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Chapter 6 Chemistry Notes

by: mkennedy24

Chapter 6 Chemistry Notes Chem 109

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These notes cover material from both lecture and the textbook. Examples and visuals are also included.
General Chemistry
Eric Malina
Class Notes
General Chemistry, Chem 109, Chemistry
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This 7 page Class Notes was uploaded by mkennedy24 on Thursday March 3, 2016. The Class Notes belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 43 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.


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Date Created: 03/03/16
Chapter 6: Thermochemistry Section 6.1: Hand warmers  Thermochemistry: Study of relationships between chemistry and energy This is the reaction that 4Fe(s) + 3O (2)  Fe O2(3) occurs in the hand warmer packages when used. But where does the heat come from?? o There is certain energy in each component of the reaction but the overall energy on the right side (reactants) is not equal to the overall energy on the left side (products), therefore: 4Fe(s) + 3O (2)  Fe O2(3) + HEAT Section 6.2: Terms and Definitions  Energy: Ability to do work o Energy  something objects or sets of objects possess o Heat & Work  ways that objects or sets of objects exchange energy The blue ball has As the balls collide, After exchanging energy, kinetic energy while they exchange because energy is not the purple ball has energy created nor destroyed, the potential blue ball exchanged energy with the purple ball forcing the purple ball to move  Kinetic Energy versus Potential Energy o Kinetic Energy: Due to motion  Thermal Energy  associated with temperature.  Type of kinetic energy  Has to do with motion of atom/ molecules in substance o Potential Energy: Due to the amount of energy stored/ built up. Due to position or composition 8 Raising an object or a set of objects against gravitational pull increases potential energy  Chemical Energy  Associated with position of electrons to the nucleus  Type of potential energy  The atom/molecules has stored energy unless it is released due to a reaction  Atom  Electron Configurations  Molecules  Bonds present (covalent, polar, non- polar, etc.)  Mixtures  Intermolecular Forces  Law of Conservation of Energy o Energy cannot be created nor destroyed o Energy can be transferred or transformed o To help study energy. . .  System: Thing(s) under study  Surroundings: Everything else in the universe  Units of Energy applied∈KE= mv1 2 2 o Joule(J): 2 kg∙m →¿ s2 o Calorie(cal) or “Thermodynamic calorie”: Energy required to raise 1 gram of water by 1 ℃→4.184 J=1cal o Food calories (Cal): Kilocalorie = 1000 cal o 1 Kilowatt-hour (kWh)= 3.6 x 10 J 5 Section 6.3: First Law of Thermodynamics  First Law of Thermodynamics: The Total Energy of the Universe is **The Law of Conservation of Energy is basically describing the First Law of Thermodynamics in different words**  Internal Energy ( ∆ E ):Sum of all kinetic energy and all potential energy in a system o State Function: Only dependent upon current state   Example: The professor stands in front of class on Wednesday. He flies to Omaha later that day. On Friday, the professor is back in the same spot in front of class. The professor’s total position is 0 and therefore is a state function  Money is not a state function. o ∆ E=E Final Initial o C(s) + O (2)  CO (g2 + energy y C(s), O2(g) CO 2 r e → product E ∆ E>0(positve∆E) a CO2(g) r C(s), O2(g) t →reactants I System Energy Flow Surroundin gs  If reactants have higher internal energy than the products, ∆ Esysis negative and energy flows out of the system into surroundings  If the reactants have lower internal energy than the products, ∆E sys is positive and energy flows into system from surroundings o Example for better understanding:  Imagine this as a checking account. Taking money out of the account or withdrawing always has a negative sign on the numbers on the transaction. Depositing money into the account is accompanied by a positive sign or is a positive amount. In chemistry, changes in internal energy is mostly due to heat and work.  