Question

The nitrite ion \(\left(\mathrm{NO}_{2}^{-}\right)\) in soil is oxidized to the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right)\) by the bacterium Nitrobacter agilis in the presence of oxygen. The half-reactions are: \(\mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{NO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=0.42 \mathrm{~V}\) $$\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=1.23 \mathrm{~V}$$ Calculate the yield of ATP synthesis per mole of nitrite oxidized.

Short answer

Approximately 10 moles of ATP are generated per mole of nitrite oxidized.

Step-by-step solution

Identify Overall Reaction

To solve for the yield of ATP synthesis, the overall reaction must be identified. The half reactions given are:1. \( \mathrm{NO}_{3}^{-} + 2\mathrm{H}^{+} + 2e^{-} \longrightarrow \mathrm{NO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O} \), with \( E^{\circ} = 0.42 \, \mathrm{V}\).2. \( \mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \), with \( E^{\circ} = 1.23 \, \mathrm{V}\).To balance the electrons, multiply the first reaction by 2 to match the electrons in the second half-reaction.

Balance the Half Reactions

Multiply the first half-reaction by 2:\[ 2(\mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + 2 e^{-} \longrightarrow \mathrm{NO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O}) \]This yields:\[ 2 \mathrm{NO}_{3}^{-} + 4 \mathrm{H}^{+} + 4 e^{-} \longrightarrow 2 \mathrm{NO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O} \]Now, you can add the balanced half reactions together.

Combine Balanced Half Reactions

Combine the balanced equations:\[(2 \mathrm{NO}_{3}^{-} + 4 \mathrm{H}^{+} + 4 e^{-} \longrightarrow 2 \mathrm{NO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O}) + (\mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 e^{-} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O})\]The overall reaction becomes:\[ 2 \mathrm{NO}_{2}^{-} + \mathrm{O}_{2} + 4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{NO}_{3}^{-} + 2 \mathrm{H}_{2}\mathrm{O}\]

Determine Cell Potential

Calculate the cell potential (\(E_{\text{cell}}\)) using \(E^{\circ}\) values:\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)\(E^{\circ}_{\text{cell}} = 1.23 \, \mathrm{V} - 0.42 \, \mathrm{V} = 0.81 \, \mathrm{V}\)

Calculate Gibbs Free Energy Change

Use \( \Delta G = -nFE \) to find Gibbs free energy change \(\Delta G\):\(\Delta G = -4 \times 96,485 \times 0.81 \, \mathrm{J/mol}\)\(\Delta G = -312,296 \, \mathrm{J/mol}\) or \(-312.3 \, \mathrm{kJ/mol}\)

Calculate ATP Yield

Each mole of ATP synthesized requires approximately \(30.5 \, \mathrm{kJ/mol}\). The yield of ATP per mole of nitrite is calculated by dividing the absolute value of \(\Delta G\) by the energy per mole of ATP:\[ \text{ATP yield} = \frac{312.3 \, \mathrm{kJ/mol}}{30.5 \, \mathrm{kJ/mol}} \approx 10.23 \]Thus, approximately 10 moles of ATP can be synthesized per mole of nitrite oxidized.

Finalize the Answer

Round the ATM yield to an appropriate significant figure given the precision of the constants used (typically two significant figures): Approximately 10 moles of ATP are generated per mole of nitrite oxidized.

Key concepts

Nitrogen Cycle

The nitrogen cycle is an essential process taking place in the ecosystem where nitrogen is converted into multiple chemical forms as it circulates among the atmosphere, terrestrial, and marine ecosystems. This cycle is crucial because nitrogen is a vital element for life, being a major part of amino acids and nucleic acids, which are building blocks of proteins and DNA respectively.
Understanding the nitrogen cycle involves the study of different processes such as fixation, nitrification, denitrification, and ammonification.

• Fixation: This is where atmospheric nitrogen (N₂) is converted into ammonia (NH₃). This is often carried out by soil bacteria known as diazotrophs.
• Nitrification: This is a two-step process where ammonia is first converted to nitrite ions ( NO₂⁻), and then into nitrate ions ( NO₃⁻) by bacteria such as Nitrosomonas and Nitrobacter.
• Denitrification: In this step, nitrate is converted back into N₂ gas, returning it to the atmosphere.
• Ammonification: The process of conversion of organic nitrogen into ammonia by the decomposition of dead matter by microorganisms.

Each of these stages plays a critical role in ensuring the availability of nitrogen in a form that can be readily used by living organisms. The balanced cycle also prevents the accumulation of excess nitrogen compounds that could be harmful to ecosystems.

Electrochemistry

Electrochemistry deals with the study of chemical processes that cause electrons to move, generating an electric current. This branch of chemistry is crucial for understanding the working of batteries, fuel cells, and even processes like corrosion.
At the core of electrochemistry is the concept of redox reactions, where oxidation refers to the loss of electrons and reduction pertains to the gaining of electrons. During such reactions, electrons are transferred between substances which can lead to the flow of current if done in a U-tube or similar setup.

• Oxidation-Reduction Reactions: These reactions involve the transfer of electrons with one substance gaining electrons (reduction) and the other losing them (oxidation).
• Galvanic Cells: Also known as voltaic cells, these are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions.
• Electrolytic Cells: These cells require electrical energy to induce non-spontaneous chemical reactions.
• Standard Electrode Potentials: The voltage of a cell can be predicted using standard electrode potentials ( E° values), showing how much a substance wants to be reduced compared to the need to physically attach an electron to it.

Understanding these key tools in electrochemistry enables better manipulation in technological applications like rechargeable batteries and electroplating.

Half-Reaction Balancing

Balancing half-reactions is a method used in electrochemistry to separate the oxidation and reduction parts of a redox reaction, which simplifies the balancing of electrons lost and gained. Each half-reaction represents either oxidation or reduction, and they must be balanced separately before they can be properly combined.
The steps in balancing half-reactions involve:

• Writing the O2 and H2O separately: Ensure all atoms other than O and H balance first.
• Balancing the Oxygen atoms: Add H₂O molecules to balance the O atoms.
• Balancing the Hydrogen atoms: Use H⁺ ions to balance H atoms in acidic conditions, or H₂O + OH⁻ in basic conditions.
• Balancing charges: Add electrons ( e⁻) to equalize the charge in both half-reactions.

By handling these stepwise, one can combine the half-reactions properly without mismatched electrons, leading to a fully balanced chemical equation for the entire reaction. This helps in accurate predictions of chemical behavior and energy yields.

Gibbs Free Energy

The concept of Gibbs Free Energy ( ΔG) is fundamental in understanding chemical reactions and their spontaneity. This thermodynamic quantity indicates how much energy is available to do work during a chemical process. When dealing with electrochemical cells, Gibbs Free Energy tells us the efficiency and capacity of the process to produce or absorb energy.
Key Elements in understanding Gibbs Free Energy include:

• Spontaneity of Reactions: If ΔG is negative, a reaction is spontaneous, meaning it can occur without external energy. A positive ΔG means the process requires energy input.
• Relation with Cell Potential: ΔG is related to the cell potential ( E°) and the number of moles of electrons, through the formula: ΔG = -nFE°, where F is Faraday's constant, equating amount of charge carried per mole of electrons.
• ATP Synthesis: In the context of ATP production, Gibbs Free Energy indicates how much chemical energy is converted into ATP. Each mole of ATP requires about 30.5 kJ/mol, thus helping determine how many moles of ATP can be produced per mole of reactant.

Understanding ΔG allows scientists and engineers to design processes and predict the feasibility and efficiency of biochemical and industrial reactions.