Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
Question
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Chapter 12, Problem 120QRT

(a)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of NOBr is given, for this reaction overall stoichiometric equation has to be given.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

  Step1:NO(g)+Br2(g)NOBr2(g)Step2:NOBr2(g)+NO(g)2NOBr(g)

Write the stoichiometric equation by cancelling species that are present in both sides and add up the remaining species,

  NO(g)+Br2(g)NOBr2(g)NOBr2(g)+NO(g)2NOBr(g)_2NO2(g)+Br2(g)2NOBr(g)

(b)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of NOBr is given, for this reaction the equilibrium constant in terms of the forward and reverse rate constant has to be given.

Concept Introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

  aAbB

  Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

  [B]b[A]a=kfkr=Kc

Where,

    kf is the rate constant of the forward reaction.

    kr is the rate constant of the reverse reaction.

    Kc is the equilibrium constant.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

  Step1:NO(g)+Br2(g)NOBr2(g)Step2:NOBr2(g)+NO(g)2NOBr(g)

The rate of forward reaction is equals to the rate of reverse reaction at equilibrium.

Write the equilibrium constant expression for the first step,

  rateforward,step1=ratereverse,step1k1[NO][Br2]=k-1[NOBr2]therefore,KC,step1=k1k-1=[NOBr2][NO][Br2]

Write the equilibrium constant expression for the second step,

  rateforward,step2=ratereverse,step2k2[NOBr2][NO]=k-2[NOBr]2therefore,KC,step1=k2k-2=[NOBr]2[NOBr2][NO]

(c)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of NOBr is given, for this reaction the overall stoichiometric equilibrium constant in terms of the forward and reverse rate constant has to be given.

Concept Introduction:

Refer part (b)

(c)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

  Step1:NO(g)+Br2(g)NOBr2(g)Step2:NOBr2(g)+NO(g)2NOBr(g)

Write the stoichiometric equation, by cancelling species that are present in both sides and add up the remaining species,

  NO(g)+Br2(g)NOBr2(g)NOBr2(g)+NO(g)2NOBr(g)_2NO2(g)+Br2(g)2NOBr(g)

Write the overall stoichiometric equilibrium constant expression,

  KC,overall=[NOBr]2[NO]2[Br2]=([NOBr2][BO][Br2])([NOBr]2[NO][NOBr2])=(k1k-1)(k2k-2)=k1k2k-1k-2

(d)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of NOBr is given, for this reaction the overall stoichiometric equilibrium constant in terms of the forward and reverse rate constant was found this has to be checked with the given statement.

Concept Introduction:

Refer part (b)

(d)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

  Step1:NO(g)+Br2(g)NOBr2(g)Step2:NOBr2(g)+NO(g)2NOBr(g)

Write the stoichiometric equation, by cancelling species that are present in both sides and add up the remaining species,

  NO(g)+Br2(g)NOBr2(g)NOBr2(g)+NO(g)2NOBr(g)_2NO2(g)+Br2(g)2NOBr(g)

Write the overall stoichiometric equilibrium constant expression,

  KC,overall=[NOBr]2[NO]2[Br2]=([NOBr2][BO][Br2])([NOBr]2[NO][NOBr2])=(k1k-1)(k2k-2)=k1k2k-1k-2

Therefore, the result obtained in part (c) confirms the statement that the equilibrium constant can be obtained by taking the product of the rate constants for all forward steps (here, k1k2) and dividing by the product of the rate constants for all reverse steps (here, k–1k–2).

