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Chemistry 170 Exam 4 study guide

by: Joseph Notetaker

Chemistry 170 Exam 4 study guide 46657 - CHM 170 - 002

Marketplace > Missouri State University > 46657 - CHM 170 - 002 > Chemistry 170 Exam 4 study guide
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About this Document

This is the material that will be on exam four. Most of it is already on the notes from last week, but there is some material that has been added
General Chemistry 2
Mark M Richter
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This 4 page Study Guide was uploaded by Joseph Notetaker on Monday April 4, 2016. The Study Guide belongs to 46657 - CHM 170 - 002 at Missouri State University taught by Mark M Richter in Spring 2016. Since its upload, it has received 11 views.


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Date Created: 04/04/16
CHM 170 Chapter 17 Part 1 H-Enthalpy S-Entropy G-Gibbs Free Energy Energy changes that occur during a chemical reaction are due to the making and breaking of chemical bonds The first law of thermodynamics states that energy is conserved ΔETotal = Δ Eystem + Δ SurEoundings = 0 If the system absorbs energy it’s represented by a positive value (+) If the system releases energy it’s represented by a negative value (-) The second law of thermodynamics states that energy is most stable in a randomized state This means that a reaction can never be 100% efficient because some energy will always be lost due to entropy Thermodynamics predicts whether a process will occur under given conditions -spontaneous reactions (Thermodynamically favorable) do not require external stimuli -nonspontaneous reactions require external stimuli In all reactions one direction is spontaneous and the other is nonspontaneous The direction of spontaneity can be determined by comparing the potential energy of the system at the start and at the end Spontaneity is determined by comparing the chemical potential before the reaction with the potential energy after the reaction. Spontaneity ≠ speed of the reaction A nonspontaneous process can occur by coupling it to a process that is spontaneous or by supplying energy from an external source Spontaneous processes release energy Most go from higher energy to lower energy (Exothermic) But some go from lower energy to higher energy (Endothermic) Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases S = klnW -23 k=1.38 * 10 Joules/Kelvin W= # of energetically equivalent ways to arrange the compounds of the system Macrostates are concerned with the entire system Microstates are concerned with the actual arrangements (this is the actual W measurement) ΔS = S final Sinitial ΔS > 0 for spontaneous reactions ΔS > 0 for more disorder ΔS < 0 for more order Entropy can be increased by increasing particles or increasing temperature N 2 3H =22NH 3 This is a system of decreasing entropy because a set of four moles has to be combined into a set of two moles When the entropy change in the system is unfavorable (negative) the entropy change in the surroundings must be positive ΔS surroundingsΔH systemtemperature Gibbs Free energy Free energy refers to energy that is available to do work ΔG=ΔH-TΔS If the answer is negative the reaction is spontaneous In the example ΔS O refers to the standard state of entropy RXN The third law of thermodynamics states that In a perfect crystal at absolute zero the entropy is zero This means ΔS is always positive Entropy is affected by -State of substance -Gas has higher entropy than liquid -Molar mass of the substance -Larger the molar mass the greater the entropy -The particular Allotrope (Elemental arrangement) -The less constrained the greater the entropy -Molecular complexity More complexity = More vibrations = More entropy -Extent of dissolution -dissolved solids have larger entropy The equation for ΔSORXNis as follows: ∆ S = n ∆S O(product)− n ∆S (reactants) RXN ∑ p ∑ r N= Stoichiometric Coefficient P= products R= reactants O The values of ΔS will be provided for you as the are determined experimentally ΔG ORXNefers to the standard state of Gibbs Free Energy there are 3 ways to calculate this: O O O 1. ∆ GRXN=∆H RXN−T ∆S RXN ∆ GO = n ∆G O(product)− n ∆G (reactants) 2. RXN ∑ p ∑ r ∆ GRXN=∆G Reaction1 Reaction2. 3. However Usually a reaction doesn’t occur under standard conditions and for that problem you require this equation: O ∆ GRXN∆G RXNRTlnQ R= Gas constant(8.314 J/mole) [C][D] Q= Reaction Quotient [ ][B]


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