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The normal boiling point of Br2(l) is 58.8C, and its molar enthalpy of vaporization is ΔHvap=29.6kJ/mol (a) When Br2(l) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8C .

Short Answer

Expert verified
During the boiling of Br2(l) at its normal boiling point, the entropy increases. The entropy change (∆S) when 1.00 mol of Br2(l) is vaporized at 58.8°C is 89.1 J/(mol·K).

Step by step solution

01

Part (a): Determining if Entropy Increases or Decreases

When a substance undergoes a phase change, such as boiling, the particles become more disordered and more dispersed. In the case of liquid bromine boiling and becoming gaseous bromine, the increased freedom of movement and spacing between the particles contribute to a higher entropy, which is a measure of disorder in a system. Thus, during the boiling of Br2(l) at its normal boiling point, the entropy increases.
02

Part (b): Calculate the Entropy Change (∆S)

Next, we calculate the entropy change (∆S) when 1.00 mol of Br2(l) is vaporized at its boiling point. We use the equation: ΔS=ΔHvapT We are given the molar enthalpy of vaporization ΔHvap=29.6kJ/mol and the boiling point temperature T=58.8C. First, we need to convert the temperature from Celsius to Kelvin: T(K)=T(C)+273.15 T(K)=58.8+273.15=331.95K Now, we apply the equation: ΔS=29.6kJ/mol331.95K Note that we need to convert kJ to J by multiplying with 1000: ΔS=29.6×1,000J/mol331.95K=89.1JmolK Therefore, the entropy change, ∆S, when 1.00 mol of Br2(l) is vaporized at its normal boiling point is 89.1 J/(mol·K).

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Key Concepts

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

Entropy
Entropy is a crucial concept in thermodynamics referring to the degree of disorder or randomness in a system. When a substance experiences a phase change, such as boiling, its entropy tends to increase. This is because the molecules in the gaseous state have more freedom to move compared to their arrangement in the liquid state.
For example, as bromine transitions from liquid to vapor, its molecules spread out becoming more disordered, thus leading to a rise in entropy. It's this tendency toward greater disorder that fundamentally drives many natural processes.
  • Bromine in liquid form is more structured than in vapor form.
  • When it boils, the increased molecular freedom results in higher entropy.
Molar Enthalpy
Molar enthalpy, specifically the molar enthalpy of vaporization, is the amount of energy required to vaporize one mole of a substance at its boiling point. It's an important measure of the energy change involved in phase changes.
For bromine (Br2(l)), the molar enthalpy of vaporization is given as 29.6 kJ/mol. Understanding this helps in calculating the change in entropy, as it directly relates to the energy absorbed during the boiling process.
  • It indicates the energy input needed for the phase change from liquid to gas.
  • The knowledge of this value aids in further thermodynamic calculations, such as entropy change.
Phase Change
Phase change, or phase transition, refers to the transformation from one state of matter to another. In the context of the exercise, bromine undergoes a phase change from liquid (Br2(l)) to gas (Br2(g)).
This process involves breaking intermolecular forces and requires specific energy inputs, as represented by the molar enthalpy of vaporization. The transition also involves changes in entropy, as the system gains disorder in the gaseous state.
  • Phase changes are significant in studying energy transfer and entropy changes.
  • Each type of phase change, such as melting or vaporization, has unique energy and volume implications.
Boiling Point
The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. For bromine (Br2(l)), this occurs at 58.8°C, where it transitions to vapor.
Understanding the boiling point is critical to evaluating how temperature influences the energy and entropy during phase changes. It allows us to pinpoint the conditions under which a substance changes its state from liquid to gas and enables precise thermodynamic calculations.
  • Indicates the temperature for the liquid-to-gas transition.
  • Influences both the enthalpy and entropy calculations of a substance.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of 4.2J/K; therefore, the entropy change of the surroundings must be 4.2J/K

From the values given for ΔH and ΔS, calculate ΔG for each of the following reactions at 298 K . If the reaction is not spontaneous under standard conditions at 298 K , at what temperature (if any) would the reaction become spontaneous?  (a) 2PbS(s)+3O2(g)2PbO(s)+2SO2(g)ΔH=844kk;ΔS=165J/K  (b) 2POCl3(g)2PCl3(g)+O2(g)ΔH=572kJ;ΔS=179J/K

A certain reaction has ΔH=+23.7kJ and ΔS= +52.4J/K . (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?(c) Calculate ΔG for the reaction at 298 K . (d) Is the reaction spontaneous at 298 K under standard conditions?

The potassium-ion concentration in blood plasma is about 5.0×103M, whereas the concentration in muscle-cell fluid is much greater (0.15M). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to K+. (a) What is ΔG for the transfer of 1 mol of K+ from blood plasma to the cellular fluid at body temperature 37C? (b) What is the minimum amount of work that must be used to transfer this K+?

Use data in Appendix C to calculate ΔH,ΔS, and ΔG at 25C for each of the following reactions.  (a) 4Cr(s)+3O2(g)2Cr2O3(s) (b) BaCO3(s)BaO(s)+CO2(g)  (c) 2P(s)+10HF(g)2PF5(g)+5H2(g) (d) K(s)+O2(g)KO2(s)

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