Chem 222 week 11 notes
Chem 222 week 11 notes Chem 222
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This 3 page Class Notes was uploaded by Leslie Pike on Friday April 15, 2016. The Class Notes belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 20 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.
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Date Created: 04/15/16
Chem 222 notes Week 11 The entropy of a solid is less than the entropy of a liquid, which is much less than the entropy of a gas. A gas has very, very high entropy. If the change in entropy is positive, the system is becoming more disordered. If the change in entropy is negative, the system is becoming more ordered. A positive ΔS means that more disorder has been added to the system. A negative ΔS means that disorder has been lost (the system has become more ordered). Entropy can be calculated by: S=k∗ln ?(W) Here, k is a constant. W is the number of microstates of a system. As W increases, so does entropy. To illustrate: I have two pairs of shoes at school with me. I’m not the most organized person and my shoes tend to get strung out all over the room, but since I only have four shoes total, they don’t make much of a mess. I have four microstates (four shoes) and low entropy. My roommate has lots of shoes—let’s say 7 pairs. She is not the most organized person either, but her shoes make a considerable mess on account of there being so many of them. She has fourteen microstates (fourteen shoes) and high entropy. Basically, the more stuff you have, the bigger potential mess you can make, and the higher the entropy. Thus, bigger molecules will have higher entropies than smaller ones (bigger molecules have more protons, neutrons, and electrons, and thus more microstates). At absolute zero, the entropy is zero. This is because, at absolute zero, there is only one microstate (everything is completely still) and the natural log of one is zero. Adding more heat to a system will cause an increase in entropy (ΔS will be positive) because things move around more when they are heated, meaning you get more microstates. A phase change of a solid to a liquid or gas gives ΔS > 0 because liquids and gases are more disordered than solids, and vice versa. Any element in its elemental state has an enthalpy of formation (ΔH ) of f. Thus, graphite, O , and other standard states have no ΔH and they can be ignored in 2 calculations of ΔH .rxnese elements in their standard states DO have a ΔS, because ΔS and ΔH are not the same thing. ΔH is energy gained or lost (aka heat, thus the capital H), and ΔS is disorder gained or lost. You can remember that disorder is symbolized with a capital S because the word disorder contains the letter “s” and the word “heat” doesn’t. For a given reaction, increase in the number of moles of gas means a positive ΔS. Decrease in the number of moles of gas means a negative ΔS. When no gas is involved, ΔS will be small. (Remember this if we get a test question that asks which reaction has the least ΔS; the correct answer will be the reaction which involves no gas molecules.) ΔS rxnis the standard entropy change for a reaction that takes place at 1 atm and 298 K. It can be calculated by the sum of the ΔS of the products, minus the sum of the ΔS of the reactants. Remember to multiply by the balancing coefficients. To determine whether or not a reaction is spontaneous at a given temperature, we calculate the Gibbs free energy for that reaction. Gibbs free energy is calculated with the following equation: ∆ G=∆H−T ∆S If the ΔG is negative, that means that energy is released by the reaction when it proceeds, and thus the reaction is spontaneous. Imagine a child on a sliding board. The child slides down spontaneously (without any outside work). The child loses energy in the act of sliding down (potential energy of the child at the top is transferred to the slide and the child’s shirt as heat). If ΔG is positive, that means that energy has to be added in order for the reaction to proceed. The child is not going to slide up the sliding board unless somebody is pushing him or her. If ΔG is zero, the reaction is at equilibrium and it does not proceed either way. A child sitting at the bottom of a sliding board won’t go up or down. ΔG 0fis the standard free energy of formation of one mole of a material. It is calculated by summing the ΔG’s of one mole of all the elements that make up the material. Superscript zeros mean standard conditions, temperature 298 K and pressure 1 atm. 0 0 0 Sample problem: Calculate ΔG for a given reaction. ΔS is 42 J/(K*mol) and ΔH is 35 kJ/mol. First step: figure out what you have, what you need, and what equation you can use to get from what you have to what you need. You have ΔS and ΔH, and you need ΔG. So, you can use the following equation: ∆G=∆H-T∆S This looks fine, except you weren’t given T. That’s okay, you don’t need to be given T. The superscript zeroes should clue you in to the fact that you are at standard conditions, which means that T is 298 K. So, you can just start plugging in numbers now, right? WRONG!!!! Before you plug anything in, make sure that it is in the proper units. ∆S was given to you with J, and ∆H with kJ. Convert kJ to J before plugging in your numbers. ∆G=∆H-T∆S=35000-298*42=22484 J/mol The equilibrium constant can be calculated from the Gibbs free energy with the following equation: ∆ G =−RT∗ln(K) Here, R=8.314 J/(mol*K). Common pitfall: be sure to convert C to K and kJ to J before you start plugging numbers in. 0 If K is greater than 1, the natural log will be positive, ∆G will be negative, and the reaction proceeds spontaneously to the right. If K is less than 1, the natural log will be negative, ∆G will be positive, and the reaction will not proceed spontaneously to the right (in fact, it will proceed spontaneously to the left). Chapter 19: Electrochemistry Electrochemistry deals with redox reactions. Electrochemistry deals with one of two situations: a reaction making electricity, or electricity being used to force a reaction to go in a certain direction. It is necessary that you understand oxidation states in order to understand redox reactions. An oxidation state is the electrical charge on an atom or molecule. Free elements (Pd , (s) 2 (g)have an oxidation state of 0 because they have no charge. Monatomic ions have an oxidation state equal to their superscript number (Fe3+ has an oxidation state of +3) Alkali metals (in solution or in compounds) have an oxidation state of +1 Alkali earth metals (in solution or in compounds) have an oxidation state of +2 Fluorine (unless it is a gas) ALWAYS, ALWAYS, ALWAYS has an oxidation state of -1. All other halogens also have an oxidation state of -1, unless they are bonded to fluorine. Oxygen (unless it is a gas) has an oxidation state of -2. Hydrogen has an oxidation state of -1, unless bonded with a halogen, then it is +1. All other elements: it depends. Oxidation states do not have to be whole numbers.
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