Question

To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin \(\left(\mathrm{HbO}_{2}\right)\) 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 \(37^{\circ} \mathrm{C}\). (The reaction is first order in \(\mathrm{Hb}\) and \(\mathrm{O}_{2} .\) ) For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} M\) 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 \(1.4 \times 10^{-4} \mathrm{M} / \mathrm{s}\) during exercise to meet the demand of increased metabolism rate. Assuming the Hb concentration to remain the same, what must be the oxygen concentration to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

Short answer

The rate of formation and consumption of O2 is \(2.5 \times 10^{-5}\) M/s. The necessary O2 concentration to sustain the given HbO2 formation rate during exercise is approximately \(8.3 \times 10^{-6}\) M.

Step-by-step solution

Calculate rate of formation of HbO2

To find the rate of formation, we need to know the rate law for this reaction. The reaction is said to be second order and is first order in both reactants Hb and O2. Thus, the rate law is: Rate = k[Hb][O2]. Substitute given values into rate law. Rate = (2.1 × 10^6 /M.s) * (8.0 × 10^-6 M) * (1.5 × 10^-6 M). Solving this will give you the rate of formation of HbO2.

Calculate rate of consumption of O2

In most reactions, the rate of consumption of a reactant is equal to the rate of formation of the product. Thus, the rate of consumption of O2 is equal to the rate of formation of HbO2 calculated in Step 1.

Calculate oxygen concentration needed for specific HbO2 formation rate

For the reaction rate of 1.4 × 10^-4 M/s to occur assuming the same Hb concentration, we first rearrange the rate law to solve for [O2]: [O2] = Rate / (k[Hb]). We then substitute the values: [O2] = (1.4 × 10^-4 M/s) / ((2.1 × 10^6 /M.s) * (8.0 × 10^-6 M)). This will yield the necessary O2 concentration for the given rate of HbO2 formation during exercise.

Key concepts

Understanding Rate Law

The rate law is a mathematical expression that relates the speed of a chemical reaction to the concentrations of the reactants. It's a way to quantify how fast a reaction happens under certain conditions. In our hemoglobin-oxygen binding example, the rate law can be expressed as:

Rate = k[Hb][O2]

This formula might look simple, but it has a lot packed into it. The symbol 'k' stands for the rate constant, which is unique to each reaction and changes with temperature. In this case, we know the rate constant is a significant number: \(2.1 \times 10^{6} / M \cdot \text{s}\) at \(37^\circ \text{C}\). The ['Hb'] and ['O2'] represent the concentrations of hemoglobin and oxygen respectively in moles per liter (M).

When we calculate the rate using the given concentrations provided in the problem, we're finding out how quickly oxyhemoglobin is formed under specific circumstances. It’s like a snapshot of the reaction speed at a particular moment, assuming all conditions are constant.

Discovering Reaction Order

Reaction order is a concept in chemical kinetics that provides insight into the relationship between the concentration of reactants and the rate of the reaction. For our reaction, the order is first order in hemoglobin and first order in oxygen, which makes it a second-order reaction overall since we simply add the exponents: 1 (for Hb) + 1 (for O2) = 2. A first-order reaction means that the rate is proportional to the concentration of a single reactant. In our textbook problem, the rate will change linearly with changes in concentration of either Hb or O2 alone.

It’s important for students to grasp this concept because it helps in predicting how a change in concentration affects the overall rate. If a reaction is zero order in a reactant, changing its concentration won't affect the rate; and if it’s second order in a reactant, doubling the concentration of that reactant quadruples the rate. These relationships are foundational to understanding chemical processes.

Exploring Hemoglobin-Oxygen Binding

The binding of oxygen to hemoglobin is a critical biological process. Hemoglobin is a protein in red blood cells that's responsible for transporting oxygen from the lungs to the rest of the body. The reaction of hemoglobin combining with oxygen to form oxyhemoglobin, which we see in our exercise, is a model of physiological importance.

Usually, this binding is affected by a variety of factors like pH, temperature, and the presence of other substances like carbon dioxide. Under typical conditions, this binding is reversible, allowing hemoglobin to pick up and release oxygen as needed. But when we exercise, our muscles need more oxygen, which means hemoglobin has to bind with more oxygen. As shown in the problem, the rate of oxyhemoglobin formation increases, and we can calculate the new oxygen concentration needed to meet the metabolic demands during exercise – illustrating how the principles of chemical kinetics apply to our everyday life.