Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 15, Problem 15.46QP

(a)

Interpretation Introduction

Interpretation: The equilibrium constant Kc for the given reaction at a given temperature is 5×1012 . On the basis of this information given questions are to be answered.

Concept introduction: The equilibrium constant (Kc) is expressed as,

Kc=[C]c[D]d[A]a[B]b

To determine: The equilibrium constant (Kc2) for the given reaction at same temperature.

(a)

Expert Solution
Check Mark

Answer to Problem 15.46QP

Solution

The equilibrium constant (Kc2) for the given reaction at same temperature is 2.23×106_ .

Explanation of Solution

Explanation

The given reactions are,

2NO(g)+O2(g)2NO2(g)NO(g)+12O2(g)NO2(g)

The equilibrium constant Kc1 for the first above reaction and at constant temperature is 5×1012 .

For a simple reaction,

aA+bBcC+dD

The equilibrium constant (Kc) is expressed as,

Kc=[C]c[D]d[A]a[B]b

Where,

  • [C] is the molar concentration of C .
  • [D] is the molar concentration of D .
  • [B] is the molar concentration of B .
  • [A] is the molar concentration of A .
  • c is the molar coefficient of C .
  • d is the molar coefficient of D .
  • b is the molar coefficient of B .
  • a is the molar coefficient of A .

The expression for the equilibrium constant of first (Kc1) given reaction is,

Kc1=[NO2(g)]2[NO(g)]2[O2(g)]1 (1)

The expression for the equilibrium constant of second (Kc2) given reaction is,

Kc2=[NO2(g)]1[NO(g)]1[O2(g)]12 (2)

Where,

  • [NO(g)] is the molar concentration of NO(g) .
  • [NO2(g)] is the molar concentration of NO2(g) .
  • [O2(g)] is the molar concentration of O2(g) .

Divide equation (1) with (2).

Kc1Kc2=[NO2(g)]2[NO(g)]2[O2(g)]1[NO2(g)]1[NO(g)]1[O2(g)]12Kc1Kc2=[NO2(g)]2[NO(g)]2[O2(g)]1×[NO(g)]1[O2(g)]12[NO2(g)]1Kc1Kc2=[NO2(g)]21[NO(g)]21[O2(g)]112Kc1Kc2=[NO2(g)]1[NO(g)]1[O2(g)]12 (3)

Compare equation (3) with (2).

Kc1Kc2=Kc2(Kc2)2=Kc1Kc2=Kc1 (4)

Substitute the value of Kc1 in equation (4).

Kc2=5×1012=2.23×106

Thus, the equilibrium constant (Kc2) for the given reaction at same temperature is 2.23×106_ .

(b)

Interpretation Introduction

To determine: The equilibrium constant (Kc2') for the given reaction at same temperature.

(b)

Expert Solution
Check Mark

Answer to Problem 15.46QP

Solution

The equilibrium constant (Kc2') of the given reaction at same temperature is 2.0×10-13_ .

Explanation of Solution

Explanation

The given reactions are,

2NO(g)+O2(g)2NO2(g)NO(g)+12O2(g)NO2(g)

The equilibrium constant Kc1 for the first above reaction and at constant temperature is 5×1012 .

For a simple reaction,

aA+bBcC+dD

The equilibrium constant (Kc) is expressed as,

Kc=[C]c[D]d[A]a[B]b

Where,

  • [C] is the molar concentration of C .
  • [D] is the molar concentration of D .
  • [B] is the molar concentration of B .
  • [A] is the molar concentration of A .
  • c is the molar coefficient of C .
  • d is the molar coefficient of D .
  • b is the molar coefficient of B .
  • a is the molar coefficient of A .

The expression for the equilibrium constant of second (Kc2') given reaction is,

Kc2'=[NO(g)]2[O2(g)][NO3(g)]2 (5)

Where,

  • [NO(g)] is the molar concentration of NO(g) .
  • [NO2(g)] is the molar concentration of NO2(g) .
  • [O2(g)] is the molar concentration of O2(g) .

Multiply equation (1) with (5).

Kc1×Kc2'=[NO2(g)]2[NO(g)]2[O2(g)]1×[NO(g)]2[O2(g)][NO3(g)]2Kc1×Kc2'=1Kc2'=1Kc1 (6)

Substitute the value of Kc1 in equation (6).

Kc2'=15×1012Kc2'=2.0×1013

Thus, the equilibrium constant (Kc2') for the given reaction at same temperature is 2.0×10-13_ .

(c)

Interpretation Introduction

To determine: The equilibrium constant (Kc2'') for the given reaction at same temperature.

(c)

Expert Solution
Check Mark

Answer to Problem 15.46QP

Solution

The equilibrium constant (Kc2'') for the given reaction at same temperature is 4.4×10-7_ .

Explanation of Solution

Explanation

The given reactions are,

2NO(g)+O2(g)2NO2(g)NO(g)+12O2(g)NO2(g)

The equilibrium constant Kc1 for the first above reaction and at constant temperature is 5×1012 .

For a simple reaction,

aA+bBcC+dD

The equilibrium constant (Kc) is expressed as,

Kc=[C]c[D]d[A]a[B]b

Where,

  • [C] is the molar concentration of C .
  • [D] is the molar concentration of D .
  • [B] is the molar concentration of B .
  • [A] is the molar concentration of A .
  • c is the molar coefficient of C .
  • d is the molar coefficient of D .
  • b is the molar coefficient of B .
  • a is the molar coefficient of A .

The expression for the equilibrium constant of second (Kc2'') given reaction is,

Kc2''=[NO(g)]1[O2(g)]12[NO2(g)]1 (7)

Where,

  • [NO(g)] is the molar concentration of NO(g) .
  • [NO2(g)] is the molar concentration of NO2(g) .
  • [O2(g)] is the molar concentration of O2(g) .

Multiply equation (1) with (7).

Kc1×Kc2''=[NO2(g)]2[NO(g)]2[O2(g)]1×[NO(g)]1[O2(g)]12[NO2(g)]1Kc1×Kc2''=[NO2(g)]21[NO(g)]21[O2(g)]112Kc1×Kc2''=[NO2(g)]1[NO(g)]1[O2(g)]12 (8)

Compare equation (8) with (7).

Kc1×Kc2''=1Kc2(Kc2'')2=1Kc1Kc2''=1Kc1 (9)

Substitute the value of Kc1 in equation (9).

Kc2''=15×1012=0.2×1012=4.4×107

Thus, the equilibrium constant (Kc2'') for the given reaction at same temperature is 4.4×10-7_ .

Conclusion

  1. a. The equilibrium constant (Kc2) for the given reaction at same temperature is 2.23×106_ .
  2. b. The equilibrium constant (Kc2') of the given reaction at same temperature is 2.0×10-13_ .
  3. c. The equilibrium constant (Kc2'') for the given reaction at same temperature is 4.4×10-7_ .

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Chapter 15 Solutions

Chemistry

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