Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 16, Problem 45E
Interpretation Introduction

(a)

Interpretation:

The direction of the equilibrium with the increase in the concentration of [H+] is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as between the chemical reaction. It is denoted as Keq. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

Keq=[Productconcentration]a[Reactantconcentration]b

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Interpretation Introduction

(b)

Interpretation:

The direction of the equilibrium with the decrease in the concentration of [OH] is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as between the chemical reaction. It is denoted as Keq. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

Keq=[Productconcentration]a[Reactantconcentration]b

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Interpretation Introduction

(c)

Interpretation:

The direction of the equilibrium with the increase in the pH is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as between the chemical reaction. It is denoted as Keq. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

Keq=[Productconcentration]a[Reactantconcentration]b

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Interpretation Introduction

(d)

Interpretation:

The direction of the equilibrium with the decrease in the pH is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as between the chemical reaction. It is denoted as Keq. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

Keq=[Productconcentration]a[Reactantconcentration]b

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

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AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N
4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)
2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)

Chapter 16 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

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