Question

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

Short answer

(a) \(2.52 \times 10^{-5} \, M/s\); (b) \(2.52 \times 10^{-5} \, M/s\); (c) \([\text{O}_2] = 8.33 \times 10^{-6} \, M \).

Step-by-step solution

Understand the rate law

The rate law for this reaction is given by the expression \( ext{Rate} = k[ ext{Hb}][ ext{O}_2] \), where \( k \) is the rate constant, and \( [ ext{Hb}] \) and \( [ ext{O}_2] \) are the concentrations of hemoglobin and oxygen, respectively.

Calculate rate of HbO2 formation

Using the rate expression, we substitute the given values: \( k = 2.1 \times 10^6 \, M^{-1}\, s^{-1} \), \( [\text{Hb}] = 8.0 \times 10^{-6} \, M \), and \( [\text{O}_2] = 1.5 \times 10^{-6} \, M \). Compute:\[\text{Rate of HbO}_2 \text{ formation} = 2.1 \times 10^6 \, M^{-1} \cdot s^{-1} \times 8.0 \times 10^{-6} \, M \times 1.5 \times 10^{-6} \, M \]\[= 2.52 \times 10^{-5} \, M/s\] .

Calculate rate of O2 consumption

The rate of consumption of \( \text{O}_2 \) is equal to the rate of formation of \( \text{HbO}_2 \) because the stoichiometry of the reaction is 1:1. Therefore, the rate of \( \text{O}_2 \) consumption is also \( 2.52 \times 10^{-5} \, M/s \).

Determine required O2 concentration during exercise

To find the oxygen concentration needed to achieve a rate of \( 1.4 \times 10^{-4} \, M/s \), use the rate law:\[\ 1.4 \times 10^{-4} \, M/s = 2.1 \times 10^6 \, M^{-1} \cdot s^{-1} \times 8.0 \times 10^{-6} \, M \times [\text{O}_2] \]Solve for \([\text{O}_2]\):\[\ [\text{O}_2] = \frac{1.4 \times 10^{-4} \, M/s}{2.1 \times 10^6 \, M^{-1} \cdot s^{-1} \times 8.0 \times 10^{-6} \, M} \]\[\ = 8.33 \times 10^{-6} \, M \].

Key concepts

Hemoglobin

Hemoglobin (Hb) is a protein found in red blood cells. It plays a crucial role in transporting oxygen from the lungs to other body parts. Hemoglobin binds oxygen molecules, allowing the blood to carry them efficiently throughout the body.

The structure of hemoglobin is a complex and beautiful one. It consists of four subunits, each containing an iron atom that can bind to an oxygen molecule. This ability to bind oxygen makes hemoglobin indispensable for respiration and overall metabolism.

• When oxygen levels are high, as in the lungs, hemoglobin binds oxygen easily.
• When oxygen levels are low, as in body tissues, hemoglobin releases the oxygen.

This dynamic function allows for effective oxygen transport to meet the changing metabolic demands of different tissues.

Without hemoglobin, our bodies could not efficiently distribute the oxygen necessary for cellular energy production and survival.

Oxyhemoglobin

Oxyhemoglobin forms when hemoglobin binds to oxygen molecules. It is represented by the formula \(\mathrm{HbO}_2\). This compound is crucial in carrying oxygen in the bloodstream.

Once hemoglobin picks up oxygen in the lungs, it becomes oxyhemoglobin and transports the oxygen to tissues where it is released to support cellular activities. This reversible binding is vital for maintaining the oxygen balance in the body.
The formation of oxyhemoglobin can be expressed as a chemical reaction:\[\mathrm{Hb}(aq) + \mathrm{O}_2(aq) \rightarrow \mathrm{HbO}_2(aq)\] This reaction is a part of the complex processes that supply oxygen to cells and remove carbon dioxide.
Oxyhemoglobin, therefore, acts as a carrier molecule, ensuring continuous oxygen delivery needed for energy production. Its levels can vary depending on the body's oxygen demand. For example, during exercise, more oxyhemoglobin is required to support increased energy needs.

Oxygen Concentration

Oxygen concentration refers to the amount of oxygen present in a solution, such as blood. It is measured in molarity (M), indicating moles of oxygen per liter of blood.

In the context of the hemoglobin and oxyhemoglobin reaction, oxygen concentration is a crucial factor influencing the rate of the reaction. Higher concentrations can increase the reaction rate, allowing more oxygen to be transported as oxyhemoglobin.

• At rest, the body requires less oxygen, resulting in lower rates of oxyhemoglobin formation.
• During exercise, oxygen concentration must increase to meet higher metabolic demands.

Understanding oxygen concentration is essential for analyzing how effectively our respiratory system responds to various physiological needs, whether during rest or physical exertion. This balance ensures that tissues receive enough oxygen to function properly without causing harm from excessive oxygen levels.

Rate Constant

The rate constant, denoted by \(k\), is a crucial factor in chemical reactions, representing the speed of the reaction under specified conditions. For the oxyhemoglobin formation, the rate constant is given as \(2.1 \times 10^6 \ M^{-1}\cdot s^{-1}\) at \(37^{\circ} C\).

The rate constant is part of the rate law equation:\[\text{Rate} = k[\text{Hb}][\text{O}_2]\] In this equation, \([\text{Hb}]\) and \([\text{O}_2]\) are the concentrations of hemoglobin and oxygen, respectively.
The rate constant value indicates how quickly or slowly a reaction occurs:

• If \(k\) is large, the reaction proceeds rapidly, meaning oxyhemoglobin forms quickly.
• If \(k\) is small, the reaction is slower, and therefore, less oxyhemoglobin is produced over the same time.

The rate constant depends on factors such as temperature and the presence of catalysts. In physiological conditions, it helps us understand how efficiently hemoglobin can transport oxygen, especially under varying conditions like rest or exercise.