“Equilibrium constant value” should be identified. Concept introduction: The value of Equilibrium constant determines the rate of forward and rate of backward reaction. If the value goes more than 1 then rate of forward reaction favours else it favours backward reaction. Given: Hb+O 2 ⇌ HbO 2 , K=2 × 10 12 Hb+CO ⇌ HbO , K=1 × 10 14 HbO 2 +CO 2 ⇌ HbCO+O 2

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Answer 1

Question

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

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Answer 2

Question

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

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Answer 3

Question

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

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Answer 4

Question

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

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Answer 5

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

Answer 6

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

Answer 7

Chapter 23, Problem 56E

Interpretation Introduction

Interpretation:

“Equilibrium constant value” should be identified.

Concept introduction:

The value of Equilibrium constant determines the rate of forward and rate of backward reaction.

If the value goes more than 1 then rate of forward reaction favours else it favours backward

reaction.

Given:

$\begin{matrix} {Hb+O}_{2} ⇌ {HbO}_{2}, K=2 \times 10^{12} \\ Hb+CO ⇌ HbO, K=1 \times 10^{14} \\ {HbO}_{2} {+CO}_{2} ⇌ {HbCO+O}_{2} \end{matrix}$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 23 - What is the electron configuration of the Cu+ ion?...Ch. 23 - Prob. 2SAQ Ch. 23 - Prob. 3SAQ Ch. 23 - Prob. 4SAQ Ch. 23 - Prob. 5SAQ Ch. 23 - Prob. 6SAQ Ch. 23 - Prob. 7SAQ Ch. 23 - Estimate the crystal field splitting energy (in...Ch. 23 - Use crystal field theory to determine the number...Ch. 23 - Prob. 10SAQ

Solution Summary: The author explains that the Equilibrium constant determines the rate of forward and backward reaction. The equilibrium constant for HbO 2 is reverse of what is given.

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Chapter 23 Solutions