/*! 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 19 Consider the following hypotheti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 19

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\). 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 and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) 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: 0 min: 0 mol 10 min: 0.014 mol 20 min: 0.023 mol 30 min: 0.029 mol 40 min: 0.034 mol (b) Average rate of disappearance of A (in M/s) for each 10-min interval: 0-10 min: -2.33 x 10^-5 M/s 10-20 min: -1.5 x 10^-5 M/s 20-30 min: -1.0 x 10^-5 M/s 30-40 min: -8.33 x 10^-6 M/s (c) Average rate of appearance of B between t=0 min and t=30 min (in M/s): 1.61 x 10^-4 M/s

Step by step solution

01

(a) Calculate the number of moles of B at each time

Since the reaction is A 鈫� B, the number of moles of B at each time can be determined by subtracting the number of moles of A remaining from the initial number of moles of A (0.065 mol). At time 0 min: Moles of B = Initial moles of A - Moles of A = 0.065 - 0.065 = 0 mol At time 10 min: Moles of B = 0.065 - 0.051 = 0.014 mol At time 20 min: Moles of B = 0.065 - 0.042 = 0.023 mol At time 30 min: Moles of B = 0.065 - 0.036 = 0.029 mol At time 40 min: Moles of B = 0.065 - 0.031 = 0.034 mol
02

(b) Calculate the average rate of disappearance of A for each 10-min interval

To find the average rate of disappearance of A, calculate the change in concentration of A (鈭哰A]) divided by the change in time (鈭唗) for each time interval. Concentration is obtained by dividing moles by volume (in L). Note: 100.0 mL = 0.100 L At 0-10 min: Average rate = (0.051 - 0.065) mol / (0.100 L * 10 min * (1/60) h) = -0.014 mol / (1/6 h) = -0.084 M/h At 10-20 min: Average rate = (0.042 - 0.051) mol / (0.100 L * 10 min * (1/60) h) = -0.009 mol / (1/6 h) = -0.054 M/h At 20-30 min: Average rate = (0.036 - 0.042) mol / (0.100 L * 10 min * (1/60) h) = -0.006 mol / (1/6 h) = -0.036 M/h At 30-40 min: Average rate = (0.031 - 0.036) mol / (0.100 L * 10 min * (1/60) h) = -0.005 mol / (1/6 h) = -0.030 M/h Convert the average rates from M/h to M/s by multiplying by (1/3600) s/h: At 0-10 min: -0.084 M/h * (1/3600) s/h = -2.33 x 10^-5 M/s At 10-20 min: -0.054 M/h * (1/3600) s/h = -1.5 x 10^-5 M/s At 20-30 min: -0.036 M/h * (1/3600) s/h = -1.0 x 10^-5 M/s At 30-40 min: -0.030 M/h * (1/3600) s/h = -8.33 x 10^-6 M/s
03

(c) Average rate of appearance of B between t=0 min and t=30 min

To find the average rate of appearance of B, calculate the change in concentration of B (鈭哰B]) divided by the change in time (鈭唗) for the time interval 0-30 min. 鈭哰B] = (0.029 mol - 0.000 mol) / 0.100 L = 0.29 M 鈭唗 = 30 min * (1/60) h = 0.5 h Average rate of appearance of B = 鈭哰B] / 鈭唗 = 0.29 M / 0.5 h = 0.58 M/h Convert the average rate from M/h to M/s by multiplying by (1/3600) s/h: 0.58 M/h * (1/3600) s/h = 1.61 x 10^-4 M/s

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Moles and Molarity Calculations
Moles and molarity are key concepts in chemistry to quantify substances and their concentration in solutions. Here, molarity is defined as moles of solute per liter of solution. For instance, the initial solution has 0.065 mol of substance \( \mathrm{A} \) in 0.100 L, giving a molarity of 0.65 M.
Moles of substance \( \mathrm{B} \) can be calculated by subtracting the moles of \( \mathrm{A} \) remaining from the initial amount. For example, at 10 minutes, this would be 0.065 mol (initial) - 0.051 mol (remaining) = 0.014 mol of \( \mathrm{B} \).
By repeating this for all time intervals, you can track how \( \mathrm{B} \) forms over time, showing direct conversions from \( \mathrm{A} \) due to the simple one-to-one reaction. This approach helps understand how substances react and change concentration over time.
Rate of Reaction
The rate of reaction tells us how quickly reactants turn to products. It's commonly expressed as concentration change per unit time. To find the average rate of disappearance of \( \mathrm{A} \), you calculate the change in molarity of \( \mathrm{A} \) over each time period, then convert to seconds for better precision.
The calculation between 0 and 10 minutes:
  • Change in moles = 0.051 mol - 0.065 mol = -0.014 mol
  • Change in concentration = \(-0.014\) mol / 0.100 L = \(-0.140\) M
  • Rate = \(-0.140\) M / 10 min = \(-0.084\) M/h, converted to \(-2.33 \times 10^{-5}\) M/s
This method gives a clear picture of how fast reactants diminish and products form, crucial for understanding and controlling reactions.
Chemical Equations Balance
Balancing chemical equations ensures that you follow the conservation of mass, stating that matter cannot be created or destroyed. In our reaction \( \mathrm{A} \rightarrow \mathrm{B} \), each molecule of \( \mathrm{A} \) converts into one molecule of \( \mathrm{B} \).
This simple one-to-one balance means for every decrease in \( \mathrm{A} \), we see an increase of \( \mathrm{B} \) by the same amount. It's a perfect demonstration of stoichiometry鈥攁 foundational concept in chemistry.
Understanding this balance in reactions allows you to calculate how much of each reactant is needed or how much product can be expected, essential for practical applications like synthesizing compounds or predicting the outcome of chemical processes.
Phase Changes in Reactions
Reactions involving phase changes often involve understanding how the physical state of a substance can impact the reaction. While the task focuses on an aqueous phase, it's crucial to note how different states (solid, liquid, gas) can affect reaction processes.
In the context of our aqueous reaction, both \( \mathrm{A} \) and \( \mathrm{B} \) remain in the same phase throughout, as denoted by \((aq)\). This consistency simplifies calculations and observations as there's no energy absorbed or released due to phase change.
However, when phase changes are involved, energy changes can become significant, influencing reaction rates and equilibrium. An understanding of phase-related behavior enhances predictability in complex reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm} .\) (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min} .\) Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

(a) The gas-phase decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right), \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(300^{\circ} \mathrm{C}\) the half-life for this process is two and a half days. What is the rate constant at this temperature? (b) At \(400^{\circ} \mathrm{C}\) the rate constant is \(0.19 \mathrm{~min}^{-1}\). What is the half-life at this temperature?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (c) Which factor is most sensitive to changes in temperature-the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.