Ch. 16 Reaction Energy

Thermochemistry  Thermochemistry: the study of the transfers of energy as heat that accompany chemical reactions and physical changes.  Calorimeter: an instrument to measure the energy absorbed or released as heat in a chemical or physical change.

Thermochemistry  Temperature: a measure of the average kinetic energy of the particles in a sample of matter. Measured in Celsius, Fahrenheit, or Kelvin  Heat: the amount of energy exchanged between samples of different temperatures. Measured in Joules  Joule (J): the SI unit of heat as well as other forms of energy. 16-2

Thermochemistry  Temperature in this chapter will be used with the Kelvin Scale. ALL TEMPS MUST BE KELVIN OR CONVERTED TO IT K= ºC 26ºC is room Temperature  K= = 299K  299K is room temperature 

Thermochemistry  What is the temperature in Kelvin for the following?  53ºC  273ºC  0ºC  -273ºC---> What is special about this one?

Thermochemistry (JUST LISTEN)  What is the difference between HEAT AND TEMPERATURE??  Temperature is just the ENERGY of the sample. How hot something is its temperature. We say hot things have a lot of ENERGY. THINK OF AIR.  Heat is the ENERGY transferred between two samples of different TEMPERATURES. THINK OF A COLD WATER BOTTLE IN YOUR HAND!  Thermochem is basically the study of HEAT FLOW between samples with different TEMPERATURES. VIDEO!!! WOOD BLOCK

Specific Heat  Specific Heat(c p ): the amount of energy required to raise the temperature of one gram of a substance by one degree (C or K) THINK OF THE OCEAN q = c p x m x ΔT heat = specific heat x mass x change in temp. unit  J J/gK g K ΔT = Tf - Ti 16-3

Example q = c p x m x ΔT Example: A 4.0g sample of glass was heated from 274K to 314 K, and was found to have absorbed 32 J of energy as heat. What is the specific heat of this glass?

Practice q = c p x m x ΔT 1) Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 K to 313 K. 96J = ( c p )(35g)(20K) c p = 0.14 J/gK 16-5

Enthalpy (SIMILAR to ΔE)  Enthalpy of Reaction: the quantity of energy transferred as heat during a chemical reaction. ΔH = H products – H reactants 16-6

Enthalpy Exothermic Rxn (-ΔH): energy is released during the reaction. CO (g) + ½O 2(g) → CO 2(g) ΔH = −283.0 kJ Endothermic Rxn(+ΔH): energy is absorbed during the rxn. C(s)+2S(s) → CS2(l) ΔH = kJ

Enthalpy  ΔH = standard enthalpy of a overall reaction.  ΔH f = standard enthalpy of formation. (elements in their standard state have ΔH f = 0, compounds with positive values are unstable) ΔH c = standard enthalpy of combustion. (the energy change that occurs during the complete combustion of one mole of a substance.) 16-12

Enthalpy  Hess’s Law: the overall enthalpy change in the net rxn is equal to the sum of enthalpy changes for the individual steps in the process.  Hess’s Law let’s us take step by step reactions and manipulate them to look like the overall reaction we want!  It also let’s us calculate the overall heat exchange ΔH from the individual steps

Hess’s Law  Example: What would we have to do to make the steps of this reaction look like the overall reaction? Calculate ΔH o for NO(g) + ½O 2 (g) → NO 2 (g) Rxn1) ½N 2 (g) + ½O 2 (g) → NO(g) ΔH o f = kJ Rxn2) ½N 2 (g) + O 2 (g) → NO 2 (g) ΔH o f = kJ

Hess’s Law  Rules for applying Hess’s Law: 1) If an equation is reversed, the sign of ΔH is also reversed. 2) If you multiply an equation by a number, you must multiply the ΔH by the same number 3) YOU CAN CANCEL (SUBTRACT) THINGS OUT FROM THE REACTANT TO PRODUCT SIDES! 4) YOU CAN ADD THINGS TOGETHER ON THE SAME SIDE OF THE ARROW!

Hess’s Law Ex. Calculate ΔH o for NO(g) + ½O 2 (g) → NO 2 (g) Solve by combining known eq. Rxn1) ½N 2 (g) + ½O 2 (g) → NO(g) ΔH o f = kJ Rxn2) ½N 2 (g) + O 2 (g) → NO 2 (g) ΔH o f = kJ NO(g) + ½O 2 (g) → NO 2 (g) Rxn 1 needs to be reversed

Hess’s Law Rxn1) NO(g) → ½N 2 (g) + ½O 2 (g) ΔH o f = kJ (reversed) Rxn2) ½N 2 (g) + O 2 (g) → NO 2 (g) ΔH o f = 33.2 kJ NO(g) + ½O 2 (g) → NO 2 (g) ΔH o = ( kJ) + (33.2 kJ)= kJ 16-16

