The following reaction is used in the industrial synthesis of polyvinyl chloride (PVC) polymer:Cl2(g) + H2C=CH2(g) → ClCH2CH2Cl(l) ∆H = –52 kcal/mol (–218 kJ/mol)a. Is ∆S positive or negative for this process?

Ch.7 Chemical Reactions: Energy, Rate and Equilibrium

McMurry - Fundamentals of General, Organic, and Biological Chemistry 8th Edition

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McMurry 8th Edition

Ch.7 Chemical Reactions: Energy, Rate and Equilibrium

Problema 40a

Capitolo 7, Problema 40a

The following reaction is used in the industrial synthesis of polyvinyl chloride (PVC) polymer:
Cl₂(g) + H₂C=CH₂(g) → ClCH₂CH₂Cl(l) ∆H = –52 kcal/mol (–218 kJ/mol)
a. Is ∆S positive or negative for this process?

Guida verificata passo dopo passo

Step 1: Begin by analyzing the reaction. The reactants are chlorine gas (Cl₂) and ethene (H₂C=CH₂), both in the gaseous state, while the product is 1,2-dichloroethane (ClCH₂CH₂Cl), which is in the liquid state. This phase change is important for determining the entropy change (∆S).

Step 2: Recall that entropy (S) is a measure of disorder or randomness in a system. Gases have higher entropy compared to liquids because gas particles are more disordered and have more freedom of movement.

Step 3: Observe that the reaction starts with two gaseous molecules (Cl₂ and H₂C=CH₂) and produces one liquid molecule (ClCH₂CH₂Cl). This indicates a decrease in the number of particles and a transition from a gaseous state to a liquid state, both of which typically result in a decrease in entropy.

Step 4: Consider the overall system. The reduction in the number of particles and the phase change from gas to liquid suggest that the system becomes more ordered, leading to a negative ∆S value.

Step 5: Conclude that ∆S for this process is negative because the reaction results in fewer particles and a transition to a more ordered liquid state, reducing the overall entropy of the system.

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