Heat(q)  Thermal Energy (temperature aspects) o If heat is given off, heat flows out of system (-q value) o If heat flows in system from surroundings (+q value) o With heat energy is high to low until thermal equilibrium is Heat from coffee cup will warm your hand (energy will flow from the cup into your hand) until both the coffee cup and your hand are equal in temperreached  Work(w): Expansion or contraction (reaction takes more volume when its done it expands, and vice versa) o Work is done to the system (+w value) o System does work (-w value)  ∆ E o Negative: Initial energy was higher than final energy; energy flow out of the system o Positive: Final energy was higher than internal energy; energy flows into the system o ∆ E=q+w Section 6.4: Quantification of Heat and Work  Heat: Transfer of thermal energy  Thermal Equilibrium: NO more temperature change; In terms of heat, NO more heat transfer  Objective: Calculate heat involved based on temperature changes o Heat Capacity: Heat required to change an objects q=C ∙∆T Extensive property: Depends temperature by 1℃ on amount of o Specific Heat Capacity: Heat required to change 1 gram of Intensive property: substant to 1℃ Depend on kind of substance being o Molar Heat Capacity: Per mole heated not amount q=nc∆T q c= Intensive property: M∙∆ T Depend on kind of Example: A reaction releases 15,000J of energy into 500.0g H O 2 substance being and the temperature increased by 7.17 ℃ . Assuming no heated not amount energy lost in transfer, what is the heat capacity and the specific heat capacity of H2O?  Positive change for the heat in reference to the water (water being the system and the surrounding is what is heating the water). Therefore, Joules are a +15,000 because the water is gaining energy Heat Capacity: 15,000J= C(7.17)  C= 2090J/ ℃ Specific Heat Capacity: q=mc∆T →15,000J= 500.0g)c )7.17→c=4.184J/℃  Work: force through a distance Distance Force  Since F is defined with pressure:  Substitute: Area Pressure W=(P∙A)∙D  Distance is the change in height or an amount of length ( h ) W=P∙A∙∆ h  Volume of a cylinder is area of its base times the height then A∙∆h=∆V W=−P∙∆V Example: Calculate work for balloon going from .1L to 1.85L with 1 atm external pressure. 101.3J W=−P∙∆V →− 1a(m 1.)(L−0.1L =−1)75L∙atmx 1L∙atm =−1.77J If ∆ V is negative then the w-value is positive Section 6.5: Measure ∆E for reaction  ∆ E=q+w  ∆ E=q Fixed volume: ∆ V=0 and Calorimetry: Measure the thermal energy Example: 1.01g of sucrose burned in a bomb calorimeter with C=4.90 kJ/ ℃ . The temperature raises from 24.92 to 28.33 ℃ . How much energy change? −(4.90kJ/℃)(28.33−24.92)→16.7kJ Section 6.6  Enthalpy( ∆ H¿ : Heat of reaction; sum of its internal energy and the products of its pressure and volume o Also a state function o ∆ H=∆E+P ∆V  Endothermic: Heat absorbed (+ ∆ H¿  Exothermic: Heat released (−∆H)  Objective: Use enthalpy in stoichiometric calculations P,V,T P,V,grams moles Mole Mole moles grams  Example: Typical BBQ grill has 13.2kg C H in 3 8ank. How much heat is released if the whole tank is burned? 1000 g 1 molC3 H 8 −2044kJ 13.2kg x x x = -6.12 x 1kg 44.09g 1 molC3 H 8 10 kJ Section 6.7 & 6.8: Relations with ∆ H  If reaction is multiplied by factor: o C H3+ 8O  3C2 + 4H O2 2 ∆ H=−2044kJ  Multiply by 2 o 2C H 3 80O  6CO2+ 8H O 2 2 ∆ H=2(−2044kJ)  Whatever the reaction is multiplied by, the enthalpy is also multiplied by the same factor  If reaction is reversed: o 3CO + 2H O  C2H + 5O3 8 2 ∆ H=+2044 kJ  Just flip the sign on the enthalpy  If reaction are added up, heat is also added up (Hess’ Law) ∆ H Example: Find RXN for C (s) + H 2(g)  CO (g) + H (g)2knowing. . .  C (s) + O 2g)  CO (g2 ∆ H=−393.5kJ  2CO(g) + O (g2  CO (g)2 ∆ H=−566.0kJ  2H 2g) + O (2)  2H O(2) ∆ H=−483.6 kJ Solution: Keep the first given the same: C(s) + O (g) 2 CO (g) 2 ∆ H=393.5kJ Flip the third given in order to get O on2the left side and multipy by 1 ½ : H 2  ½ O + 2 2 ∆ H= (483.6kJ ) 2 Flip second given and multiply by ½ : CO  CO 2 ½ O 2 ∆ H= 1(−566.0kJ ) 2 Section 6.9: Standard Enthalpy of Formation  Objective: Use standard enthalpy of formation to calculate ∆ H RXN o Standard state:  Gas: 1 atm (partial pressure)  Liquid and Solid: Pure, most stable form (1 atm and 25 ℃ )  Aqueous: 1M (one molarity) o Standard Enthalpy of reaction ( ∆ H ¿ RXN  All reactants and products at standard conditions o Standard enthalpy of formation ∆ H°F  Enthalpy change for producing 1 mole of substance for its elements (all at standard state)  Pure compound: The change in enthalpy when 1 mole of the compound forms from its constituent elements in their standard states  Pure element in standard state: ∆ H° F0  Example: Calculate ∆ H° RXN N 2 4l) + N 2 (4)  2N O2g) + 2H O(g2 Reactants  elements Alike terms are highlighted Elements  Products because those can be canceled out since there is the same N 2 4 N +22H - 2 ∆ H amount of each side of the N 2 4 N +22O - ∆2H 2N 2 O  2N O (2( ∆ H¿¿ 2H 2 O  2H O (2( ∆ H¿¿ Shortcut. . . ∆ H° RXN=Σ∆H F(products) F(reac)antsere Σ means These numbers sum being multiplied Values from above: by the moles Products: come from the 2 mol(81.6kJ/mol)+2 mol(-241.8kJ/mol)=- result in the 320.4kJ/mol example above Reactants: and table 6.5 in 1mol(50.6kJ/mol)+ 1 mol(11.1 kJ/mol)=61.7kJ/mol the textbook. Products – Reactants  (-320.4) – (61.7) = (-382.1 kJ/mol)


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