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Chapter 9: Problem 121

The commercial production of 1,2 -dichloroethane, a solvent used in dry cleaning, involves the reaction of ethylene with chlorine: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(l)$$ Is \(\Delta S\) for this reaction likely to be positive or negative? Explain.

Short Answer

Expert verified
\( \Delta S \) is likely negative due to the decrease in moles of gas and transition to a liquid state.

Step by step solution

01

Understand the Reaction

The reaction involves ethylene gas \( C_{2}H_{4}(g) \) and chlorine gas \( Cl_{2}(g) \) combining to form one mole of 1,2-dichloroethane liquid \( C_{2}H_{4}Cl_{2}(l) \). This means we are starting with two gaseous reactants and ending with one liquid product.
02

Analyze the State Changes

Since gases have higher entropy compared to liquids, the transition from gases (\( C_{2}H_{4}(g) \) and \( Cl_{2}(g) \)) to a liquid (\( C_{2}H_{4}Cl_{2}(l) \)) generally indicates a decrease in entropy. This is because the molecular randomness and freedom of motion decrease when gases combine to form a liquid.
03

Predict the Sign of \( \Delta S \)

Given that the system moves from having two moles of gas to one mole of liquid, there's a significant decrease in the number of moles of gas, leading to a decrease in disorder. Therefore, we expect \( \Delta S \) to be negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy
Entropy is a measure of the disorder or randomness in a system. In chemistry, it is used to predict how spontaneous a process or reaction might be.
  • A system with high entropy is more disordered and has more possible arrangements for its particles.
  • On the other hand, low entropy corresponds to a more ordered state with fewer possible arrangements.
In this specific reaction where ethylene and chlorine gases become a liquid (1,2-dichloroethane), the change in entropy (\( \Delta S \)) is considered. Because gases have more molecular motion and freedoms than liquids, the transition from gas to liquid usually signals a decrease in entropy. This means the particles in the liquid are more ordered than when they are in gaseous form, leading to a more negative entropy change in this reaction.
State Changes in Chemical Reactions
In chemical reactions, state changes often significantly impact entropy. The conversion of substances from one state of matter to another鈥攕uch as gas to liquid鈥攁lters the arrangement and energy of particles.
  • Gases typically have high energy and freedom of movement, resulting in high entropy.
  • Liquids, conversely, have less entropy due to restricted molecular motion and more ordered particle arrangement.
The reaction \( \mathrm{C}_{2}\mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}\mathrm{Cl}_{2}(l) \) highlights how a state change from gaseous reactants to a liquid product leads to lower entropy. This decrease is typical when gases condense to form liquids.
Chemical Reaction Analysis
Analyzing chemical reactions involves examining the changes in states, enthalpy, and entropy among the reactants and products.In the given reaction, both reactants start in the gaseous state and end as a liquid product. This process is crucial:
  • It allows chemists to predict whether a reaction is favorable in terms of spontaneity by examining \( \Delta S \) and overall energy change.
  • This reaction specifically goes from two moles of gas to one mole of liquid, suggesting a notable drop in entropy and potential energy minimization.
This analysis helps in predicting the behavior of chemical reactions in industrial processes like the synthesis of solvents, ensuring efficient applications in commercial products. Understanding such dynamics is essential in optimizing reaction conditions and maximizing outputs.

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Most popular questions from this chapter

Water gas is the name for the mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) prepared by reaction of steam with carbon at \(1000^{\circ} \mathrm{C}\) : $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$"Water gas" The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information to calculate \(\Delta H^{\circ}\) in kilojoules for the water-gas reaction: $$\begin{aligned}\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} \\ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-566.0 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ}=-483.6 \mathrm{~kJ} \end{aligned}$$

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), the most widely produced chemical in the world, is made by a two-step oxidation of sulfur to sulfur trioxide, \(\mathrm{SO}_{3}\), followed by reaction with water. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{SO}_{3}\) in \(\mathrm{kJ} / \mathrm{mol}\), given the following data: $$\begin{array}{ll}\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-296.8 \mathrm{~kJ} \\ \mathrm{SO}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H^{\circ}=-98.9 \mathrm{~kJ} \end{array}$$

Ethyl alcohol has \(\Delta H_{\text {fusion }}=5.02 \mathrm{~kJ} / \mathrm{mol}\) and melts at \(-114.1^{\circ} \mathrm{C}\). What is the value of \(\Delta S_{\text {fusion }}\) for ethyl alcohol?

The boiling point of a substance is defined as the temperature at which liquid and vapor coexist in equilibrium. Use the heat of vaporization \(\left(\Delta H_{\text {vap }}=30.91 \mathrm{~kJ} / \mathrm{mol}\right)\) and the entropy of vaporization \(\left[\Delta S_{\text {vap }}=93.2 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\right]\) to calculate the boiling point \(\left({ }^{\circ} \mathrm{C}\right)\) of liquid bromine.

A \(110.0 \mathrm{~g}\) piece of molybdenum metal is heated to \(100.0^{\circ} \mathrm{C}\) and placed in a calorimeter that contains \(150.0 \mathrm{~g}\) of water at \(24.6^{\circ} \mathrm{C}\). The system reaches equilibrium at a final temperature of \(28.0^{\circ} \mathrm{C}\). Calculate the specific heat of molybdenum metal in \(\mathrm{J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\). The specific heat of water is \(4.184 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).
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