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Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\). (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix C. Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements Web site (www.webelements.com).

Short Answer

Expert verified
The estimated normal boiling point of Br$_2$ using Trouton's rule is 340.45 K, while the actual boiling point is 332 K, resulting in a difference of about 8 K. Possible sources of error in the calculation include the approximations made by Trouton's rule, the assumption that ∆H$_\text{vap}$ remains constant with temperature, and potential errors in the data provided in Appendix C.

Step by step solution

01

Find the enthalpy of vaporization for Br_2 from the data

Using Appendix C, we can find the enthalpy of vaporization for bromine. The Appendix gives the following information: ∆H of vaporization for Br_2 = 29.96 kJ/mol Now, we need to determine the actual boiling point of Br_2.
02

Use Trouton's rule to estimate the normal boiling point

According to Trouton's rule, the standard molar entropy of vaporization for many liquids at their normal boiling points is 88 J/mol-K. We can now relate this value to the enthalpy of vaporization we found earlier to estimate the boiling point: ∆S_vap = 88 J/mol-K ∆H_vap = 29,960 J/mol (convert kJ to J) Solving for the normal boiling point, we can use the following equation: T = ∆H_vap / ∆S_vap T = 29,960 J/mol / 88 J/mol-K T = 340.45 K So, the estimated normal boiling point of Br_2 is 340.45 K. #b) Compare the calculated value with the actual value and identify possible sources of error#
03

Look up the actual boiling point of Br_2

Referring to a chemistry handbook or a reliable online source like WebElements, find the actual normal boiling point of Br_2. According to WebElements, the actual normal boiling point of Br_2 is 332 K.
04

Compare and identify sources of error

Our estimated boiling point using Trouton's rule was 340.45 K, while the actual boiling point is 332 K. The difference between them is about 8 K. Possible sources of error in the calculation include: 1. Trouton's rule is an approximation that may not hold perfectly for all liquids. 2. Assumption that ∆H_vap remains constant with temperature, which might not be true in reality. 3. Errors in the data provided in Appendix C. As we can see, Trouton's rule provides a reasonable, but not perfectly accurate, estimate of the normal boiling point for Br_2.

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Key Concepts

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

Entropy of Vaporization
The concept of entropy is key in understanding why substances change states, like from liquid to gas. Entropy of vaporization refers to the amount of disorder or randomness in a system when a liquid evaporates into a gas. When a liquid becomes a gas, the molecules move from being closely packed to being spread out, increasing the disorder.

Trouton's Rule provides a handy estimate for this increase in disorder. It states that the standard molar entropy of vaporization for many liquids is approximately 88 J/mol-K at their normal boiling point.

This constant is derived from observed values in various substances and offers a useful way to estimate the boiling point when the enthalpy of vaporization is known. It’s worth noting that this value is an average and might not be perfectly accurate for all liquids, as different substances may have unique intermolecular forces affecting their entropy of vaporization.

Understanding the entropy of vaporization helps in predicting how much energy a substance will require to transform from liquid to vapor, providing insights into molecular interactions during phase transitions.
Enthalpy of Vaporization
Enthalpy of vaporization, also known as heat of vaporization, is the energy required to convert a liquid into a gas at a constant pressure. This value is crucial for calculating boiling points and understanding how substances absorb heat during phase changes.

For instance, in the exercise, the enthalpy of vaporization for bromine ( Br_2 ) was given as 29.96 kJ/mol. To make further calculations using Trouton's Rule, it’s typically converted into joules, resulting in 29,960 J/mol.

In context, this energy input allows bromine molecules to overcome intermolecular forces, making the transition into the gaseous phase possible. The enthalpy of vaporization remains relatively constant with temperature for many substances, which simplifies the process of estimating the normal boiling point when combined with the entropy of vaporization.
  • Reflects the "strength" of intermolecular forces
  • Provides a basis for calculating boiling points under constant pressure
  • Converted into joules for easy integration with entropy equations
The assumption of consistency in the enthalpy of vaporization helps in making practical calculations, although real-world deviations might occur due to varying conditions.
Boiling Point Estimation
Estimating the boiling point of a liquid involves understanding its enthalpy and entropy of vaporization. Trouton's Rule can be applied to make this estimation easier.

This rule states that the ratio of the enthalpy of vaporization (delta H_{vap}) to the entropy of vaporization (delta S_{vap}) yields the boiling point in Kelvin. In mathematical terms, it can be represented as:

\[T = \frac{\Delta H_{vap}}{\Delta S_{vap}}\]
Substituting the known values for bromine, like 29,960 J/mol for \Delta H_{vap} and 88 J/mol-K for \Delta S_{vap}, the estimated boiling point is calculated.

One must consider potential errors, as seen in the example where the estimate was slightly off from the actual boiling point of bromine (332 K). Such discrepancies might arise due to:
  • Inaccuracies in entropy/enthalpy values
  • Approximation limitations of Trouton's Rule
  • Pressure or temperature variations not accounted for in standard calculations
Despite these potential sources of error, Trouton's Rule gives a good approximation for practical purposes, explaining why it continues to be a valuable tool in thermodynamics.

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Most popular questions from this chapter

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2} .\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ},\) as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H\) is positive. Is this process an exception to Bertholet's generalization that all spontaneous changes are exothermic? Explain.

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