CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
bartleby

Videos

Question
Book Icon
Chapter 16, Problem 16.35P

(a)

Interpretation Introduction

Interpretation:

The order with respect to each reactant has to be found.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(a)

Expert Solution
Check Mark

Explanation of Solution

The reaction rate of the chemical reaction:

Givenreaction: A (g) + B(g)+C(g)D(g)Reaction Rate = k [A]m[B]n[C]p,where 'm, n and p' are orders of the reactants.

Let’s find the order of reactant [A] by comparing first and second experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments1and2,rate1=[A]1m[B]1n[C]1p, rate1 = 6.25 ×10-3mol/L.srate2 = k[A]2m[B]2n[C]2p, rate2 = 1.25×102 mol/L.srate2rate1=k [A]2m[B]2n[C]2pk [A]1m[B]1n[C]1p1.25×102mol/L.s6.25 ×10-3 mol/L.s=(0.10)m(0.05)n(0.01)p(0.05)m(0.05)n(0.01)p 2 = (2)m(2)1(2)mm = 1 

Let’s find the order of reactant [B] by comparing second and third experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments2and3,rate2=[A]2m[B]2n[C]2p, rate2 = 1.25 ×10-2mol/L.srate3 = k[A]3m[B]3n[C]3p, rate3 = 5.00×102 mol/L.srate3rate2=k [A]3m[B]3n[C]3pk [A]1m[B]1n[C]1p5.00×102 mol/L.s1.25 ×10-2 mol/L.s=(0.100)m(0.100)n(0.01)p(0.100)m(0.05)n(0.01)p 4 = (2)n(2)2(2)nn = 2 

Let’s find the order of reactant [C] by comparing first and fourth experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments1and4,rate1=[A]1m[B]1n[C]1p, rate1 = 6.25 ×10-3mol/L.srate4 = k[A]4m[B]4n[C]4p, rate4 = 6.25×103 mol/L.srate4rate1=k [A]4m[B]4n[C]4pk [A]1m[B]1n[C]1p6.25×103 mol/L.s6.25 ×10-3mol/L.s=(0.05)m(0.05)n(0.02)p(0.05)m(0.05)n(0.01)p 1 = (2)p(1)2(2)plog(1.0)=plog(2.0)log(1)=0p = 0 

In order to figure out the reaction equation the order of the reactants needed, which is calculated by comparing any two experiments where the concentration of two reactants are constant and another varies, and in vice-versa. Hence, Rate equation is

Reaction rate = k [A]1[B]2[C]0.

The respective order of reactant (C) is ZERO. Hence, the reaction rate becomes,

Reaction rate = k [A]1[B]2

(b)

Interpretation Introduction

Interpretation:

The rate law of the given reaction has to be written.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(b)

Expert Solution
Check Mark

Explanation of Solution

The reaction rate of the chemical reaction:

Givenreaction: A (g) + B(g)+C(g)D(g)Reaction Rate = k [A]m[B]n[C]p,where 'm, n and p' are orders of the reactants.

Let’s find the order of reactant [A] by comparing first and second experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments1and2,rate1=[A]1m[B]1n[C]1p, rate1 = 6.25 ×10-3mol/L.srate2 = k[A]2m[B]2n[C]2p, rate2 = 1.25×102 mol/L.srate2rate1=k [A]2m[B]2n[C]2pk [A]1m[B]1n[C]1p1.25×102mol/L.s6.25 ×10-3 mol/L.s=(0.10)m(0.05)n(0.01)p(0.05)m(0.05)n(0.01)p 2 = (2)m(2)1(2)mm = 1 

Let’s find the order of reactant [B] by comparing second and third experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments2and3,rate2=[A]2m[B]2n[C]2p, rate2 = 1.25 ×10-2mol/L.srate3 = k[A]3m[B]3n[C]3p, rate3 = 5.00×102 mol/L.srate3rate2=k [A]3m[B]3n[C]3pk [A]1m[B]1n[C]1p5.00×102 mol/L.s1.25 ×10-2 mol/L.s=(0.100)m(0.100)n(0.01)p(0.100)m(0.05)n(0.01)p 4 = (2)n(2)2(2)nn = 2 

