Thermochemistry. Chapter 6. Dec 19 8:52 AM. Thermochemistry. Energy: The capacity to do work or to produce heat

Size: px

Start display at page:

Download "Thermochemistry. Chapter 6. Dec 19 8:52 AM. Thermochemistry. Energy: The capacity to do work or to produce heat"

Belinda Harmon
6 years ago
Views:

1 Chapter 6 Dec 19 8:52 AM Intro vocabulary Energy: The capacity to do work or to produce heat Potential Energy: Energy due to position or composition (distance and strength of bonds) Kinetic Energy: Energy due to the motion of the object and depends on the mass of the object and its velocity (KE = 1/2mv 2 ) Law of Conservation of Energy: Energy can be converted from one form to another but can be neither created nor destroyed (Several different forms!) Temperature: A property that reflects the random motions of the particles in a particular substance (measurement of kinetic energy) Heat: The transfer of energy between two objects due to a temperature difference (Total Energy both potential and kinetic) Dec 19 8:53 AM 1

Chemical Energy: System: The part of the universe on which we focus on Surroundings: Include everything else in the universe Exothermic: Energy flows out of the system as heat A + B C + D +

2 Chemical Energy: System: The part of the universe on which we focus on Surroundings: Include everything else in the universe Exothermic: Energy flows out of the system as heat A + B C + D + Energy(heat) Endothermic: Heat flows into the system A + B + Energy(heat) C + D Dec 19 9:00 AM Energy in bonds (but really NOT in bonds) When atoms are not bonded (or apart from each other) it is said they have ZERO potential energy The lowest point (the most negative) potential energy is where the bond is formed. ENERGY IS RELEASED when bonds formed (Whether Inter or Intra). This is why bond formation is EXOTHERMIC. Dec 18 1:07 PM 2

3 CH 4 + 2O 2 CO 2 + 2H 2 O Let's calculate the energy for this reaction using bond energies. List of average bond energies is listed below and you will need to draw the Lewis dot structures to see ALL the bonds present. Show which side of the reaction the energy would be (product exo or reactant endo) Average Bond Energies (kj/mol) H C 413 H O 467 O=O 495 Remember: C=O 745 ΔH rxn = ΣBonds reactants ΣBonds products Dec 19 9:04 AM Consider the following reaction: CH 4 + 2O 2 CO 2 + 2H 2 O + Heat Potential The energy required to break the bonds in the reactants and the energy released when the bonds in the products are formed (means the bonds in the reactants are WEAKER (on average) compared to the bonds in the products) Dec 19 9:13 AM 3

4 potential The amount of energy to break the bonds in the reactants is higher than that of the energy released from the forming of the bonds in the products (means the bonds in the products are WEAKER (on average) compared to the bonds in the reactants) Dec 19 9:19 AM First Law of Thermodynamics: The energy of the universe is constant Enthalpy (H): The energy (both kinetic and potential) flow of the system as heat (at constant pressure). Change in enthalpy is also called Heat of reaction q p = ΔH Dec 19 9:24 AM 4

5 In terms of a reaction the enthalpy change is expressed by this equation: ΔH = H products H reactants (bonds formed) (bonds broken) The reaction is endothermic when the products have a higher enthalpy than the reactants this means heat was absorbed by the system the reaction is exothermic when the products have a lower enthalpy than the reactants this means heat was released by the system Note: this is not the same equation as Bond Energy calculations. This equation looks at it from a different perspective (potential energy (bond energy) versus total energy (enthalpy). Dec 19 10:52 AM When 1 mole of methane (CH 4 ) is burned at constant pressure, 890 kj of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure (Hint: calculate your molar enthalpy (kj/mol) first). We did this type of math in Heating Curves! Is enthalpy an intensive or extensive property? Dec 19 10:52 AM 5

