   Chapter 17, Problem 89AE

Chapter
Section
Textbook Problem

# Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O2. Consider the following reactions and approximate standard free energy changes: Hgb + O 2 → HgbO 2         Δ G ∘ = − 70   KJ Hgb + CO → HgbCO         Δ G ∘ = − 80   KJ Using these data, estimate the equilibrium constant value at 25°C for the following reaction: HgbO 2 + CO   ⇌ HgbCO   + O 2

Interpretation Introduction

Interpretation: The formation reaction of HgbO2,HgbCO and their ΔG° value is given. The value of K is to be calculated at given temperature for the reaction between HgbO2 and CO .

Concept introduction: Equilibrium constant K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

The expression for standard free energy change is,

ΔG°=RTln(K)

Explanation

Explanation

Given

The stated reaction is,

HgbO2+COHgbCO+O2

For the reaction,

Hgb+O2HgbO2

The value of ΔG° is 70kJ .

The reverse reaction of the above equation is,

HgbO2Hgb+O2 (1)

The value of ΔG° is 70kJ .

For the reaction,

Hgb+COHgbCO (2)

The value of ΔG° is 80kJ .

Addition of equation (1) and (2) will give the required stated reaction. The resultant value of ΔG° from equation (1) and (2) is,

(80+70)kJ=10kJ_

The value ΔG° for the given reaction is 10kJ .

Given temperature is 25°C .

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 10kJ into joule is,

10kJ=(10×103)J=10×103<

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