Let’s find the order of reactant [C] by comparing first and fourth experiments as follows,

Findorderofthereaction:_ Comparingfirsttwoexperiments1and4,rate1=[A]1m[B]1n[C]1p, rate1 = 6.25 ×10-3mol/L.srate4 = k[A]4m[B]4n[C]4p, rate4 = 6.25×103 mol/L.srate4rate1=k [A]4m[B]4n[C]4pk [A]1m[B]1n[C]1p6.25×103 mol/L.s6.25 ×10-3mol/L.s=(0.05)m(0.05)n(0.02)p(0.05)m(0.05)n(0.01)p 1 = (2)p(1)2(2)plog(1.0)=plog(2.0)log(1)=0p = 0 

In order to figure out the reaction equation the order of the reactants needed, which is calculated by comparing any two experiments where the concentration of two reactants are constant and another varies, and in vice-versa. Hence, Rate equation is

Reaction rate = k [A]1[B]2[C]0.

The respective order of reactant (C) is ZERO. Hence, the reaction rate becomes,

Reaction rate = k [A]1[B]2

(c)

Interpretation Introduction

Interpretation:

The value of ‘k’ of the given reaction has to be found.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(c)

Expert Solution
Check Mark

Explanation of Solution

The value of rate constant:

Reaction rate = k [A]1[B]2

Consider, any one of the given experimental value,

Experiment 1: [A] = 0.05   ;  [B] = 0.05  ; [C]  =0.05;Reactionrate = 6.25×103

Substituting the above values into the reaction rate, the rate constant ‘k’ value is obtained as follows,

Reaction rate = k [A]1[B]2.6.25×10-3 mol/L.s = k (0.05)1(0.05)2k = 6.25×10-3 mol/L.s (0.05mol/L)1(0.05mol/L)2k=50.0 ×103 L2/mol2.s.

The rate constant value is obtained as shown above. By substituting the any one of the concentrations of reactants and the initial rate into the reaction equation obtained at first.

Hence, the value of rate constant is 50.0 ×103 L2/mol2.s.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 16 Solutions

