Problem 76 - Chemistry: Principles and Reactions

Question

Consider the following reactions at \(25^{\circ} \mathrm{C}\) : $$ \begin{aligned} &\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O} \quad \Delta G^{\circ}=-2870 \mathrm{~kJ} \\ &\mathrm{ADP}(a q)+\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O} \\ &\Delta G^{\circ}=31 \mathrm{~kJ} \end{aligned} $$ Write an equation for a coupled reaction between glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), and ADP in which \(\Delta G^{\circ}=-390 \mathrm{~kJ}\).

Step-by-Step Solution

Answer

Question: Find a coupled reaction between glucose and ADP that results in a standard Gibbs free energy change of -390 kJ. Solution: Combine 3 glucose reactions with 260 ADP reactions to achieve the desired standard Gibbs free energy change of -390 kJ. The final coupled equation is: $$ 3\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}+260\mathrm{ADP}+260\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}+18\mathrm{O}_{2}+520\mathrm{H}^{+} \longrightarrow 260\mathrm{ATP}+18\mathrm{CO}_{2}+525\mathrm{H}_{2}\mathrm{O} $$

Step 1: Identify the given information

We are given two separate reactions: 1. The glucose reaction with oxygen with \(\Delta G^{\circ} = -2870 \, \mathrm{kJ}\) 2. The ADP reaction with \(\Delta G^{\circ} = 31 \, \mathrm{kJ}\) And we are asked to find a combined reaction between glucose and ADP with \(\Delta G^{\circ} = -390 \, \mathrm{k} \mathrm{J}\).

Step 2: Determine the scale of reactions to achieve the required Gibbs free energy change

Let's say the number of glucose reactions required is \(x\), and the number of ADP reactions required is \(y\). We can write the following equation to find the ratio between these numbers: $$-2870x + 31y = -390$$

Step 3: Determine the ratio between the reactions

For simplicity, let's rearrange the equation from step 2 and solve for y in terms of x: $$y = \frac{-390 + 2870x}{31}$$ Now, we need to find an integer solution, as we cannot have reactions in fractions. By testing different values of \(x\), we can find an integer solution: For \(x = 1\), \(y = \frac{-390 + 2870}{31} = \frac{2480}{31}\), which is not an integer. For \(x = 2\), \(y = \frac{-390 + 5740}{31} = \frac{5350}{31} = 172.58...\), which is not an integer. For \(x = 3\), \(y = \frac{-390 + 8610}{31} = 260\), which is an integer. So, we find that for \(x = 3\) and \(y = 260\), we can achieve the desired Gibbs free energy change.

Step 4: Write the final coupled equation

As we determined that 3 glucose reaction needs to be combined with 260 ADP reactions, we can write the final equation for a coupled reaction between glucose and ADP as: $$ \begin{aligned} 3(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a\mathrm{q})+6 \mathrm{O}_{2}(g) \longrightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2}\mathrm{O})+260(\mathrm{ADP}(a\mathrm{q})+\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}(a\mathrm{q})+2 {\mathrm{H}^{+}(a\mathrm{q})} \longrightarrow \mathrm{ATP}(a\mathrm{q})+\mathrm{H}_{2} \mathrm{O}) \\\ \longrightarrow 3\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}+260\mathrm{ADP}+260\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}+18\mathrm{O}_{2}+520\mathrm{H}^{+} \longrightarrow 260\mathrm{ATP}+18\mathrm{CO}_{2}+525\mathrm{H}_{2}\mathrm{O} \end{aligned} $$

Key Concepts

Coupled Reactions

In the realm of chemistry, coupled reactions are an incredibly important concept. They occur when two or more chemical reactions are linked together, with the outcome from one being used in the other. The classic example is when one reaction releases energy, making it exergonic, and a second reaction requires energy, making it endergonic.
This method is beneficial because it allows otherwise unfavorable reactions to proceed by coupling them with energetically favorable ones.

• An example from biology is the coupling of cellular respiration with the synthesis of adenosine triphosphate (ATP), where the energy released from breaking down glucose drives the formation of ATP.
• To calculate the overall change in Gibbs Free Energy ( abla G^{ ext{∘}}), we sum the abla G^{ ext{∘}} values of each individual reaction.

It's essential to note that only the sum of abla G^{ ext{∘}} has to be negative for the coupled process to be spontaneous. Understanding this allows us to design reactions in laboratories and industrial processes that are both efficient and energy-saving.

Thermodynamics

Thermodynamics is the cornerstone of understanding chemical reactions, focusing on the energy and heat changes they involve. It encompasses several laws which help predict how reactions proceed. At the core lies Gibbs Free Energy ( abla G), a critical parameter to analyze.
abla G helps determine if a reaction is spontaneous. For a process to occur without the input of external energy, abla G must be negative. This ties into the concept of coupled reactions, allowing us to combine reactions to create an overall spontaneous process.

• The First Law of Thermodynamics tells us that energy cannot be created or destroyed; it can only change forms.
• The Second Law states that the entropy of a closed system will never decrease over time.

By applying these principles, we can better understand reaction feasibility, which is instrumental in both theoretical and practical chemistry applications. Thermodynamics helps predict whether reactions will move forward under given conditions and what energy changes can be expected.

Chemical Equilibrium

Chemical equilibrium describes the state where the forward and reverse reactions occur at equal rates. This results in the concentration of reactants and products remaining constant over time. In essence, it's when a reaction is perfectly balanced.
For a reaction at equilibrium, Gibbs Free Energy change ( abla G) is zero, meaning the reaction has no net gain or loss of energy. However, the standard Gibbs Free Energy change ( abla G^{ ext{∘}}) provides insight into the position of equilibrium. By using the equation: abla G = abla G^{ ext{∘}} + RT ext{ln} ext{Q} , we can explore how changes in pressure, temperature, or concentration will affect the system.

• A reaction poised at equilibrium can be pushed in either direction by altering conditions such as temperature and pressure, a principle known as Le Chatelier's Principle.
• At the molecular level, the rates of the forward and reverse reactions are equal, even though individual molecules are still actively reacting.

Understanding this balance is essential in predicting and controlling chemical reactions, as well as in industrial processes where product yield optimizations are crucial.