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General Chemistry Thermochemistry

by: Morgan Walker

General Chemistry Thermochemistry Chem 1314

Marketplace > Oklahoma State University > Chemistry > Chem 1314 > General Chemistry Thermochemistry
Morgan Walker
OK State
GPA 3.2

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These notes cover the remainder of chapter 6, Thermochemistry.
General Chemistry
Dr. Jimmie Weaver
Class Notes
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This 3 page Class Notes was uploaded by Morgan Walker on Saturday March 12, 2016. The Class Notes belongs to Chem 1314 at Oklahoma State University taught by Dr. Jimmie Weaver in Winter 2016. Since its upload, it has received 9 views. For similar materials see General Chemistry in Chemistry at Oklahoma State University.


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Date Created: 03/12/16
Chapter 6 Thermochemistry cont Specific Heat  Measure of a substances intrinsic ability to absorb heat  Specific heat capacity- the amount of hear energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius  Molar hear capacity- the amount of heat energy required to raise the temperature of 1 mole of a substance by 1 degree Celsius Specific Heat of Water  Water can absorb a lot of heat energy without a large increase in its temperature due to its high specific heat capacity  The large amount of water absorbing heat from the air keeps beaches cool in summer Heat Transfer  Low to high, stops when the temperatures are equal  Heat lost by hot equals heat gained by cold  Block of metal at 35 degrees Celsius added to water at 8 degrees Celsius  Thermal energy goes from metal to water  Exact temperature change depends on o Mass of metal o Mass of water o Specific heat capacity of metal and water Pressure- volume work  PV work is caused by a volume change against an external pressure  When gas expands, Δvolume is positive, but system is doing work, so w is negative gas  As long as external pressure is kept constant o - wgasExternal Pressure * change in Δvolume  w=-P ΔV  to convert units to joules o 101.3 J= 1 atm(L) Bomb calorimeter  Used to find ΔE rxnmeasured in kJ/mole Given: 1.550g C 6 14 Ti 28.87 °C T=f38.13°C C = cal3 kJ/°C ???????????????? ΔE rxn???????????? ???? ???? ΔT= 38.13°C - 28.87°C= 12.26°C 6 14 −70.25???????? ΔE rxn= qcalC calT q cal.73 kJ/°C(12.26°C)= 70.25k .018????????????????6 14 3 ΔE rxn3.903 X10 kJ/mol qrxn-q ca q rxn70.25Kj 1.550g C 6 1486.06 g C H 6 1418 mol C H 6 14 Enthalpy (H)  H is the sum of internal energy if the system o H= E + PV  Change in enthalpy (ΔH) is heat involved in a reaction at constant pressure o When negative heat is lost by system (exothermic) o When positive hear is absorbed by system (endothermic) Exothermic  Surrounding temperature rises due to thermal energy being released  During reaction current bonds are broken and new ones are formed Endothermic  Surrounding temperature drops due to thermal energy being absorbed  During reaction old bonds broken new ones are formed Examples  Sweat evaporating from skin- endothermic o Since sweat is evaporating the skin is cooling, therefore its endothermic  Water freezing- exothermic  Wood burning- exothermic Enthalpy of Reaction  Extensive property, the more reactants there are the larger the change Measuring ΔH  Reaction done in aqueous reaction at constant pressure  q reaction= -q solution o = - mass of solution * heat capacity of solution * change in temp  ΔH = q (at constant pressure) = q rxn rxn o Per mole Relationships  When reaction is multiplied by a factor rxnis multiplied by the same factor  If reaction is reversed then the sign orxnchanges Hess’s Law  Change of enthalpy for a stepwise process is the sum of the enthalpy changes if steps o Sum of heat in step one plus sum of heat in step two equal total sum  ΔH 1 ΔH =2ΔH 3 Standard conditions  Standard state- the state of a material at a definite set of conditions  Standard enthalpy change- the change in enthalpy when all products and reactants are in their standard states  Standard enthalpy formation- the change in enthalpy for a reaction forming one mole of a pure compound Formation Reaction  Reactions of elements in their standard state to form one mole of a pure compound Writing C(g)O 2 (g)CO 2 (g)  The elements must be in their standard state o C (g) 2 (g)CO 2 (g) o Several forms of solid carbon, but its ΔH=0 so it has to be graphite o Oxygen standard state is the diatomic gas  Equation must be balanced but the coefficient of the product compound must be one Calculating  Any reaction can be written as the sum of formation reactions (reversed formation reaction) for reactants and products  ΔH for the reaction is the sum of the ΔHffor the component reaction  ΔH rxn nΔH(prfducts) - ∑ nΔH(reaftants) o n= coefficient of the reaction o 2O 2=2


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