   Chapter 21, Problem 81AE

Chapter
Section
Textbook Problem

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

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction is to be calculated.

Concept introduction:

Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature, pressure etc.

Explanation

The standard free energy change ΔGο for the reaction is -10 kJ_ .

The given reactions are,

Hb+O2HbO2                   ΔG1ο=70 kJ (1)

Hb+COHbCO                 ΔG2ο=80 kJ (2)

The equation (1) is reversed and added with equation (2) and the resultant equation is,

HbO2+COHbCO+O2 (3)

The standard free energy change ΔGο for equation (3) is calculated by the formula,

ΔGο=ΔG1ο+ΔG2ο

Where,

• ΔG1ο is the standard free energy change for equation (1).
• ΔG2ο is the standard free energy change for equation (2).

Reversing the equation (1) will also reverse the sign of ΔG1ο value.

Substitute the values of ΔG1ο and ΔG2ο in the above equation

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