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CH 111, Week 7 Notes

by: Jordyn Meekma

CH 111, Week 7 Notes CH111

Jordyn Meekma

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About this Document

This covers the introductory material of thermochemistry
General Chemistry 1
Dr. Tom Getman
Class Notes
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This 2 page Class Notes was uploaded by Jordyn Meekma on Monday October 10, 2016. The Class Notes belongs to CH111 at Northern Michigan University taught by Dr. Tom Getman in Summer 2016. Since its upload, it has received 3 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.


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Date Created: 10/10/16
CH 111 Notes: Week 7 Chapter 5: Thermochemistry Involves energy changes during a chemical process Kinetic Energy: (KE) the energy of movement o KE = ½ mv 2 m = mass of molecules v = average velocity of molecules o KE is proportional to temperature (higher temperature = higher KE) Potential Energy: aka. Electrostatic Potential Energy (E ) el o Eel (Q x1Q )/2 Q 1 charge of species 1 Q 2 charge of species 2 d = distance between species o Too much distance = no attraction too little distance = repulsion o –Eeleans more stability  Chemical Systems o Open System: can exchange matter and energy with its surroundings o Closed System: can exchange energy but not matter with its surroundings o Isolated System: cannot exchange matter or energy with its surroundings  Purely hypothetical, but we approximate it over short periods of time  Internal Energy: (E) the sum of KE and E el o ∆E = E – E (change in internal energy) final initial o *The sign (+ or -) is important*  Units of Energy o calorie: (cal.) the amount of energy required to raise the temperature of 1 gram of water by 1ᵒC o Joule: (J) 4.184 J = 1 cal  ∆E = q + w q = heat change w = work = -P∆V P = pressure ∆V = change in volume o At constant pressure (open system) q = q p qp= ∆H  Change in Enthalpy: (∆H) refers to the amount of energy released or absorbed in a chemical reaction o ∆H = q =pq + P∆V o Exothermic: gives off heat, forms stronger bonds in the products  ∆H = (-) o Endothermic: absorbs heat, forms stronger bonds in the reactants  ∆H = (+)  A gas forming reaction produces 12.0 L of a gas against a constant pressure of 1.05 atm. Calculate the work done by the gas in joules. o w = -P∆V = -(1.05 atm)(12.0L – 0L) = -12.6 L*atm  101 J = 1 L*atm  -12.6 L*atm (101 J/1 L*atm) = -1.27x10 J3  Heating Curve for Water 100ᵒC ∆H vaporization0.67 kJ/mol ∆H condensation 40.67 ∆H fusion6.01 kJ/mol kJ/mol 0ᵒ ∆H solidification01 C kJ/mol  Molar heat capacity: (c ) (upits: J/mol*ᵒC) amount of energy required to raise 1 mole of a substance by 1 ᵒC o q = n*c p∆T n = # of moles ∆T = T final Tinitial)  Specific heat: (c )s(units: J/g*ᵒC) amount of energy required to raise 1 gram of a substance by 1ᵒC o q = m*c *∆T m = mass (grams) s  Heat capacity: (C ) (pnits: J/ᵒC) amount of energy required to raise the temperature of a piece of equipment by 1ᵒC o q = C p∆T  Molar heat capacity of water o cp(ice) = 37.1 J/mol*ᵒC o cp(liq. H 2) = 75.3 J/mol*ᵒC o cp(steam) = 33.6 J/mol*ᵒC  Calculate the amount of energy required to warm 58.0 g of ice from -10ᵒC to 25ᵒC o (convert grams to moles and J to kJ to keep units consistent) 58.0 g H 2 (1 mol H O/28.02 g) = 3.219 mol H O 2 cp(ice) = 0.0371 kJ/mol*ᵒC cp(liq H2O) = 0.0753 kJ/mol*ᵒC o q total q ice+ q fusion q liquid o q = [n*c pice)*∆T] + [n*∆H ] fusn*c (liqpH O)*∆2] o q = (3.219 mol)(0.0371 kJ/mol*ᵒC)(0ᵒC – (-10ᵒC)) + (3.219 mol)(6.01 kJ/mol) + (3.219 mol)(0.0753 kJ/mol*ᵒC)(25ᵒC - 0ᵒC) = 26.6 kJ  Calorimetry: the method of measuring a heat change of a chemical process o Calorimeter: apparatus used to measure heat changes during chemical processes


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