Coupled Reactions

#Thermodynamically Unfavorable Reactions & Coupled Reactions

#Introduction: Spontaneity and Energy

Most of the time, we talk about reactions that happen on their own—thermodynamically favorable reactions, where ΔG° < 0 and K > 1. But what about reactions that don't want to happen? These are thermodynamically unfavorable reactions (ΔG° > 0, K < 1), and they won't go without a little push. Let's explore how we can make these reactions happen.

Understanding thermodynamically favorable and unfavorable reactions is crucial, as it ties into spontaneity, equilibrium, and energy changes. Expect to see this concept in both multiple-choice and free-response questions.

#External Energy Sources

Nonspontaneous reactions need an external energy source to get going. A common source is electricity. Think of it like jump-starting a car. By using electrical energy, we can force nonspontaneous redox reactions to occur in electrolytic cells. These cells are like the opposite of batteries, using electricity to drive a reaction that wouldn't happen on its own. Imagine charging your phone—that's an electrolytic cell in action! 🔋

Caption: An electrolytic cell uses an external power source to drive a non-spontaneous redox reaction.

#Coupled Reactions: Making the Unfavorable Favorable

Another way to make nonspontaneous reactions happen is through coupled reactions. This is like using a strong reaction to pull a weaker reaction along. Coupled reactions involve a nonspontaneous reaction and a spontaneous reaction that share a common intermediate. Remember from kinetics, an intermediate is a substance produced in one step of a reaction mechanism and consumed in a later step. It's like a baton passed between runners in a relay race.

Caption: In this reaction mechanism, 'O' is an intermediate.

#Example: Extracting Copper

Let's look at an example. The reaction Cu2S → 2Cu + S (ΔG° = +86.2 kJ) is nonspontaneous. However, we can couple it with the spontaneous reaction S + O2 → SO2 (ΔG° = -300.1 kJ). By adding these reactions together, we get:

#Cu2S → 2Cu + S (ΔG° = +86.2 kJ) S + O2 → SO2 (ΔG° = -300.1 kJ)

Cu2S + O2 → 2Cu + SO2 (ΔG° = -213.9 kJ)

The overall reaction is now spontaneous because the large negative ΔG° of the second reaction more than compensates for the positive ΔG° of the first reaction. This is like combining a small debt with a large income to get a positive balance.

#Biological Significance

Coupled reactions are super important in biology. For example, the conversion of ATP to ADP, which releases energy, is often coupled with other reactions in biological systems. This allows cells to perform work by using the energy released from ATP hydrolysis. The image below shows a similar concept in organic chemistry. Don't worry about the specific molecules if they're unfamiliar.

Caption: An example of coupled reactions in organic chemistry, involving ATP and ADP.

#Practice Problem

Let's try a practice problem to solidify our understanding of coupled reactions. Given:

Fe2O3 → 2Fe + 3/2O2 (ΔG° = +742.2 kJ) CO + 1/2O2 → CO2 (ΔG° = -283.5 kJ)

Calculate ΔG° for the reaction: Fe2O3 + 3CO → 2Fe + 3CO2

Solution:

1. Multiply the second reaction by 3 to cancel out the O2:

   3CO + 3/2O2 → 3CO2 (ΔG° = -283.5 kJ * 3 = -850.5 kJ)

2. Add the two reactions and their ΔG° values:

   #Fe2O3 → 2Fe + 3/2O2 (ΔG° = +742.2 kJ) 3CO + 3/2O2 → 3CO2 (ΔG° = -850.5 kJ)

   Fe2O3 + 3CO → 2Fe + 3CO2 (ΔG° = -108.3 kJ)

#Final Exam Focus

• Key Concepts:
  • Spontaneous vs. nonspontaneous reactions (ΔG° and K)
  • Electrolytic cells and their relation to nonspontaneous redox reactions
  • Coupled reactions and common intermediates
  • Hess’s Law for ΔG° calculations
• Common Question Types:
  • Calculating ΔG° for coupled reactions
  • Identifying spontaneous and nonspontaneous processes
  • Understanding the role of external energy sources
  • Applying Hess’s Law to find ΔG°
• Exam Tips:
  • Remember that ΔG° values are additive for coupled reactions.
  • Pay attention to the direction of the reaction and the sign of ΔG°.
  • Practice problems involving Hess’s Law and coupled reactions.

#

Practice Question

Multiple Choice Questions

1. Which of the following statements is true regarding a thermodynamically unfavorable reaction? (A) It occurs spontaneously. (B) It has a ΔG° < 0. (C) It requires an external energy source to proceed. (D) It has an equilibrium constant K > 1. 2. In a coupled reaction, what is the role of the common intermediate? (A) It is a catalyst that speeds up the reaction. (B) It is a product of one reaction and a reactant in another. (C) It is a reactant in both the spontaneous and nonspontaneous reactions. (D) It is a product in both the spontaneous and nonspontaneous reactions.

2. Given the reactions: A → B (ΔG° = +50 kJ) B → C (ΔG° = -100 kJ)

   What is the ΔG° for the overall reaction A → C? (A) -150 kJ (B) -50 kJ (C) +50 kJ (D) +150 kJ

Free Response Question

Consider the following reactions:

Reaction 1: N2(g) + O2(g) → 2NO(g) ΔG° = +173.1 kJ Reaction 2: 2NO(g) + O2(g) → 2NO2(g) ΔG° = -70.0 kJ

(a) Is Reaction 1 spontaneous or nonspontaneous under standard conditions? Explain. (b) Calculate the ΔG° for the reaction: N2(g) + 2O2(g) → 2NO2(g). (c) If Reaction 1 is coupled with another reaction, what characteristic must the other reaction have to make the overall process spontaneous? (d) Describe how an electrolytic cell could be used to drive Reaction 1. Answer Key and Scoring Breakdown

Multiple Choice Answers:

1. (C)
2. (B)
3. (B)

Free Response Question Scoring:

(a) (1 point) Reaction 1 is nonspontaneous because its ΔG° is positive.

(b) (2 points) * Add the two reactions: N2(g) + O2(g) → 2NO(g) ΔG° = +173.1 kJ 2NO(g) + O2(g) → 2NO2(g) ΔG° = -70.0 kJ -------------------------------- N2(g) + 2O2(g) → 2NO2(g) * Add the ΔG° values: ΔG° = +173.1 kJ + (-70.0 kJ) = +103.1 kJ

(c) (1 point) The other reaction must have a negative ΔG° value that is large enough to make the overall ΔG° for the coupled reaction negative.

(d) (2 points) * An electrolytic cell uses an external power source to drive a nonspontaneous reaction. * The external power source would provide the energy needed to push the reaction forward, converting N2 and O2 to NO.