6 Calorimetry q = mcδt Heat Capacity (C): the measurement of how much energy must be added to raise the temperature of a substance. Specific heat capacity: the heat capacity per gram of a substance J/C*g or J/K*g Molar heat capacity: the heat capacity per mole of a substance J/C*mol or J/K*mol Again, we have already used this! Calorimetry Let's say we mix 50.0 ml of 1.0M HCl at 25.0 with 50.0 ml of 1.0M NaOH also at 25.0 in a calorimeter. After the reactants are mixed by stirring, the temp. is observed to increase to 31.9 (C H2O = 4.18 J/ *g). First identify the surroundings and system!!! 6

7 Jan 7 10:13 AM Calorimetry When 1.00L of 1.00M Ba(NO 3 ) 2 solution at 25.0 is mixed with 1.00L of 1.00M Na 2 SO 4 solution at 25.0 in a calorimeter, the white solid BaSO 4 forms and the temperature of the mixture increases to (C H2O = 4.18 j/c*g). Calculate the molar enthalpy of BaSO 4 formed. Surroundings? System? Net ionic equation Molar enthalpy 7

8 List of possible metals for Lab (J/g*c) Copper Lead Tin Zinc Aluminum Nickel Jan 11 8:09 AM Bomb Calorimetry Suppose we wish to measure the energy of combustion of octane (C 8 H 18 ), a component of gasoline. A g sample of octane is placed in a bomb calorimeter known to have a heat capacity of 11.3 kj/. (This means that 11.3 kj of energy is required to raise the temperature of the water and other parts of the calorimeter by 1 ). The octane is ignited in the presence of excess oxygen, and the temperature increase of the calorimeter is q = ΔT x C 8

9 Calorimetry It has been suggested that hydrogen gas might be a substitute for natural gas. To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kj/. When a 1.50 g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3. When a 1.15 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was Calculate the energy of combustion (per gram for hydrogen and methane) Energy released 1.5 g methane = (11.3 kj/c)(7.3c) = 83 kj Energy released 1.15 g hydrogen = (11.3 kj/c)(14.3c) = 162 kj Energy released 1 g Methane = 83 kj/1.5g = 55 kj/g Energy released 1 g hydrogen = 162/1.15g = 141 kj/g Characteristics of Enthalpy Changes 1. If a reaction is reversed, the sign of ΔH is also reversed N 2 (g) + 2O 2 (g) 2NO 2 (g) 2NO 2 (g) N 2 (g) + 2O 2 (g) ΔH rxn = + 68 kj ΔH rxn = 68 kj 2. the magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. N 2 (g) + 2O 2 (g) 2NO 2 (g) 2N 2 (g) + 4O 2 (g) 4NO 2 (g) ΔH rxn = + 68 kj ΔH rxn = kj 1/2N 2 (g) + O 2 (g) NO 2 (g) ΔH rxn = + 34 kj 9

10 Hess's Law: Going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps Let's look at the oxidation of nitrogen to form nitrogen dioxide. It can be written in one step: N 2 (g) + 2O 2 (g) 2NO 2 (g) ΔH rxn = 68 kj However the reaction can be carried out in two distinct steps N 2 (g) + O 2 (g) 2NO(g) ΔH 1 = 180 kj 2NO(g) + O 2 (g) 2NO 2 (g) ΔH 2 = 112 kj ΔH 2 + ΔH 3 = 180 kj + ( 112 kj) = 68 kj = ΔH 1 This is one of two ways to calculate using Hess's Law Two forms of carbon are graphite, the soft, black, slippery material used in "lead" pencils, and diamond, the brilliant, hard gemstone. using the enthalpies of combustion for graphite ( 394 kj/mole) and diamond ( 396 kj/mole), calculate ΔH for the conversion of graphite to diamond. C graphite (s) + O 2 (g) CO 2 (g) C diamond (s) + O 2 (g) CO 2 (g) ΔH = 394 kj ΔH = 396 kj 10

11 Diborane (B 2 H 6 ) is highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program. calculate ΔH for the synthesis of diborane from its elements, according to the equation Use the following data: 2B(s) + 3H 2 (g) B 2 H 6 (g) (a) 2B(s) + 3/2O 2 (g) B 2 O 3 (s) (b) B 2 H 6 (g) + 3O 2 (g) B 2 O 3 (s) + 3H 2 O(g) (c) H 2 (g) + 1/2O 2 (g) H 2 O(l) (d) H 2 O(l) H 2 O(g) ΔH = 1273 kj ΔH = 2035 kj ΔH = 286 kj ΔH = 44 kj Hints for Hess's Law 1. Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal 2. reverse any reactions to give the required reactants and products 3. multiply the reactions to give the correct numbers of reactants and products. 11