CHEMISTRY >CUSTOM<

Ch. 16.4 - Substance X (black) changes to substance Y (red)...Ch. 16.4 - Prob. 16.6BFPCh. 16.4 - Prob. 16.7AFPCh. 16.4 - Prob. 16.7BFPCh. 16.5 - Prob. 16.8AFPCh. 16.5 - Prob. 16.8BFPCh. 16.5 - Prob. 16.9AFPCh. 16.5 - Prob. 16.9BFPCh. 16.6 - The mechanism below is proposed for the...Ch. 16.6 - Prob. 16.10BFPCh. 16.6 - Prob. 16.11AFPCh. 16.6 - Prob. 16.11BFPCh. 16.7 - Prob. B16.1PCh. 16.7 - Aircraft in the stratosphere release NO, which...Ch. 16.7 - Prob. B16.3PCh. 16 - Prob. 16.1PCh. 16 - Prob. 16.2PCh. 16 - A reaction is carried out with water as the...Ch. 16 - Prob. 16.4PCh. 16 - Prob. 16.5PCh. 16 - Prob. 16.6PCh. 16 - Prob. 16.7PCh. 16 - Prob. 16.8PCh. 16 - Prob. 16.9PCh. 16 - Prob. 16.10PCh. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - By what factor does the rate in Problem 16.27...Ch. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Give the overall reaction order that corresponds...Ch. 16 - Phosgene is a toxic gas prepared by the reaction...Ch. 16 - How are integrated rate laws used to determine...Ch. 16 - Define the half-life of a reaction. Explain on the...Ch. 16 - For the simple decomposition reaction AB(g) ⟶A(g)...Ch. 16 - For the reaction in Problem 16.41, what is [AB]...Ch. 16 - The first-order rate constant for the reaction A...Ch. 16 - The molecular scenes below represent the...Ch. 16 - In a first-order decomposition reaction, 50.0% of...Ch. 16 - A decomposition reaction has a rate constant of...Ch. 16 - In a study of ammonia production, an industrial...Ch. 16 - Prob. 16.48PCh. 16 - Prob. 16.49PCh. 16 - Prob. 16.50PCh. 16 - Prob. 16.51PCh. 16 - Prob. 16.52PCh. 16 - Prob. 16.53PCh. 16 - Prob. 16.54PCh. 16 - Prob. 16.55PCh. 16 - Assuming the activation energies are equal, which...Ch. 16 - For the reaction A(g) + B(g) ⟶AB(g), how many...Ch. 16 - Prob. 16.58PCh. 16 - Prob. 16.59PCh. 16 - Prob. 16.60PCh. 16 - The rate constant of a reaction is 4.7×10−3 s−1 at...Ch. 16 - The rate constant of a reaction is 4.50×10−5...Ch. 16 - Prob. 16.63PCh. 16 - For the reaction A2 + B2 → 2AB, Ea(fwd) = 125...Ch. 16 - Prob. 16.65PCh. 16 - Prob. 16.66PCh. 16 - Prob. 16.67PCh. 16 - Explain why the coefficients of an elementary step...Ch. 16 - Is it possible for more than one mechanism to be...Ch. 16 - What is the difference between a reaction...Ch. 16 - Why is a bimolecular step more reasonable...Ch. 16 - Prob. 16.72PCh. 16 - If a fast step precedes a slow step in a two-step...Ch. 16 - Prob. 16.74PCh. 16 - Prob. 16.75PCh. 16 - In a study of nitrosyl halides, a chemist proposes...Ch. 16 - Prob. 16.77PCh. 16 - Consider the reaction . Does the gold catalyst...Ch. 16 - Does a catalyst increase reaction rate by the same...Ch. 16 - In a classroom demonstration, hydrogen gas and...Ch. 16 - Prob. 16.81PCh. 16 - Prob. 16.82PCh. 16 - Prob. 16.83PCh. 16 - Consider the following reaction energy...Ch. 16 - Prob. 16.85PCh. 16 - Prob. 16.86PCh. 16 - A slightly bruised apple will rot extensively in...Ch. 16 - Prob. 16.88PCh. 16 - Prob. 16.89PCh. 16 - Prob. 16.90PCh. 16 - Prob. 16.91PCh. 16 - The citric acid cycle is the central reaction...Ch. 16 - Prob. 16.93PCh. 16 - Prob. 16.94PCh. 16 - Prob. 16.95PCh. 16 - Prob. 16.96PCh. 16 - For the reaction A(g) + B(g) ⟶ AB(g), the rate is...Ch. 16 - The acid-catalyzed hydrolysis of sucrose occurs by...Ch. 16 - At body temperature (37°C), the rate constant of...Ch. 16 - Is each of these statements true? If not, explain...Ch. 16 - For the decomposition of gaseous dinitrogen...Ch. 16 - Prob. 16.102PCh. 16 - Suggest an experimental method for measuring the...Ch. 16 - Prob. 16.104PCh. 16 - Many drugs decompose in blood by a first-order...Ch. 16 - Prob. 16.106PCh. 16 - Prob. 16.107PCh. 16 - Prob. 16.108PCh. 16 - Prob. 16.109PCh. 16 - Prob. 16.110PCh. 16 - Prob. 16.111PCh. 16 - Prob. 16.112PCh. 16 - Prob. 16.113PCh. 16 - Prob. 16.114PCh. 16 - Prob. 16.115PCh. 16 - Prob. 16.116PCh. 16 - Prob. 16.117PCh. 16 - The growth of Pseudomonas bacteria is modeled as a...Ch. 16 - Prob. 16.119PCh. 16 - Prob. 16.120PCh. 16 - Prob. 16.121PCh. 16 - Prob. 16.122PCh. 16 - Prob. 16.123PCh. 16 - Prob. 16.124PCh. 16 - Human liver enzymes catalyze the degradation of...Ch. 16 - Prob. 16.126PCh. 16 - Prob. 16.127P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Kinetics: Chemistry's Demolition Derby - Crash Course Chemistry #32; Author: Crash Course;https://www.youtube.com/watch?v=7qOFtL3VEBc;License: Standard YouTube License, CC-BY