Chemistry Study guide for Chapter 19
Chemistry Study guide for Chapter 19 CHEM 1080
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This 4 page Study Guide was uploaded by Rachel Notetaker on Sunday February 28, 2016. The Study Guide belongs to CHEM 1080 at Tulane University taught by James Donahue in Spring 2016. Since its upload, it has received 34 views. For similar materials see General Chemistry II in Chemistry at Tulane University.
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Date Created: 02/28/16
Chapter 19 Spontaneous process- a process that occurs in a system left to itself Nonspontaneous process- will not occur unless some external action is continuously applied o Spontaneous does not mean fast Entropy- the thermodynamic property related to the way in which the energy of a system is distributed among the available microscopic energy levels o In some cases, if it is positive, that means that the reaction is spontaneous Microstates o The more states a given number of particles can accupt, the more microstates the system has o The particular way a number of particles is distributed among these states o S (entropy) = k (constant) ln W (microstates) This is Boltzmann’s equation It provides a connection between the macroscopic description of a system(P, V, T) and the microscopic state, which involves knowledge of the position and velocity of every particle in the system Entropy change o Based on heat and temperature o ∆S = q reversible Entropy is directly proportional to heat because the higher the heat, the greater microstates The heat must be reversible to specify the pathway that the system takes. This is because to be a state function (which S is), the system must be independent of the path by which heat is lost o Situations where there is an increase in entropy Pure liquids are formed from solids Gases are formed from either solids or liquids The number of molecules of gas increases as a result of a chemical reaction The temperature of a substance increase Turn to p. 826 for an example o Entropy changes are on a per mole basis Trouton’s Rule: a useful generalization For many liquids at their normal boiling points, the standard molar entropy of vaporization has the value of about 87 J/mol*K ∆S vapas the same value for the two but the entropy of vapor produced by the vaporization of solvent form the ideal solution is also greater than the entropy of the vapor obtained from the pure solvent Absolute Entropy o We look for a condition in which the substance is in its lowest possible energy state We then evaluate entropy changes as the substance is brought to other conditions of temperature and pressure-we add together these entropy changes and obtain the absolute entropy o Third law of thermodynamics: the entropy of a perfect crystal at 0 K is zero o Over temperature ranges in which there are no transitions ∆S⁰ values are obtained from measurments of specific heats as a function of temperature o S⁰- standard molar entropies of substances at 25⁰ C can be used to calculate the entropy change of a reaction— ∆S⁰= [∑v p⁰(products) - ∑v S⁰r(reactants)] v being the stoichiometric coefficients o Explanation of why NO ha2 more entropy than NO Because entropy increases when a substance absorbs heat and some of this heat goes into raising the average translational kinetic energies of molecules NO 2as more types of motion (3 as opposed to NO’s 1) Vibrational motion can happen between 3 molecules instead of just 2 and each has a different direction that it can go More possible ways to distribute energy among NO 2 molecules Second law of thermodynamics o Two entropy changes must always be considered simultaneously- the entropy of the system and the change of the surroundings ∆S = ∆S system∆S surroundings If both changes are positive, it is spontaneous and If both are negative it is nonspontaneous Water, has a negative entropy change in the system and a positive entropy change in the surroundings (absorbing the heat) o The freezing of water below 0 ⁰ C is spontaneous o All spontaneous processes produce an increase in the entropy of the universe o Gibbs energy ∆S surroundingsH systemT This makes it easier to solve for ∆S total because finding the ∆S surroundings is really hard This is because ∆H for the system is just the reverse for it for the surroundings, so if you make it negative, it works! If you then divide everything by T and adjust the signs you get o -T∆Suniv = ∆Hsys - T∆Ssys so everything is in terms of the system ∆G (gibbs energy change) = ∆H - T∆S and so ∆G = -T∆S ∆G is negative when ∆S is positive is spontaneous ∆G is positive, process is nonspontaneous ∆G= 0, process is at equilibrium Look at ttable 19.1 on page 834 to see all rules of spontaneity If H and S are both negative, it is usually spontaneous at lower temperatures If H and S are both positive, it is usually spontaneous at higher temperatures Gibbs energy and work If an exothermic reaction increases in entropy, the amount of energy available to do work in the surroundings is greater than - ∆H If entropy decreases the amount of energy available to do work is less that - ∆H The amount of work that can be extracted from a chemical process is - ∆G o - ∆G represents the energy freely availale for doing work o Possible to do work in an endothermic process if T∆S exceeds ∆H o Standard gibbs energy change ∆G⁰ This corresponds to the reactants and products in their standard states Leads to values of zero for the gibbs energies of formation of the elements in their reference forms ∆G changes sign when a process is reversed Subtracting reactants from products can only be used at standard conditions whereas ∆G⁰ = ∆H⁰ - T∆S⁰ can be used at any temperature o @ Equilibrium If ∆G = 0 the forward and reverse reactions are just as likely to occur As the temperature increases, the magnitude of ∆G decreases At a higher temp, the magnitude of T∆S exceeds the magnitude of ∆H, therefore ∆G is negative and spontaneous On a graph, the point at which the lines for ∆H and T∆S intersect is when the system is at equilibrium Always pay attention to what the temperature is because some reactions with a positive ∆G mightbe more more likely to occur at a different temp (water vaporizes spontaneously even though ∆G=+but only at high temp or dif. P o Nonstandard conditions S=S⁰ - RlnP ∆G = ∆G⁰ + RT lnQ can use if the temp and P are constant Q=K at equilibrium and ∆G=0 so ∆G⁰ = -RT ln K Activities A = the effective concentration of a substance in the system/the effective concentration of that substance in a standard reference state Ex. Activity of .1 M solution of HCl in water is .1M/1M = .1 ∆G⁰ and K as functions of temp o Assume ∆H⁰ is independent of temp and ∆S⁰ o BUT T∆S is strongly temp dependent which makes ∆G⁰ dependent on temp o So if you are trying to find a temperature when K is a certain value, you must use both ∆G⁰ = ∆H⁰ - T∆S⁰ and ∆G⁰= -RT ln K to find T b/c H and S are temp dependent and you already know the K you want
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