AP Chemistry

AP Chemistry

AP Chemistry Unit 5 Thermochemistry Chemical reactions inherently involve a transfer of energy 1) From one substance to another 2) From one type to another a) Kinetic to potential b) Potential to kinetic

3) To reorder the system Remember that energy is always conserved The net flow of heat is from warmer to cooler objects Heat will flow until equilibrium is reached Where: Q = heat transferred

CP is the specific heat at constant pressure T is the change in temperature Every substance has a unique value of specific heat Calorimetry is the measuring of the heat change for a chemical or physical process

Bomb Calorimeter Why constant pressure? The internal energy (U) of a system increases whenever heat (q) is added or work (w) is done on the system

We now define enthalpy (H) Enthalpy changes whenever internal energy or volume changes Since internal energy, pressure, and volume are state functions, enthalpy is also a state function The value of a state function depends only on the current conditions and not the path taken Sign convention 1)

2) 3) 4) 5) When heat is added to a system this is positive When work is done to a system this is positive When heat is removed from a system this is negative When work is done by a system this is negative A net increase of energy, through either heat or work, is positive

6) A net decrease of energy, through either heat or work, is negative A positive value for enthalpy is endothermic A negative value of enthalpy is exothermic The enthalpy difference between reactants and products in a chemical reaction is the enthalpy of reaction Enthalpy values for many elements and compounds can be found in reference tables Enthalpy values are typically listed as enthalpies of

formation An enthalpy of formation is the enthalpy change that would result when the substance was formed from its constituent elements The enthalpy of formation of an element in its standard state is defined to be zero Standard Enthalpies of Formation, , at 298 K Substance Substance Acetylene

Acetylene Ammonia Ammonia Benzene Benzene Calcium Carbonate Calcium Carbonate Calcium oxide Calcium oxide Carbon dioxide Carbon dioxide

Carbon monoxide Carbon monoxide Diamond Diamond Ethane Ethane Ethanol Ethanol Ethylene Ethylene Glucose

Glucose Hydrogen bromide Hydrogen bromide Formula Formula C2H2(g) C2H2(g) NH3(g) NH3(g) C6H6(l)

C6H6(l) CaCO3(s) CaCO3(s) CaO(s) CaO(s) CO2(g) CO2(g) CO(g) CO(g) C(s) C(s)

C2H6(g) C2H6(g) C2H5OH(l) C2H5OH(l) C2H4(g) C2H4(g) C6H12O6(s) C6H12O6(s) HBr(g) HBr(g)

(kJ/mol) -26.7 -26.7 -46.19 -46.19 49.04 49.04 -1207.1 -1207.1 -635.5 -635.5

-393.5 -393.5 -110.5 -110.5 1.88 1.88 -84.68 -84.68 -277.7 -277.7 52.30

52.30 -1260 -1260 236.23 236.23 Substance Substance Hydrogen chloride Hydrogen chloride Hydrogen fluoride

Hydrogen fluoride Hydrogen iodide Hydrogen iodide Methane Methane Methanol Methanol Propane Propane Silver chloride Silver chloride

Sodium bicarbonate Sodium bicarbonate Sodium carbonate Sodium carbonate Sodium chloride Sodium chloride Sucrose Sucrose Water Water Water vapor

Water vapor Formula Formula HCl(g) HCl(g) HF(g) HF(g) HI(g) HI(g) CH4(g)

CH4(g) CH3OH(l) CH3OH(l) C3H8(g) C3H8(g) AgCl(s) AgCl(s) NaHCO3(s) NaHCO3(s) Na2CO3(s) Na2CO3(s)

NaCl(s) NaCl(s) C12H22O11(s) C12H22O11(s) H2O(l) H2O(l) H2O(g) H2O(g) (kJ/mol) -92.30

-92.30 -268.6 -268.6 25.9 25.9 -74.85 -74.85 -238.6 -238.6 -103.85 -103.85

-127.0 -127.0 -947.7 -947.7 -1130.9 -1130.9 -411.0 -411.0 -2221 -2221 -285.8

-285.8 -241.8 -241.8 Consider ethyne (acetylene) which is used in a cutting (welders) torch. Determine the enthalpy of reaction when one mole of ethyne burns in excess oxygen to form carbon dioxide and water vapor C2H2(g) + O2(g) 2CO2(g) + H2O(g)

The reaction in the previous problem was combustion C2H2(g) + O2(g) 2CO2(g) + H2O(g) Combustion is very useful in thermochemistry since it is a very exothermic reaction and produces copious amounts of heat that is easy to measure enthalpy of combustion is nothing more than the enthalpy of reaction where the reaction is combustion

For the previous problem, we could have written There would have been absolutely no difference in how the problem would have been worked When you are given a problem that mentions the enthalpy of combustion of XYZ(s), that means to react XYZ with oxygen to form the most common oxides

Enthalpy of combustion and enthalpy of reaction are the same thing, combustion is just a specific type of reaction Enthalpy of formation is also just another type of reaction If a problem mentions to enthalpy of formation of XYZ(s), it means X(s) + Y(g) + Z(g) XYZ(l) In many cases a formation reaction is impossible

to actually do: the elements simply dont react that way This does not mean the enthalpy of formation cannot be determined or is unimportant If enthalpies of formation cannot be determined experimentally, they can be calculated from other reactions Enthalpies of formation are what are tabulated and used in all other problems NOTE: the words heat and enthalpy are often

used interchangably such as enthalpy of reaction and heat of reaction When 25.0 g of octane are burned in excess oxygen, 1199.6 kJ of heat is liberated. Determine the heat of formation of octane We are asked for H for 8C(gr) + 9H2(g) C8H18(l) But we are given data for C8H18(l) + O2(g) 8CO2(g) + 9H2O(l)