12 N 2 H 4 (l) + H 2 (g) 2NH 3 (g) N 2 H 4 (l) + CH 4 O(l) CH 2 O(g) + N 2 (g) + 3H 2 (g) N 2 (g) + 3H 2 (g) 2NH 3(g) CH 4 O(l) CH 2 O(g) + H 2(g) ΔH = 37 kj ΔH = 46 kj ΔH = 65 kj Jan 14 8:19 AM Zn(s) + 1/8S 8 (s) + 2O 2 (g) ZnSO 4 (s) Zn(s) + 1/8S 8 (s) ZnS(s) 2ZnS(s) + 3O 2 (g) 2ZnO(s) + 2SO 2 (g) 2SO 2 (g) + O 2 (g) 2SO 3 (g) ZnO(s) + SO 3 (g) ZnSO 4 (s) ΔH = kj ΔH = kj ΔH = kj ΔH = kj Jan 14 8:27 AM 12

13 Standard Enthalpies of Formation The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. (ΔH o f) The degree symbol (ΔH o ) means that the reaction was carried out under standard conditions Standard State: A precisely defined reference state Gaseous substances at exactly 1 atm pressure Pure substances in a condensed state (solid or liquid) is the pure liquid or solid Solutions have a concentration of exactly 1 M. Elements is the standard state of the element at 1 atm and 25C (oyxgen is a gas, sodium is a solid, bromine is a liquid, etc.) Elements have a ΔH o f of 0 kj, since they do not need to be formed! Standard Enthalpies of Formation (examples) 1/2N 2 (g) + O 2 (g) NO 2 (g) ΔH o f = 34 kj/mol C(s) + 2H 2 (g) + 1/2O 2 (g) CH 3 OH(l) ΔH o f = 239 kj/mol Notice that nitrogen and oxygen are both in their standard state and that the reaction is for 1 mole of product. 13

14 Standard Enthalpies of Formation We use Hess's Law to determine the heat of formation for a reaction CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) C(s) + 2H 2 (g) CH 4 (g) C(s) + O 2 (g) CO 2 (g) H 2 (g) + 1/2O 2 (g) H 2 O(l) ΔH o f = 75 kj/mol ΔH o f = 394 kj/mol ΔH o f = 286 kj/mol Standard Enthalpies of Formation (Second way to calculate using Hess's Law) ΔH Reaction = Σn p ΔH o f (Products) Σn r ΔH o f (reactants) 4NH 3 (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O(l) Compound ΔH o f (kj/mol) NH 3 (g) 46 NO 2 (g) 34 H 2 O(l) 286 Al 2 O 3 (s) 1676 Fe 2 O 3 (s) 826 CO 2 (g) 394 CH 3 OH(l) 239 C 8 H 18 (l)

15 Standard Enthalpies of Formation Compound ΔH o f (kj/mol) NH 3 (g) 46 NO 2 (g) 34 H 2 O(l) 286 Al 2 O 3 (s) 1676 Fe 2 O 3 (s) 826 CO 2 (g) 394 CH 3 OH(l) 239 C 8 H 18 (l) 269 2Al(s)+ Fe 2 O 3 (s) Al 2 O 3 (s) + 2Fe(s) Standard Enthalpies of Formation Methanol (CH 3 OH) is often used as a fuel in high performance engines in race cars. using the data below, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds but assume for this problem that gasoline is pure liquid octane (C 8 H 18 ) Compound ΔH o f (kj/mol) NH 3 (g) 46 NO 2 (g) 34 H 2 O(l) 286 Al 2 O 3 (s) 1676 Fe 2 O 3 (s) 826 CO 2 (g) 394 CH 3 OH(l) 239 C 8 H 18 (l) 269 2CH 3 OH(l) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(l) 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O(l) 15

