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Ch.15 - Chemical Equilibrium
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All textbooksTro 4th EditionCh.15 - Chemical EquilibriumProblem 73

Chapter 15, Problem 73

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ⇌ HbCO(aq) + O2(aq) a. Use the reactions and associated equilibrium constants at body temperature given here to find the equilibrium con- stant for the reaction just shown. Hb(aq) + O2(aq) ⇌ HbO2(aq) Kc = 1.8 Hb(aq) + CO(aq) ⇌ HbCO(aq) Kc = 306

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Hello everyone today. We are being given the following statement and asked to determine the equilibrium constant for the above reaction. So we have the following reaction is a vital step in the metabolism of glucose. We have glucose plus 80 p. gives us glucose six phosphate and ADP. And then we have subsequent half reactions of that. So first we want to write down what our overall reaction would be and that would be glucose plus A. T. P. Gives us glucose six fast feet plus A. D. People notice how I didn't include the phosphate groups, the inorganic phosphate, the P. I. That's because they're on opposite sides of that equation. So they cancel each other out with this information. We essentially want to take both of their K. Values and multiply them by each other to get the overall K value. So we're gonna take that 3.114 times 10 to the negative three And multiply it by 2.716 times 10 to the 5th And get an end value of 845 0.76. I hope this helped. And until next time.
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