Problem 115 - Chemistry and Chemical Reactivity

Question

A Living organisms derive energy from the oxidation of food, typified by glucose. $$ \mathrm{C}_{0} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) $$ Electrons in this redox process are transferred from glucose to oxygen in a series of at least 25 steps. It is instructive to calculate the total daily current flow in a typical organism and the rate of energy expenditure (power). (See T. P. Chirpich: Journal of Chemical Education, Vol. \(52,\) p. 99 1975.) (a) The molar enthalpy of combustion of glucose is \(-2800 \mathrm{kJ} / \mathrm{mol}\) -nan. If you are on a typical daily diet of 2400 Cal (kilocalories), what amount of glucose (in moles) must be consumed in a day if glucose is the only source of energy? What amount of \(\mathrm{O}_{2}\) must be consumed in the oxidation process? (b) How many moles of electrons must be supplied to reduce the amount of \(\mathrm{O}_{2}\) calculated in part (a)? (c) Based on the answer in part (b), calculate the current flowing, per second, in your body from the combustion of glucose. (d) If the average standard potential in the electron transport chain is \(1.0 \mathrm{V},\) what is the rate of energy expenditure in watts?

Step-by-Step Solution

Answer

About 3.59 moles of glucose and 21.52 moles of \(\mathrm{O}_2\) are needed. The current flowing is approximately 96.05 A, and the power is 96.05 W.

Step 1: Convert Daily Caloric Intake to Joules

The daily caloric intake is 2400 Cal. Since 1 Cal = 4.184 kJ, we can convert this into Joules: \[ 2400 \text{ Cal} \times 4.184 \text{ kJ/Cal} = 10041.6 \text{ kJ} \]

Step 2: Calculate Moles of Glucose

The molar enthalpy of combustion of glucose is \(-2800 \text{ kJ/mol}\). To find the moles of glucose needed for a daily energy of 10041.6 kJ, use:\[ \frac{10041.6 \text{ kJ}}{-2800 \text{ kJ/mol}} \approx 3.586 \text{ mol} \] (we take the absolute value as energy needed is positive in physics context)

Step 3: Determine Moles of Oxygen Required

The balanced equation shows that each mole of glucose requires 6 moles of \(\mathrm{O}_2\). Therefore, for 3.586 moles of glucose:\[ 3.586 \text{ mol glucose} \times 6 \text{ mol } \mathrm{O}_2/\text{mol glucose} = 21.516 \text{ mol } \mathrm{O}_2 \]

Step 4: Calculate Moles of Electrons Needed

The reduction of one molecule of \(\mathrm{O}_2\) involves 4 electrons:\[ 21.516 \text{ mol } \mathrm{O}_2 \times 4 \text{ mol electrons/mol } \mathrm{O}_2 = 86.064 \text{ mol electrons} \]

Step 5: Compute Current in Amperes

Using Faraday's constant, \(96485 \text{ C/mol}\), the charge in Coulombs is:\[ 86.064 \text{ mol electrons} \times 96485 \text{ C/mol} = 8301241.64 \text{ C} \]Then divide by the number of seconds in a day (86400 s) to find current:\[ \frac{8301241.64 \text{ C}}{86400 \text{ s}} \approx 96.05 \text{ A} \]

Step 6: Calculate Power in Watts

Power is the product of current and potential difference: \[ P = I \times V = 96.05 \text{ A} \times 1.0 \text{ V} = 96.05 \text{ W} \]

Key Concepts

Calorimetry

Calorimetry is a branch of thermodynamics dedicated to finding the amount of heat involved in chemical reactions or physical changes. It is particularly useful when calculating the energy content of food, as in our example involving glucose. Calorimeters can measure the heat generated or absorbed during a reaction, thereby providing insights into the energy change. In this problem, the calorimetric concept helps us compute the energy produced from glucose combustion, crucial for understanding how organisms use food for energy. This heat energy is quantified by converting standard energy units—Cal to Joules—to make precise calculations. By knowing the caloric content of glucose, we can determine the number of moles consumed for daily energy requirements, effectively linking the abstract concept of calorimetry to real-world energy metabolism in living beings.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between two species. These reactions are essential in energy production within biological systems. In our exercise, the conversion of glucose to carbon dioxide and water represents a redox reaction, where glucose donates electrons (is oxidized) and oxygen accepts electrons (is reduced). This electron transfer is a key methodological step in the biochemical energy production chain. Understanding these reactions involves grasping the significance of electron flow from the reducing agent (glucose) to the oxidizing agent (oxygen). This makes it a fundamental concept not just in chemistry but also in cellular respiration, since the release of energy during glucose oxidation helps power various biological processes.

Electron Transport Chain

The electron transport chain (ETC) is a series of proteins and other molecules in cellular membranes that facilitate redox reactions. This chain is a critical component of cellular respiration where electrons are shuttled from electron donors to acceptors. Each transfer releases energy used to pump protons across a membrane, ultimately generating ATP—our cells' energy currency. With the average potential difference in the chain noted as 1.0 V, this drives the synthesis of a significant amount of ATP during glucose metabolism. In the context of our problem, the ETC utilizes electrons released from glucose oxidation, creating the current and power necessary to sustain life. Recognizing the ETC's role underscores the intricate relationship between electron transfer and energy production in living organisms. It's a fine-tuned electric circuit within cells, harnessing the flow of electrons for biological work.