Chapter 5 Thermochemistry Notes
Chapter 5 Thermochemistry Notes Chem 111-003
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This 3 page Class Notes was uploaded by Sharneece Gary on Wednesday September 30, 2015. The Class Notes belongs to Chem 111-003 at University of South Carolina taught by Dmitry V Peryshkov in Fall 2015. Since its upload, it has received 47 views.
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Date Created: 09/30/15
Chapter 5 Thermochemistry Vocabulag Thermochemistry the study of the relationship between heat and chemical reactions Kinetic Energy is energy possessed by matter because it is in motion Thermal Energy energy in the form of random motion particles in any sample of matter Heat is energy that causes a change in the thermal energy of a sample Potential Energy is the energy derived from the position or condition of matter Chemical Energy is a form of potential energy derived from the forces that hold the atoms together in compounds Law of Conservation of energy states that the energy of the universe is constant during a chemical or physical change Enthalpy is a measure of the energy of the system Calorimetry is the experimental measurement of heat released or absorbed by a chemical reaction Specific heat Cs is the heat needed to increase the temperature of 1gram of matter by 1 K Q mCsAT AT T2T1 Hess s Law states that when two or more thermochemical equations are added the enthalpy change of the resulting equation is the sum of those for the added equations Classes of Thermochemical Reactions Exothermic when the system transfers heat to the surroundings Heat from in goes out or exits Endothermic when a reaction absorbs heat from the surroundings Heat from out comes in Change in Enthalpy The change in enthalpy AH is equal to the heat absorbed or given off by the system at constant pressure difference between the changes of energy Thermochemical Equations A thermochemical equation is an equation for which AH is given AH 1818kJ heat absorbed Endothermic reaction AH 222x13quot3 kJheat released Exothermic reaction Example Calculate the enthalpy change when 110g of C02 form from the combustion of proprane C3H8 C3H8g 502g 9 3C02g 4H20 I AH 222x10quot3 kJ Step 1 Convert 110g C02 into Moles 110g C02 x 1 mol C0244gC02 025 mol C02 Step 2 Convert moles to k using the k provided 025mol C02 x 222x10quot3kJ3mo C02 0185 x 10quot3 kJ Specific Heat How much heat must be added to 120 g of water Cs 4184 Jg degrees C to raise its temperature from 234 degrees C to 368 degrees C Step 1 Plug numbers into the qmCsAT equation Q120g x 4184 368234 O 6728 kJ Calorimetry In the reaction of 0220g of magnesium with 200 g of a HCl solution which is in excessthe temperature of the solution increases by 252 degrees C Assume a heat capacity of 418 Jg degrees C Calculate the AH in k of the following equation Mgs 2HCaq 9 MgCl2aq H2g Step 1 Plug numbers into the qmCsAT equation Q 20022g x 418 2522109 J m total mass and in this case that is 200g 0220g 2109J because it was usedburned Step 2 Covert g of Mg to moles of Mg 0220gMg x 1mol Mg243g Mg 905 x 10quot3 mol Mg Step 3 AH JMol AH 2109J905x10quot3 mol Mg 2230 KJ Hess s Law The objective is to try and reform the two equation into the new equation for the enthalpy change This can be accomplished by reversingflipping the equation multiplyingdividing it by other numbers Whatever it takes to make that position of the equation and then you drop the unnecessary parts in which they should cancel out Example Calculate AH for 3C2H2g 9 C6H6 given the thermochemical equations 1 2C2H2g 502g 9 4 C02 g 2H20l AH 1692 k 2 2C6H6 1502g 9 12C02g 6H20l AH 6339 kl Step 1 Figure out with parts ofthe equation match where In equation one the 2C2H2 matches with the 3C2H2 in the formulated equation In equation two the 2C6H6 matches with the C6H6 in the formulated equation but they are on opposite sides We will multiply it by 1 to reverse the equation make sure to note beside each equation every change you make to it Step 2 Time to get the numbers to match up In equation one you will need to multiply everything by 15 because you want the coefficient 2 on C2H2 to be 3 In equation two you will need to divide everything by 2 because you want the coefficient 2 on C6HC to be 1 Step 3 After doing the previous steps things should start to cancel and bring you to the equation of 3C2H2 9 C6H6 Step4 With the noted changes do the exact same thing to each corresponding k and then added them together Equation 1 1692 x 15 2538 kl Equation 2 6339 x1 6339 2 3170 kl 2538 3170 Enthalpy of Reaction Use the standard enthalpies of formation to calculate the enthalpy change for the reaction P4010S 6H20G 9 4H3POS SUBSTANCE AH f kJmol P4010S 1640 H20g 242 H3PO4 1279 Step 1 Know the formula n x AHf product m x AHf reactants Note n amp m are the coefficients Step2 plug in numbers 4mo x 1279 kJmol 1mol x 1640 kJmol 6mol x 242 kJmol 5116 kl 3092 kl 2030 kl