Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 16, Problem 40PS

For each of the following reactions, predict whether the equilibrium lies predominantly to the left or to the right. Explain your predictions briefly.

  1. (a) H2S(aq) + CO32−(aq) ⇄ HS(aq) + HCO3(aq)
  2. (b) HCN(aq) + SO42−(aq) ⇄ CN(aq) + HSO4(aq)
  3. (c) SO42−(aq) + CH3CO2H(aq) ⇄ HSO4(aq) + CH3CO2(aq)

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  HA(aq)+B(aq)   BH+(aq)  +    A(aq)(acid)    (base)        (conjugate     acid)   (conjugate     base)

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants (Kaand Kb) for reactants as well as of the products. The more the value of Ka for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of Kb for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in right direction predominantly.

  H2S(aq)+   CO32(aq)   HS(aq)  +    HCO3(aq) (acid)         (base)            (conjugate     base)     (conjugate     acid)

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  H2S(aq)+   CO32(aq)   HS(aq)  +    HCO3(aq) (acid)         (base)            (conjugate     base)     (conjugate     acid)

Here, the acid  H2S( Ka=1.2×107) is stronger acid as compared to conjugate acid HCO3( Ka=4.8×1011), hence  H2S will ionize faster as compared to HCO3 due to which the equilibrium will shift in left direction. So it can be stated that the given reaction is reactant favoured.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  HA(aq)+B(aq)   BH+(aq)  +    A(aq)(acid)    (base)        (conjugate     acid)   (conjugate     base)

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants (Kaand Kb) for reactants as well as of the products. The more the value of Ka for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of Kb for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in left direction predominantly.

  HCN(aq) +  SO42(aq)   HSO4(aq)  +    CN(aq) (acid)             (base)            (conjugate    acid)     (conjugate     base)

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  HCN(aq) +  SO42(aq)   HSO4(aq)  +    CN(aq) (acid)             (base)            (conjugate    acid)     (conjugate     base)

Here, the conjugate acid   HSO4( Ka=1.2×102) is stronger acid as compared to HCN( Ka=4.0×1010), hence   HSO4 will ionize faster as compared to HCN due to which the equilibrium will shift in left direction. Also SO42 conjugate base is stronger base than CN and it SO42 will also ionize faster as compared to CN. So it can be stated that the given reaction is reactant favoured.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  HA(aq)+B(aq)   BH+(aq)  +    A(aq)(acid)    (base)        (conjugate     acid)   (conjugate     base)

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants (Kaand Kb) for reactants as well as of the products. The more the value of Ka for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of Kb for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in left side predominantly.

  CH3COOH(aq) +  SO42(aq)   HSO4(aq)  +    CH3COO(aq)    (acid)                      (base)           (conjugate    acid)          (conjugate     base)

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  CH3COOH(aq) +  SO42(aq)   HSO4(aq)  +    CH3COO(aq)    (acid)                      (base)           (conjugate    acid)          (conjugate     base)

Here, the conjugate acid   HSO4( Ka=1.2×102) is stronger as compared to acid CH3COOH( Ka=1.8×105), hence HSO4 will ionize faster as compared to CH3COOH, due to which the equilibrium will shift in left direction. Also the base  SO42 ion is stronger base than conjugate base CH3COO and hence SO42 will also ionize faster as compared to CH3COO. So it can be stated that the given reaction is reactant favoured.

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Chapter 16 Solutions

Chemistry & Chemical Reactivity

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