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The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8^{\circ} \mathrm{C}\), and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{kl} / \mathrm{mol}\). (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00\) mol of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8^{\circ} \mathrm{C}\)

Short Answer

Expert verified
(a) The entropy of liquid Bromine increases when it boils at its normal boiling point. (b) The change in entropy when 1.00 mol of liquid Bromine is vaporized at 58.8°C is \(89.12 \, J/mol \cdot K\).

Step by step solution

01

Convert temperature from Celsius to Kelvin

First, we need to convert the temperature from Celsius to Kelvin using the following formula: T(K) = T(°C) + 273.15 T(K) = 58.8 + 273.15 T(K) = 331.95 K
02

Calculate the change in entropy

Now that we have the temperature in Kelvin, we can proceed to calculate the change in entropy using the formula: ΔS = ΔH_vap / T where ΔH_vap is in J/mol First, convert ΔH_vap from kJ/mol to J/mol by multiplying by 1000: \(ΔH_{vap} = 29.6 \, kJ/mol * 1000 = 29600 \, J/mol\) Now, we can plug in the values we have into the equation: ΔS = 29600 J/mol / 331.95 K ΔS = 89.12 J/(mol K)
03

Determine if the entropy increases or decreases

Since ΔS is a positive value (89.12 J/mol K), this means that the entropy of Bromine increases when it boils at its normal boiling point. #Answer#: (a) The entropy of liquid Bromine increases when it boils at its normal boiling point. (b) The change in entropy when 1.00 mol of liquid Bromine is vaporized at 58.8°C is 89.12 J/mol K.

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Key Concepts

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

Understanding Entropy in Phase Changes
Entropy is a key concept in thermodynamics, representing the degree of disorder or randomness in a system. During a phase change, such as boiling, the system's entropy changes. For instance, when
  • liquid bromine (\(\text{Br}_2(l)\)) boils, it transitions from a more ordered liquid state to a less ordered gas state.
  • Since gases have more freedom of movement, they are more disordered, leading to an increase in entropy.
This is why the change in entropy \(\Delta S\) is positive when bromine boils at its normal boiling point of 58.8°C. The calculated entropy change of 89.12 J/(mol K) confirms that there is an increase in disorder.
The Role of Enthalpy in Phase Transitions
Enthalpy is the total heat content of a system and plays a vital role during phase transitions. When bromine boils, it requires energy to overcome intermolecular forces.
  • This energy change is known as the molar enthalpy of vaporization \(\Delta H_{vap}\).
  • For bromine, \(\Delta H_{vap}\) is 29.6 kJ/mol at its normal boiling point.
This endothermic process absorbs heat from the surroundings, helping bromine transition to a gaseous phase. During boiling, heat is absorbed without changing temperature but increases the enthalpy of the system, facilitating the transition.
Phase Transition: Boiling of Bromine
Phase transitions involve a change of state, and understanding them requires considering both entropy and enthalpy. For bromine:
  • The boiling point is 58.8°C or 331.95 K, where the liquid turns to vapor.
  • During this transition, the entropy increases as the molecules spread out in the gas phase.
The calculated positive \(\Delta S\) indicates increased disorder. The process involves absorbing heat (endothermic), leading to a phase change where enthalpy rises. This dual consideration of entropy and enthalpy helps explain why boiling occurs and its energetic nature.

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Most popular questions from this chapter

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2}\) ? (c) If \(\mathrm{P}_{\mathrm{so}_{2}}=\mathrm{P}_{\mathrm{A}_{2} \mathrm{~s}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

Which of the following processes are spontaneous: (a) the melting of ice cubes at \(-10^{\circ} \mathrm{C}\) and 1 atm pressure; (b) separating a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) into two separate samples, one that is pure \(\mathrm{N}_{2}\) and one that is pure \(\mathrm{O}_{2}\) (c) alignment of iron filings in a magnetic field; (d) the reaction of hydrogen gas with exygen gas to form water vapor at room temperature; (e) the dissolution of HCl(g) in water to form concentrated hydrochloric acid?

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ}\). (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

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