We can determine the molar heat of combustion of octane When 25.0 g of octane are burned in excess oxygen, 1199.6 kJ of heat is liberated. Determine the heat of formation of octane Hess Law Since enthalpy is a state function, the enthalpy difference between reactants and products is the

same regardless of how we get there This means that if we break a reaction into steps, each with its own enthalpy of reaction, the enthalpy change for the overall reaction will be equal to the sum of the enthalpy changes for the steps Determine the heat of reaction for 2Sr(s) + 2C(gr) + 3O2(g) 2SrCO3(s) Given: SrCO3(s) SrO(s) + CO2(g) H = 234 kJ C(gr) + O2(g) CO2(g)

2Sr(s) + O2(g) 2SrO(s) H = -394 kJ H = -1184 kJ Individual reactions can be manipulated (reversed, multiplied, etc.) so that they can be added to produce the final equation Whatever is done to the reaction must also be done to the enthalpy value!

2Sr(s) + 2C(gr) + 3O2(g) 2SrCO3(s) SrCO3(s) SrO(s) + CO2(g) H = 234 kJ C(gr) + O2(g) CO2(g) 2Sr(s) + O2(g) 2SrO(s) H = -394 kJ H = -1184 kJ The first reaction will have to be reversed and doubled 2SrO(s) + 2CO2(g) 2SrCO3(s) H = -468 kJ

2Sr(s) + 2C(gr) + 3O2(g) 2SrCO3(s) 2SrO(s) + 2CO2(g) 2SrCO3(s) H = -468 kJ C(gr) + O2(g) CO2(g) 2Sr(s) + O2(g) 2SrO(s) H = -394 kJ H = -1184 kJ The second reaction will have to be doubled 2C(gr) + 2O2(g) 2CO2(g)

H = -788 kJ 2Sr(s) + 2C(gr) + 3O2(g) 2SrCO3(s) 2SrO(s) + 2CO2(g) 2SrCO3(s) H = -468 kJ 2C(gr) + 2O2(g) 2CO2(g) H = -788 kJ 2Sr(s) + O2(g) 2SrO(s)

H = -1184 kJ The third reaction is correct, now to add 2Sr(s) + 2C(gr) + 3O2(g) 2SrCO3(s) 2SrO(s) + 2CO2(g) 2SrCO3(s) H = -468 kJ 2C(gr) + 2O2(g) 2CO2(g) H = -788 kJ 2Sr(s) + O2(g) 2SrO(s)

H = -1184 kJ 2SrO(s) + 2CO2(g) + 2C(gr) + 2O2(g) + 2Sr(s) + O2(g) 2SrCO3(s) + 2CO2(g) + 2SrO(s) 2C(gr) + 3O2(g) + 2Sr(s) 2SrCO3(s) H = -2440 kJ Another approach to determining enthalpies of

reaction is bond energies Energy is released when bonds form (exothermic) Energy is required to break bonds (endothermic) The difference between the two should be the enthalpy of reaction If we think in terms of products minus reactants as usual, we would have This would give a positive value when the energy of the products is greater than the energy of the

reactants But since the formation of product bonds is exothermic (-) this gives the wrong sign For the correct sign, we need to use Which is reactants minus products Using bond energies, determine the enthalpy of reaction for the following reaction 2C2H3Br + 6O2 4CO2 + 3H2O + Br2O

C=C 2 612 = 1224 CH 6 413 = 2478 CBr 2 276 = 552 O=O 6 495 = 2970 C=O 8 799 = 6392 HO 6 463 = 2778 BrO 2 235 = 470 7224

Hrxn = 7224 9640 = 2416 kJ 9640 Using bond energies, determine the enthalpy of reaction for the following reaction 2C2H3Br + 6O2 4CO2 + 3H2O + Br2O C=C 2 612 = 1224 CH 6 413 = 2478 CBr 2 276 = 552

O=O 6 495 = 2970 C=O 8 799 = 6392 HO 6 463 = 2778 BrO 2 235 = 470 7224 Hrxn = 7224 9640 = -2416 kJ

9640 Using bond energies only gives an approximation for enthalpy of reaction The bond energy between two atoms also depends on what they are bonded to A reaction diagram plots potential energy versus reaction coordinate or progress The difference in energy between the reactants and

the activated complex must come from kinetic energy (a) represents the activation energy for the forward reaction (b) is the activation energy for the reverse reaction (c) is the enthalpy of reaction This reaction is endothermic Exothermic

Activation energy is what is necessary to initiate a reaction A spontaneous reaction is self-sustaining, not necessarily self-initiating. Bond energy is a form of potential energy Bond energy depends on the force of attraction and the distance between the atoms Coulombs Law: Bond length is the distance where potential energy is at a

minimum All moving particles possess kinetic energy K = mv2 Heat of solution Often during the solution process, the average distance between particles (think potential energy) changes Since no energy is added or removed from the system, when potential energy changes so too must kinetic As a consequence, the mixing of substances, even without a

chemical reaction, often results in a temperature change There are three steps in the solution process 1) Separation of solvent particles (Endothermic) 2) Separation of solute particles (Endothermic) 3) Coming together of solute and solvent (solvation) (Exothermic) Heat of solution will be #1 + #2 #3

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