(122.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (a) How long will it take to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in the sealed flask to \(45.0 \mathrm{~g}\) ? (b) If the decomposition is stopped after \(29.0 \mathrm{~h}\), what volume of \(\mathrm{Cl}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm is produced?

Two mechanisms are proposed for the reaction $$\begin{array}{cl}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \\ \text { Mechanism 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} & \text { (fast) } \\ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \\ \text { Mechanism 2: } \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} & \text { (slow) } \end{array}$$ Show that each of these mechanisms is consistent with the observed rate law: rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{O}_{2}\right]\).

The decomposition of ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), is a first-order reaction. It is found that it takes 212 s to decompose \(0.00839 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{6}\) to \(0.00768 \mathrm{M}\). (a) What is the rate constant for the reaction? (b) What is the rate of decomposition (in \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}\) ) when \(\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]=\) \(0.00422 \mathrm{M} ?\) (c) How long (in minutes) will it take to decompose \(\mathrm{C}_{2} \mathrm{H}_{6}\) so that \(27 \%\) remains? (d) What percentage of \(\mathrm{C}_{2} \mathrm{H}_{6}\) is decomposed after \(22 \mathrm{~min}\) ?
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