/*! 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 35 The following data were measured... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 35

The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):$ $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{BF}_{3}\right](M)} & {\left[\mathrm{NH}_{3}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

Short Answer

Expert verified
(a) Rate law: Rate = k[BF3]^{-1}[NH3] (b) Overall order of the reaction: 0 (c) Rate constant, k = 213 M鈦宦箂鈦宦� (d) Rate = 1.065 M/s when [BF3] = 0.100 M and [NH3] = 0.500 M

Step by step solution

01

Write the general form of the rate law

The general form of the rate law for the given reaction can be written as: Rate = k[BF3]^m[NH3]^n Where k is the rate constant, m and n are the orders of the reaction with respect to BF3 and NH3, respectively.
02

Analyze the effect of varying [BF3] on the initial rate

Compare experiments 3 and 5, where the concentration of NH3 is constant at 0.100 M. We can find the order of reaction with respect to BF3 by setting up the following ratio: \(\frac{\text{Rate}_{5}}{\text{Rate}_{3}} = \frac{k[BF3_{5}]^m[NH3_{5}]^n}{k[BF3_{3}]^m[NH3_{3}]^n}\) Substituting the given data into the equation: \(\frac{0.0596}{0.0682} = \frac{k(0.175)^m(0.100)^n}{k(0.200)^m(0.100)^n}\) Since the concentration of NH3 is constant, it can be canceled out: \(\frac{0.0596}{0.0682} = \frac{(0.175)^m}{(0.200)^m}\) Now, solve for the value of m: \(m = \log_{\frac{0.175}{0.200}}\left(\frac{0.0596}{0.0682}\right) = -1\) So, the order of reaction with respect to BF3 is -1.
03

Analyze the effect of varying [NH3] on the initial rate

Compare experiments 2 and 3, where the concentration of BF3 is constant at 0.250 M. We can find the order of reaction with respect to NH3 by setting up the following ratio: \(\frac{\text{Rate}_{3}}{\text{Rate}_{2}} = \frac{k[BF3_{3}]^m[NH3_{3}]^n}{k[BF3_{2}]^m[NH3_{2}]^n}\) Substituting the given data into the equation: \(\frac{0.0682}{0.1065} = \frac{k(0.200)^m(0.100)^n}{k(0.250)^m(0.125)^n}\) Since we found that m = -1, you can simplify the equation: \(\frac{0.0682}{0.1065} = \frac{(0.100)^n}{(0.125)^n}\) Now, solve for the value of n: \(n = \log_{\frac{0.100}{0.125}}\left(\frac{0.0682}{0.1065}\right) = 1\) So, the order of reaction with respect to NH3 is 1.
04

Determine the rate law, overall order and calculate the rate constant

We have found the order of reaction with respect to BF3 to be -1 and with respect to NH3 to be 1. Therefore, the rate law is: Rate = k[BF3]^{-1}[NH3] The overall order of reaction = (-1) + 1 = 0. Now, to find the rate constant (k), substitute any experiment's data into the rate law and solve for k: \(0.2130 = k(0.250)^{-1}(0.250)\) \(k \approx 213 \, \text{M}^{-1}\, \text{s}^{-1}\)
05

Calculate the rate for the given concentrations of reactants

Given, [BF3] = 0.100 M and [NH3] = 0.500 M. Substitute the values into the rate law equation and calculate the rate: Rate = \(213 \, \text{M}^{-1}\, \text{s}^{-1}(0.100)^{-1}(0.500)\) Rate = 1.065 M/s Summary of the answers: (a) Rate law: Rate = k[BF3]^{-1}[NH3] (b) Overall order of the reaction: 0 (c) Rate constant, k = 213 M鈦宦箂鈦宦� (d) Rate = 1.065 M/s when [BF3] = 0.100 M and [NH3] = 0.500 M

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!

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

When chemists are performing kinetics experiments, the general rule of thumb is to allow the reaction to proceed for 4 half-lives. (a) Explain how you would be able to tell that the reaction has proceeded for 4 half-lives. (b) Let us suppose a reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes 6 days to proceed for 4 half-lives and is first order in A. However, when your lab partner performs this reaction for the first time, he does not realize how long it takes, and he stops taking kinetic data, monitoring the loss of A, after only 2 hours. Your lab partner concludes the reaction is zero order in A based on the data. Sketch a graph of [A] versus time to convince your lab partner the two of you need to be in the lab for a few days to obtain the proper rate law for the reaction.

The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) & \longrightarrow \\ \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ was measured at several temperatures, and the following data were collected: $$ \begin{array}{ll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \boldsymbol{k}\left(\boldsymbol{M}^{-1} \mathrm{~s}^{-1}\right) \\ \hline 15 & 0.0521 \\ 25 & 0.101 \\ 35 & 0.184 \\ 45 & 0.332 \\ \hline \end{array} $$ Calculate the value of \(E_{a}\) by constructing an appropriate graph.

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

Indicate whether each statement is true or false. If it is false, rewrite it so that it is true. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

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. (b) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

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.