16 Spontaneity, Entropy, and Free Energy Chapter 16 Jan 7 10:48 AM Spontaneous Processes and Entropy Spontaneous Processes: A process that occurs without outside interventions. This can be fast or slow (Kinetics has nothing to do with Spontaneity) Thermodynamics will tell us the energy of the reaction and what direction it will occur, but not the speed of the process I.E. According to thermodynamics, a diamond should change spontaneously to graphite. Jan 22 10:17 AM 16

17 Spontaneous Processes and Entropy Spontaneous processes: A ball will roll down a hill spontaneously but will never roll back up the hill If exposed to air and moisture, steel rusts spontaneously. However, rust does not spontaneously change back to iron metal and oxygen gas A gas fills its container uniformly. it never spontaneously collects at one end of the container Heat flow always occurs from a hot object to a cooler one. the reverse process never occurs spontaneously Wood burns spontaneously in an exothermic reaction to form carbon dioxide and water, but wood is not formed when carbon dioxide and water are heated together At temperatures below 0C, water spontaneously freezes, and at temperatures above 0C, ice spontaneously melts Jan 22 10:21 AM Spontaneous Processes and Entropy At one point in time it was thought that reactions are spontaneous because it is exothermic, but the melting of ice is endothermic yet it is spontaneous Entropy (S): The measurement of molecular randomness or disorder (the driving force for spontaneous reactions) describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state This is like probability: Nature spontaneously proceeds towards the states that have the highest probabilities of existing. Jan 22 10:28 AM 17

18 Spontaneous Processes and Entropy Let's look at an ideal gas in a vacuum chamber like below? Arrangement 1 Arrangement 2 Arrangement 3 1 possible way 4 possible way 6 possible way Arrangement 4 Arrangement 5 4 possible way 1 possible way Mar 19 1:06 PM Spontaneous Processes and Entropy Which one has more entropy? Solid CO 2 or gaseous CO 2 What is the SIGN for the entropy change in the following (more randomness is positive, less randomness (more order) is negative)? Solid sugar is added to water to form a solution Iodine vapor condenses on a cold surface to form crystals Mar 19 1:05 PM 18

19 Entropy Changes in Chemical Reactions N 2 (g) + 3H 2 (g) 2NH 3 (g) We start with 4 moles of reactants and end with 2 moles of products. Therefore this lowers the "chaos" in the system leading to a negative entropy value 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) We start with 9 moles of reactants and end with 10 moles of products. Therefore this increases the "chaos" in the system leading to a positive entropy value Mar 19 1:09 PM Entropy Changes in Chemical Reactions Predict the sign of entropy for each of the following reactions CaCO 3 (s) CaO(s) + CO 2 (g) 2SO 2 (g) + O 2 (g) 2SO 3 (g) Mar 19 1:10 PM 19

20 Second law of Thermodynamics The Second Law of Thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe Nature is always moving towards the most probable state available The entropy of the universe is always increasing ΔS univ = ΔS sys + ΔS surr But, we are not typically looking at the universe, we are looking at the system, so this method does not work for our purposes. Mar 19 1:07 PM Free Energy ΔG = ΔH TΔS G = (Gibb's) Free Energy H = Enthalpy T = Temperature (K) S = Entropy of the system If ΔG is negative ( ), then the process is spontaneous (favorable) If ΔG is also zero, then the reaction is at equilibrium (this will be VERY important here and in later chapters!!!) Mar 19 1:08 PM 20

21 Free Energy ΔG = ΔH TΔS Lets look at water melting at 10C, 0C, and 10C H 2 O(s) H 2 O(l) ΔH 0 = 6.03 x 10 3 J/mol ΔS 0 = 22.1 J/K*mol T(K) ΔH 0 (J/mol) 6.03x x10 3 ΔS 0 (J/K*mol) ΔG = ΔH TΔS (J/mol) +2.2x Notice, that at the 0, ΔG is 0 (remember phase changes equilibrium x x10 2 This shows that some processes (if ΔS and ΔH are the same sign) that the process will be temperature dependent! Mar 19 1:07 PM Free Energy ΔS positive, ΔH negative ΔS positive, ΔH positive ΔS negative, ΔH negative ΔS negative, ΔH positive Spontaneous at all temps Spontaneous at high temperatures (where exothermicity is relatively unimportant Spontaneous at low temperatures (where exothermicity is dominant) Process not spontaneous at any temperature (reverse process is spontaneous at all temps) LOOK AT THE EQUATION!!! Jan 23 3:33 PM 21