Practice 2) Calculate the enthalpy of reaction for the combustion of methane, CH 4, to form CO 2(g) and H 2 O (l). CH 4 + 2O 2 → CO 2 + 2H 2 O ΔH o = ? Rnx1)C + 2H 2 → CH 4 ΔH o f = kJ Rxn2)C + O 2 → CO 2 ΔH o f = kJ Rxn3) H 2 + ½O 2 → H 2 O ΔH o f = kJ CH 4 + 2O 2 → CO 2 + 2H 2 O ΔH o = ? Rxn 1 needs to be reversed, rxn 3 needs to be doubled

Practice Rnx1) CH 4 → C + 2H 2 ΔH o f = 74.3 kJ Rxn2)C + O 2 → CO 2 ΔH o f = kJ Rxn3) 2(H 2 + ½O 2 → H 2 O) ΔH o f = 2( kJ) CH 4 + 2O 2 → CO 2 + 2H 2 O ΔH o = kJ 16-18

Practice Example: Calculate the enthalpy of formation of pentane, C 5 H 12 from the following reactions: 5C (s) + 6H 2(g) → C 5 H 12(g) ΔH o f = ? Rnx1) C 5 H 12(g) + 8O 2(g) → 5CO 2(g) + 6H 2 O (l) ΔH o c = kJ Rxn2) C (s) + O 2(g) → CO 2(g) ΔH o f = kJ Rxn3) H 2(g) + ½O 2(g) → H 2 O (l) ΔH o f = kJ 16-19

Practice Rxn1 needs to be reversed, rxn2 needs to be multiplied by 5, rxn3 needs to be multiplied by 6. Rnx1) 5CO 2(g) + 6H 2 O (l) → C 5 H 12(g) + 8O 2(g) ΔH o c = kJ Rxn2) 5( C (s) + O 2(g) → CO 2(g) ) ΔH o f = 5( kJ) Rxn3) 6(H 2(g) + ½O 2(g) → H 2 O (l) ) ΔH o f = 6( kJ) 5C (s) + 6H 2(g) → C 5 H 12(g) ΔH o f = kJ 16-20

Practice 3) Calculate the ΔH o f of butane, C 4 H 10 from the following rxns. 4C (s) + 5H 2(g) → C 4 H 10(g) ΔH o f = ? Rnx1) C 4 H 10(g) + 13/2O 2(g) → 4CO 2(g) + 5H 2 O (l) ΔH o c = kJ Rxn2) C (s) + O 2(g) → CO 2(g) ΔH o f = kJ Rxn3) H 2(g) + ½O 2(g) → H 2 O (l) ΔH o f = kJ 16-21

Practice Rxn1 needs to be reversed, rxn2 needs to be multiplied by 4, rxn3 needs to be multiplied by 5. 4C (s) + 5H 2(g) → C 4 H 10(g) ΔH o f = ? Rnx1) 4CO 2(g) + 5H 2 O (l) → C 4 H 10(g) + 13/2O 2(g) ΔH o c = kJ Rxn2) 4(C (s) + O 2(g) → CO 2(g) ) ΔH o f = 4( kJ) Rxn3) 5(H 2(g) + ½O 2(g) → H 2 O (l) ) ΔH o f = 5( kJ) 4C (s) + 5H 2(g) → C 4 H 10(g) ΔH o f = kJ 16-22

Spontaneous Reactions  Entropy (S): a measure of the degree of randomness of the particles, such as molecules, in a system.  +ΔS= MORE CHAOS/RANDOMNESS  -ΔS= MORE ORDER  In the UNIVERSE Entropy is always increasing (+ΔS) !  Think of an egg breaking…do eggs unbreak?  What would be the sign of the entropy of the following?  LIQUID  SOLID  LIQUID  GAS

Spontaneous Reactions  Free Energy Change (ΔG): the difference between the change in enthalpy and the product of the kelvin temp. and the entropy change. ΔG o = ΔH o – TΔS o ΔH= Enthaply (J) ΔS= Entropy (J/K) T= Temperature (K) ΔG = - Spontaneous ΔG = + Not spontaneous ΔG = 0 Equilibrium 16-23

Example Example: For the rxn NH 4 Cl (s) → NH 3(g) + HCl (g), ΔH o = 176 J and ΔS o = J/K. Calculate ΔG o, is the rxn spontaneous at 25.15K? 16-24

Example Example: For the rxn NH 4 Cl (s) → NH 3(g) + HCl (g), ΔG o = 184 J and ΔS o = J/K. Calculate ΔH o, 298K? 16-24

Example Example: For the rxn NH 4 Cl (s) → NH 3(g) + HCl (g), ΔG o = -165 J and ΔH o = J. Calculate ΔS o, at 25.15K? 16-24

Practice 4) For the rxn Br 2(l) → Br 2g) ΔH o = 31 J and ΔS o = 93 J/K. At what temp. will this rxn be spontaneous (-ΔG)?

Ch. 16 The End!