Question

Which of the following statements is correct for a spontaneous polymerization reaction?

• A. \(∆G<0,∆H<0,∆S<0\)
• B. \(∆G<0,∆H>0,∆S>0\)
• C. \(∆G>0,∆H<0,∆S>0\)
• D. \(∆G>0,∆H>0,∆S>0\)

Answer 1

The Correct Option is A

The spontaneity of a chemical reaction, including polymerization, is determined by the Gibbs free energy change (∆G). For a reaction to be spontaneous, ∆G must be negative:

\( \Delta G = \Delta H - T\Delta S \)

where ∆H is the enthalpy change, ∆S is the entropy change, and T is the temperature in Kelvin.

In the case of spontaneous polymerization:

• ∆H<0: The process is exothermic, releasing energy as bonds are formed between monomers.
• ∆S<0: Polymerization involves the formation of a more ordered structure from monomers, leading to a decrease in entropy.

Even though ∆S is negative, the large negative ∆H ensures that ∆G remains negative, satisfying the condition for spontaneity. Thus, the correct answer is:

\( \Delta G<0, \quad \Delta H<0, \quad \Delta S<0 \)

Answer 2

Correct Answer:

Option 1: ∆G < 0, ∆H < 0, ∆S < 0

Explanation:

1. Gibbs Free Energy (ΔG):

For a spontaneous process, ∆G must be negative (∆G < 0).

2. Enthalpy Change (ΔH):

Polymerization reactions are generally exothermic, so ∆H is usually negative (∆H < 0).

3. Entropy Change (ΔS):

Polymerization reduces the number of independent particles, leading to a decrease in entropy, so ∆S is usually negative (∆S < 0).

4. Relationship between ΔG, ΔH, and ΔS:

∆G = ∆H - T∆S

To ensure ∆G < 0, the magnitude of ∆H must be greater than T∆S.