Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J/mol‐K. b. Look up the normal boiling point of Br 2 in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

hey everyone, we're told that trodden rule states that the standard molar entropy of vaporization for many liquids at their normal boiling point is about 88 joules per mole times kelvin. What is the normal boiling point of iodine? Assuming that trout and rule applies that it's standard entropy of vaporization remains constant with temperature. So here we have the vaporization of iodine. Writing out this reaction. We have iodine in its liquid state and this will vaporize and turn into its gaseous state. As we've learned, the standard entropy of vaporization is going to be equal to our products minus our reactant. So in this case we have our standard NLP of our iodine in its gaseous state minus the standard entropy of iodine in its liquid state. Now plugging in these values we have 62.4 kg joules per mole minus 13.5 kg jewels Permal which was provided to us in our question stem this will get us to a total of 48.9 kg joules per mole. Now to calculate our boiling point, we've learned that the formula is going to be the standard entropy of vaporization divided by the standard entropy of vaporization plugging in these values we have 48.9 kg joules per mole and we want to convert this into jewels so we know that we have 1000 jewels per one kg jule. Next we're going to divide this all By our standard molar entropy of vaporization which was 88 joules per mole kelvin and when we calculate this out and cancel out all of our units, we end up with a total of 5.6 times 10 squared Calvin, which is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.

Ch.19 - Chemical Thermodynamics

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Brown 14th Edition

Ch.19 - Chemical Thermodynamics

Problem 93b

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Chapter 19, Problem 93b

Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J/mol‐K. b. Look up the normal boiling point of Br₂ in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

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Identify the normal boiling point of Br2 (bromine) from a reliable source such as a chemistry handbook or the WebElements website.

Calculate the standard molar entropy of vaporization for Br2 using Trouton’s rule, which suggests it should be approximately 88 J/mol-K at the normal boiling point.

Compare the calculated value of the standard molar entropy of vaporization with the value obtained from experimental data or literature.

Analyze possible sources of error in the calculation, such as deviations of Br2 from ideal behavior, experimental errors in measuring boiling point, or inaccuracies in the Trouton's rule approximation for certain types of liquids.

Consider any incorrect assumptions that might have been made in applying Trouton’s rule to Br2, such as ignoring molecular interactions or the specific molecular structure of bromine that could affect its entropy of vaporization.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Trouton's Rule

Trouton's Rule states that the standard molar entropy of vaporization for many liquids at their normal boiling points is approximately 88 J/mol·K. This empirical rule helps predict the entropy change associated with the phase transition from liquid to gas, providing a useful benchmark for comparing different substances. Understanding this rule is essential for analyzing the thermodynamic properties of liquids and their vaporization processes.

Normal Boiling Point

The normal boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure (1 atm). At this point, the liquid phase transitions to the gas phase. Knowing the normal boiling point is crucial for understanding the conditions under which a liquid will vaporize and for applying Trouton's Rule effectively in calculations involving entropy.

Sources of Error in Calculations

When calculating thermodynamic properties like entropy of vaporization, potential sources of error include assumptions about ideal behavior, inaccuracies in temperature measurements, and variations in atmospheric pressure. Additionally, the presence of impurities or deviations from ideal gas behavior can lead to discrepancies between calculated and observed values. Recognizing these factors is important for critically evaluating experimental results and improving the accuracy of thermodynamic predictions.

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Related Practice

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of ΔS for this dissolution process?

Textbook Question

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as CH₂F₂. An air-conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram).

During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why?

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Textbook Question

During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (e) Suppose that a house and its exterior are both initially at 31 °C. Some time after the air conditioner is turned on, the house is cooled to 24 °C. Is this process spontaneous or nonspontaneous?

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Textbook Question

(c) In general, under which condition is ΔG°_f more positive (less negative) than ΔH°_f ? (i) When the temperature is high, (ii) when the reaction is reversible, (iii) when ΔS°_f is negative.

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Textbook Question

Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

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Textbook Question

Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (ii) C₂H₆(g) + 7 Cl₂(g) → 2 CCl₄(g) + 6 HCl(g) (iii) BaO(s) + CO₂(g) → BaCO₃(s) (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

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