1. The hemoglobin in blood establishes an equilibrium with oxygen gas very quickly. The equilibrium can be represented as heme + O2 heme-O2 where "heme" stands for hemoglobin and "heme-O;" stands for the hemoglobin-oxygen complex. The value for the equilibrium constant for this reaction is about 9.2 x 105. Carbon monoxide also binds with hemoglobin by the following reaction: heme + co = heme-co This reaction has an equilibrium constant of 2.3 x 103. Which reaction's equilibrium lies farther toward products? Does your answer justify the toxicity of CO?

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
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1. The hemoglobin in blood establishes an equilibrium with oxygen gas very quickly. The
equilibrium can be represented as
heme + 02 =
heme-O2
where "heme" stands for hemoglobin and "heme-Oz" stands for the hemoglobin-oxygen
complex. The value for the equilibrium constant for this reaction is about 9.2 x 1018. Carbon
monoxide also binds with hemoglobin by the following reaction:
heme + co =
This reaction has an equilibrium constant of 2.3 x 1023. Which reaction's equilibrium lies
heme-co
farther toward products? Does your answer justify the toxicity of CO?
Transcribed Image Text:1. The hemoglobin in blood establishes an equilibrium with oxygen gas very quickly. The equilibrium can be represented as heme + 02 = heme-O2 where "heme" stands for hemoglobin and "heme-Oz" stands for the hemoglobin-oxygen complex. The value for the equilibrium constant for this reaction is about 9.2 x 1018. Carbon monoxide also binds with hemoglobin by the following reaction: heme + co = This reaction has an equilibrium constant of 2.3 x 1023. Which reaction's equilibrium lies heme-co farther toward products? Does your answer justify the toxicity of CO?
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