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General Chemistry 202 week 3 test 2

by: Catherine Montz

General Chemistry 202 week 3 test 2 Chem 202

Marketplace > University of Louisville > Chemistry > Chem 202 > General Chemistry 202 week 3 test 2
Catherine Montz
U of L

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About this Document

These notes cover chapters 44 and 45
General Chemistry 2
Dr. Kuta
Class Notes
chem 202, Study Guide, chemistry 2, Test 2
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This 4 page Class Notes was uploaded by Catherine Montz on Monday February 29, 2016. The Class Notes belongs to Chem 202 at University of Louisville taught by Dr. Kuta in Spring 2016. Since its upload, it has received 30 views. For similar materials see General Chemistry 2 in Chemistry at University of Louisville.


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Date Created: 02/29/16
Chem 202 Dr. Kuta Test 2 Chapters 44 and 45 Chapter 44 44.1 - Entropy, Qualitatively There is a simple equation for entropy S = k lnW S = entropy k = Boltzmann constant (1.381 x 10 -23J/K) without thermal energy then W can exist as 1 There is a connection between energy and temperature within this equation Entropy - is a measure of the options for the motional energy of a system at some specified temperature Exothermic - J’s leave the system Endothermic - J’s enter the system 44.2 - Entropy Quantitatively Only final and initial states count These are the equations you will need to answer any problems, it is very simplistic. o H = H final Hinitial o U = U - U final initial o S = S finalSinitial Enthalpies = J and kJ’s Entropy = J/K Standard Molar Entropies - actual values for S of substances are measureable. They are measured as a function of temperature and the energy which is required to heat a mole of sample from 0 K to that temperature All pure substances have positive S but solutes can have negative or positive Lets look at this chart and see how entropy is in different states of matter Phase solid liquid Gas Entropy Little Moderate A lot IFs stronger Somewhat strong Very little if any Dissolving of gases into water is always disfavored by entropy, along with liquids that are primarily hydrocarbon deadspace. Two points for ionics: o it is possible to assign entropies to individual ions in solution, you do this by assigning a zero value and all ions are measured relative to it. o We assume full dissociation, so we can represent the ionic solute as either a neutral compound or as separate ions Chapter 45 45.1 - Spontaneous Spontaneous- can happen on its own under certain conditions Spontaneity - capability to occur 45.2 - A Universe of Options universe = system + surroundings total = system + surroundings H sys -H surr 45.3 - The Law The law comes down to the reaction or process must result in more options overall S must be positive, if the reaction leads to negative than it cannot happen on its own under given conditions. Transfer of heat causes a change in entropy given by S = H/ T Whoever is gaining the heat is also gaining entropy. Whoever is losing heat, is losing entropy. Total change in entropy S univerives from two sources: o Reactions cause chemical identities to change, so the entropy changes and the change is designated as S sys o The reaction may cause heat to exchange with the surroundings. The gain or loss of heat by surroundings is given by S surr o This can be summarized as S univ =S sys S surr “Can happen” requires and increase in S univo it must be positive Change in entropy of the surroundings due to het transfer S surr= H surrT Energy conservation requires that H = -H surr sys So we combine these to get S surr -H sys/ T S univS sys- Hsys/ T This equation provides us with an answer if a reaction is allowed or not, however we need to go further and work with energy units so our equation becomes TS = TS - H / T multiply everyone times temperature and your univ sys sys answer will need to be in J’s or kJ. When energy is released it is negative so then we get the equation -TS univ= H sys- TSsys One more final step and we get G sys H sys- TSsys G is called Gibbs energy, this is our working equation for the law regarding whether or not a reaction can happen or not. Here is some cases you need to be familiar with CASE 1 o S univis positive which means G sysis negative o S univincreases during the process o G decreases during the process sys o It can happen by itself under given conditions o Its allowed o Its spontaneous o Its exergonic CASE 2 o S univs negative which means G syss positive o S univecreases during the process o G sysincreases during the process o The process cannot happen by itself under given conditions o Process is forbidden o It is not spontaneous o Its endergonic CASE 3 o S univs zero, so G syss zero o S stays the same univ o G sysstays the same Generally this is an equilibrium condition 45.4 - Free Energy, Qualitatively Three variations: 1 - process is exothermic and system entropy increases (-H ) - (+sys ) = sys G sys 2- process is endothermic and system entropy decreases (+H ) - (-TSsys sys = G sys 3- process is exothermic and system entropy decrease G = (­)H ­ T(­)S 4­ process is endothermic and system entropy increases  G= (+)H ­ T(+)S G = H ­ TS  under standard conditions  It is simple plug and chug for these equations to find the quantitatively 


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