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Chapter 12 Solutions

Chemistry: The Molecular Science

Ch. 12.5 - For the equilibrium 2 SO2(g) + O2(g) 2 SO3(g) Kc...Ch. 12.5 - Prob. 12.7CECh. 12.5 - Prob. 12.6PSPCh. 12.5 - Prob. 12.7PSPCh. 12.6 - Prob. 12.8CECh. 12.6 - Prob. 12.9ECh. 12.6 - Prob. 12.10CECh. 12.6 - Prob. 12.8PSPCh. 12.7 - For the ammonia synthesis reaction ⇌ Does the...Ch. 12.8 - Prob. 12.13CECh. 12 - Prob. 1QRTCh. 12 - Prob. 2QRTCh. 12 - Prob. 3QRTCh. 12 - Decomposition of ammonium dichromate is shown in...Ch. 12 - For the equilibrium reaction in Question 4, write...Ch. 12 - Indicate whether each statement below is true or...Ch. 12 - Prob. 7QRTCh. 12 - Prob. 8QRTCh. 12 - Prob. 9QRTCh. 12 - Prob. 10QRTCh. 12 - The atmosphere consists of about 80% N2 and 20%...Ch. 12 - Prob. 12QRTCh. 12 - Prob. 13QRTCh. 12 - Prob. 14QRTCh. 12 - Prob. 15QRTCh. 12 - Prob. 16QRTCh. 12 - Prob. 17QRTCh. 12 - Prob. 18QRTCh. 12 - Prob. 19QRTCh. 12 - Prob. 20QRTCh. 12 - Prob. 21QRTCh. 12 - Prob. 22QRTCh. 12 - Prob. 23QRTCh. 12 - Prob. 24QRTCh. 12 - Prob. 25QRTCh. 12 - Prob. 26QRTCh. 12 - Prob. 27QRTCh. 12 - Prob. 28QRTCh. 12 - Prob. 29QRTCh. 12 - Prob. 30QRTCh. 12 - Given these data at a certain temperature,...Ch. 12 - The vapor pressure of water at 80. C is 0.467 atm....Ch. 12 - Prob. 33QRTCh. 12 - Prob. 34QRTCh. 12 - Prob. 35QRTCh. 12 - Prob. 36QRTCh. 12 - Carbon dioxide reacts with carbon to give carbon...Ch. 12 - Prob. 38QRTCh. 12 - Prob. 39QRTCh. 12 - Prob. 40QRTCh. 12 - Nitrosyl chloride, NOC1, decomposes to NO and Cl2...Ch. 12 - Suppose 0.086 mol Br2 is placed in a 1.26-L flask....Ch. 12 - Prob. 43QRTCh. 12 - Prob. 44QRTCh. 12 - Prob. 45QRTCh. 12 - Using the data of Table 12.1, predict which of...Ch. 12 - Prob. 47QRTCh. 12 - The equilibrium constants for dissolving silver...Ch. 12 - Prob. 49QRTCh. 12 - Prob. 50QRTCh. 12 - At room temperature, the equilibrium constant Kc...Ch. 12 - Prob. 52QRTCh. 12 - Consider the equilibrium N2(g)+O2(g)2NO(g) At 2300...Ch. 12 - The equilibrium constant, Kc, for the reaction...Ch. 12 - Prob. 55QRTCh. 12 - Prob. 56QRTCh. 12 - Prob. 57QRTCh. 12 - At 503 K the equilibrium constant Kc for the...Ch. 12 - Prob. 59QRTCh. 12 - Prob. 60QRTCh. 12 - Prob. 61QRTCh. 12 - Prob. 62QRTCh. 12 - Prob. 63QRTCh. 12 - Prob. 64QRTCh. 12 - Prob. 65QRTCh. 12 - Prob. 66QRTCh. 12 - Prob. 67QRTCh. 12 - Hydrogen, bromine, and HBr in the gas phase are in...Ch. 12 - Prob. 69QRTCh. 12 - Prob. 70QRTCh. 12 - Prob. 71QRTCh. 12 - Prob. 72QRTCh. 12 - Prob. 73QRTCh. 12 - Prob. 74QRTCh. 12 - Consider the system 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) +...Ch. 12 - Prob. 76QRTCh. 12 - Predict whether the equilibrium for the...Ch. 12 - Prob. 78QRTCh. 12 - Prob. 79QRTCh. 12 - Prob. 80QRTCh. 12 - Prob. 81QRTCh. 12 - Prob. 82QRTCh. 12 - Prob. 83QRTCh. 12 - Prob. 84QRTCh. 12 - Prob. 85QRTCh. 12 - Prob. 86QRTCh. 12 - Prob. 87QRTCh. 12 - Consider the decomposition of ammonium hydrogen...Ch. 12 - Prob. 89QRTCh. 12 - Prob. 90QRTCh. 12 - Prob. 91QRTCh. 12 - Prob. 92QRTCh. 12 - Prob. 93QRTCh. 12 - Prob. 94QRTCh. 12 - Prob. 95QRTCh. 12 - Prob. 96QRTCh. 12 - Prob. 97QRTCh. 12 - Prob. 98QRTCh. 12 - Prob. 99QRTCh. 12 - Prob. 100QRTCh. 12 - Two molecules of A react to form one molecule of...Ch. 12 - Prob. 102QRTCh. 12 - In Table 12.1 (←Sec. 12-3a) the equilibrium...Ch. 12 - Prob. 104QRTCh. 12 - Prob. 105QRTCh. 12 - Prob. 106QRTCh. 12 - Prob. 107QRTCh. 12 - Which of the diagrams for Questions 107 and 108...Ch. 12 - Draw a nanoscale (particulate) level diagram for...Ch. 12 - The diagram represents an equilibrium mixture for...Ch. 12 - The equilibrium constant, Kc, is 1.05 at 350 K for...Ch. 12 - For the reaction in Question 111, which diagram...Ch. 12 - Prob. 113QRTCh. 12 - Prob. 114QRTCh. 12 - Prob. 115QRTCh. 12 - For the equilibrium...Ch. 12 - Prob. 117QRTCh. 12 - Prob. 119QRTCh. 12 - Prob. 120QRTCh. 12 - When a mixture of hydrogen and bromine is...Ch. 12 - Prob. 122QRTCh. 12 - Prob. 123QRTCh. 12 - Prob. 124QRTCh. 12 - Prob. 125QRTCh. 12 - Prob. 12.ACPCh. 12 - Prob. 12.BCPCh. 12 - Prob. 12.CCPCh. 12 - Prob. 12.DCPCh. 12 - Prob. 12.ECPCh. 12 - Prob. 12.FCP
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