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CHMA11H3 Lecture Notes - Lecture 19: Boltzmann Equation, Spontaneous Process, Thermodynamics

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School
UTSC
Department
Chemistry
Course
CHMA11H3
Professor
Nirusha Thavarajah

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CHMA11H3 Full Course Notes
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CHMA11H3 Full Course Notes
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Document Summary

Spontaneous: is the direction of the reaction that would take a reaction mixture closer to a state of chemical equilibrium: most are exothermic but some are endothermic, enthalpy is not the only reason for spontaneity. Spontaneous process: is a process that occurs without any outside interventions. The fundamental goal of thermodynamics is to predict spontaneity. 17. 2 - entropy and the second law of thermodynamics: Entropy (s): is the measure of the lack of order due to the dispersal of energy (j) and matter per unit temperature (k). Entropy is a state function, which means that it depends on where a reaction begins and finishes. The second law of thermodynamics: is any spontaneous process that causes an increase in the entropy of the universe. The third law of thermodynamics: there is no disorder in a perfect crystal at 0 k, or s = 0. The reference point for entropy is provided by this law.

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Related Questions

Model 1 – Spontaneous Processes Process Description Change in Enthalpy (∆H) Change in Entropy (∆S) Spontaneous? A Two pure → Homogeneous substances mixture ~ 0 Increasing Yes B Solute on Solute on one side of → both sides of membrane membrane ~ 0 Increasing Yes C Polypeptide → Individual chain amino acids Exothermic (negative) Increasing (positive) Yes D C3 H8 + 5O2 → 3CO2 + 4H2 O Exothermic Increasing Yes E 6CO2 + 6H2 O → C6 H12O6 + 6O2 Endothermic Decreasing No F Glucose → Starch Endothermic Decreasing No G Water → Ice Exothermic Decreasing Below 0 °C, yes Above 0 °C, no H Cold water (25 °C) → Hot water (60 °C) Endothermic Increasing Below 60 °C, no Above 60 °C, yes

Some of the processes in Model 1 are spontaneous—that is, they will occur without any additional work being done on the system. For example, a solute will diffuse across a membrane until the concentrations of both sides are equal. However, glucose will not spontaneously form from carbon dioxide and water in the atmosphere. The Second Law of Thermodynamics states that a process will be spontaneous when it results in an increase of total entropy in the universe. In other words, either the system or the surroundings must have an increase in entropy, or both. (Please note that the term “spontaneous” does not imply that the change will happen quickly. Rusting is spontaneous under the right conditions, but it can still occur very slowly.)

13. Predict the change in entropy of the surroundings for an endothermic reaction.

Answer for PART A and PART B

Entropy and the Second Law of Thermodynamics


According to the second law of thermodynamics, all reactionsproceed spontaneously in the direction that increases totalentropy. The total entropy is defined as


%u0394Stotal=%u0394Ssystem+%u0394Ssurroundings

The direction of a spontaneous change is always determined bythe sign of the total entropy change:


-The reaction is spontaneous if %u0394Stotal>0.

-The reaction is nonspontaneous if %u0394Stotal<0.

-The reaction is at equilibrium if %u0394Stotal=0.


The value of %u0394Stotal depends on %u0394Ssystem and%u0394Ssurroundings, which are defined by%u0394Ssystem=S(products)%u2212S(reactants) and%u0394Ssurroundings= %u2212%u0394Hsystem/T


Spontaneity is therefore affected by the enthalpy of the systemand the Kelvin temperature.




********PartA
CaO can be used as a drying agent. One suchapplication occurs when water is added to dry concrete or cement.The reaction that occurs is

CaO(s)+H2O(l)%u21CCCa(OH)2(s)

The product is commonly called slaked lime.

Assuming the commonly used standard-state temperature of25%u2218C,calculate %u0394Stotal for this reaction using tablefrom the table below.

Substance S%u2218
[J/(K%u22C5mol)] 		%u0394H%u2218f
(kJ/mol)
CaO(s) 39.9 %u2212635.1
H2O(l) 69.9 %u2212285.8
Ca(OH)2(s) 83.4 %u2212986.1
Express your answer to threesignificant figures and include the appropriate units.



*********PartB
A particular solvent with %u0394S%u2218vap=112.9J/(K%u22C5mol) and %u0394H%u2218vap=38.0kJ/mol is being consideredfor an experiment. In the experiment, which is to be run at 75%u2218C, the solvent must not boil. Basedon the total entropy change associated with the vaporizationreaction, would this solvent be suitable and why or why not?
The solvent would
A particular solvent with and is being considered for anexperiment. In the experiment, which is to be run at 75 , thesolvent must not boil. Based on the total entropy change associatedwith the vaporization reaction, would this solvent be suitable andwhy or why not?
be suitable because %u0394Stotal>0.
not be suitable because %u0394Stotal>0.
be suitable because %u0394Stotal<0.
not be suitable because %u0394Stotal<0.