Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 12, Problem 97QRT

(a)

Interpretation Introduction

Interpretation:

For the formation of carbonylbromide from Carbon monoxide and Bromine, the value of KP has to be calculated.

Concept Introduction:

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration.  Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants.  Each partial pressure 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 reactionkfPAa=krPBa

On rearranging,

    PBbPAa=kfkr=Kp

Where,

    kf is the rate constant of the forward reaction.

    kr is the rate constant of the reverse reaction.

    Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

  PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT.  Thus, equilibrium constant Kp is given as

  Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

  R is gas constant,

  T is absolute temperature,

  Δn is difference in moles of products to reactants in gas phase.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information,

  Formation of carbonylbromide from Carbon monoxide and Bromine,

  CO(g)+Br2(g)COBr2(g)

Equilibrium temperature is 346K, the partial pressures of COBr2,CO,andBr2 are 0.12, 1.00, and 0.65atm respectively.

Calculate the Kp value for the given reaction

  KP=PCOBr2PCOPBr2=0.12(1.0)(0.65)=0.1846=0.18

Therefore, the Kp value for the given reaction is 0.18.

(b)

Interpretation Introduction

Interpretation:

In the formation of carbonylbromide from Carbon monoxide and Bromine, the partial pressure of Bromine is decreased into 0.50, now the partial pressure of all gases has to be calculated after equilibrium is re-established.

Concept Introduction:

Le Chatelier's principle:

It states that if a system in equilibrium gets disturbed due to modification of concentration, temperature, volume, and pressure, then it reset to counteract the effect of disturbance.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information,

  Formation of carbonylbromide from Carbon monoxide and Bromine,

  CO(g)+Br2(g)COBr2(g)

Equilibrium temperature is 346K, the partial pressures of COBr2,CO,andBr2 are 0.12, 1.00, and 0.50atm respectively.

Construct ICE table for the above reaction,

  CO(g)+Br2(g)COBr2(g)_Initialpressure(atm):1.000.500.12Changeinpress.(atm):+x+x-xEquilibriumpressure(atm):1.00+x0.50+x0.12-x

At equilibrium,

  KP=PCOBr2PCOPBr2=0.12-x(1.0+x)(0.50+x)=0.1846

Construct quadratic equation,

(0.1846)(1.00)(0.50) + (0.1846)(1.00)x + (0.1846)(0.50)x + (0.1846)x2 = 0.12-x(0.09231)+(0.1846+0.09231)x+(0.1846)x2=0.12-x0.1846x2+1.2769x-0.02769=0

Find the value of x,

  x=-b±b2-4ac2a=-1.2769±(1.2769)2-4(0.1846)(-0.0276)2(0.1846)=-1.2769±1.6305+0.02040.3692=-1.2769+1.28490.3692=0.0216

Calculate the partial pressure of all gases by using the value of x,

  PCO=1.00+0.02=1.02atmPH2=0.50+0.02=0.52atmPCOBr2=0.12-0.02=0.10atm

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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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