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# UK - CHE 107 - Exam 2 - Study Guide

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UK - CHE 107 - Exam 2 - Study Guide

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##### Description: Study guide for Exam 2
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Unformatted text preview: chibo CHE 107 O Entropysland Free Energy G) * The Entropy of a substance de cero on the thermal energy and, particte arrangemente Phase changes affect Dayti dispersal the temperature chunges off the speed and thermal energy of the particles. Example: 1) N2 ,75 K ) Nola look) the nitrogen atoms are more dispersed in the gaseous Phase: Entropy will increase 2) Co, la 57 C) ta cos Is. -80 C) o as temperature decreases the molecules slow down to a more ordered, less chaotic state, Entropy will decrease More atorns per molecule and higher molecular mass give rise to higher entropy. Standard Entropy change equationi As yn = En sop products - E s (reactants) S Standard molar en tropy n and m = reaction coefficientsExample HCl(a) + Nad NaOH(s) -- Nads the Asprm =[kielle (po la -[1 mol) [186.9 km) + line = -109,2 J/k Gibbs energy of formation valves is tound using Gibbs energy of reaction equation A Guryn E. GO (products)- GP (reactants) Spontaneous reactions have a negative value for the Gibbs energy chang . \$ Enthalpy or entropy alone does not deter reaction spontaneity, but one can identify the signs of AHans As thet contribute favorably to spontaneity. Gibbs equution describes the relationshi between free energy, enthalpy and entros AGAH-TASChill Solutions and factors and Mulurity and molality reviews 4 A solvent is a component of a sovu that is present in the greatest amount 8 A solute is a component that is present in a smaller amount compared to the solvent Examples A solution containing 70.0 mb of ethanolland 40.0mL of water. Ethanol = solvent water = solute * though water is often used as a solvent, it is not alwayes a solvent. Henry's lawi gas pressure lp and solubility Is) are directly proportionali SEK P R = Henry's Law constant at particular temperature To find solubility or gas pressure of a compound if k'is constanti SL-S2 or SR P2 Examples Carbon dioxide - Si=0,035\M Pi= 785 torr and Py = 2050 tor. Sz? Post [2050 tar(6.0351 M) 52 (785 torr = 0;0917 M

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