22 Free Energy At what temperatures is the following process spontaneous at 1 atm? Br 2(l) Br 2(g) ΔH 0 = 31.0 kj/mol and ΔS 0 = 93.0 J/K*mol What is the normal boiling point of liquid bromine? Mar 19 1:09 PM Entropy Changes in Chemical Reactions Third Law of Thermodynamics: The entropy of a perfect crystal at 0K is zero This allows for calculations to be done to determine standard entropy (S 0 ) values for common substances at 298K and 1 atm (Hess's Law!) Since entropy (like enthalpy) is a state function (not pathway dependent) we can use the equation below to calculate the entropy change in a reaction ΔS 0 reaction = Σn p S 0 products Σn r S 0 reactants Jan 24 5:53 PM 22

23 Entropy Changes in Chemical Reactions Calculate the ΔS 0 at 25C for the reaction 2NiS(s) + 3O 2 (g) 2SO 2 (g) + 2NiO(s) with the following standard entropy values S 0 ΔS 0 reaction = ΣΔn p S 0 products Σn r S 0 reactants Substance S 0 (J/K*mol) SO 2 (g) 248 NiO(s) 38 O 2 (g) 205 NiS(s) 53 Jan 24 6:00 PM Entropy Changes in Chemical Reactions Calculate the ΔS 0 for the reduction of aluminum oxide by hydrogen Al 2 O 3 (s) + 3H 2 (g) 3H 2 O(g) + 2Al(s) with the following standard entropy values S 0 Substance S 0 (J/K*mol) Al 2 O 3 (s) 51 H 2 (g) 131 Al(s) 28 H 2 O(s) 189 ΔS 0 reaction = ΣΔn p S 0 products Σn r S 0 reactants Also can calculate ΔS using reactions with known ΔS values Mar 19 1:11 PM 23

24 Free Energy and Chemical Reactions Standard Free Energy Change (ΔG 0 ): The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔG 0 = 33.33kJ We cannot measure Free Energy (there is no instrument to measure free energy) Mar 19 1:11 PM Free Energy and Chemical Reactions There are several ways to calculate Free Energy For the reaction: ΔG 0 = ΔH 0 TΔS 0 C(s) + O 2 (g) CO 2 (g) The values of ΔH 0 + ΔS 0 are known to be kj and 3.05 J/K, respectively, calculate ΔG 0 at 298K Jan 28 12:59 PM 24

25 Free Energy and Chemical Reactions Consider the reaction: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Carried out at 25 0 C and 1 atm. calculate ΔH 0, ΔS 0, and ΔG 0 using the following data: Substance ΔH 0 f (kj/mol) S 0 (J/K*mol) SO 2 (g) SO 3 (g) O 2 (g) Jan 28 1:08 PM Free Energy and Chemical Reactions Since Free Energy is a State Function we can use this method as well: 2CO(g) + O 2 (g) 2CO 2 (g) ΔG 0 =? Use the following data to calculate ΔG 0 for the reaction above: 2CH 4 (g) + 3O 2 (g) 2CO(g) + 4H 2 O(g) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) ΔG 0 = 1088 kj ΔG 0 = 801 kj Mar 19 1:12 PM 25

26 Dec 18 2:50 PM 26

Entropy. Spontaneity. Entropy. Entropy mol of N 2 at 1 atm or 1 mol of N 2 at atm. process a process that occurs without intervention

Entropy. Spontaneity. Entropy. Entropy mol of N 2 at 1 atm or 1 mol of N 2 at atm. process a process that occurs without intervention Entropy Spontaneity process a process that occurs without intervention can be fast or slow Entropy (s) the measure of molecular randomness or disorder Think of entropy as the amount of chaos Entropy Predict