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Chemical Reactions involve: reactants turned into products an energy change in form of heat Energy changes accompany all chemical reactions and are due.

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Presentation on theme: "Chemical Reactions involve: reactants turned into products an energy change in form of heat Energy changes accompany all chemical reactions and are due."— Presentation transcript:


2 Chemical Reactions involve: reactants turned into products an energy change in form of heat Energy changes accompany all chemical reactions and are due to rearranging of chemical bonding.

3 The addition of energy is always a requirement for breaking of bonds but the breaking of bonds in and of itself does not release energy. Energy release occurs when NEW BONDS are formed.

4 If more energy is released when new bonds form than was required to break existing bonds, then the difference will result in an overall release of energy…. EXOTHERMIC If more energy is required to break existing bonds then is released when new bonds form, then energy is being absorbed….. ENDOTHERMIC

5 THERMODYNAMICS: the study of energy transformations To study thermodynamics (thermochemistry) situations need to be viewed a little differently…….. Surroundings : everything else that is not a part of the system System : All substances under going a physical or chemical change

6 Concentrated Sulfuric acid is added to distilled water in a beaker. H 2 SO 4 (l) + H 2 O(l)  HSO 4 - (aq) + H 3 O + (aq) System: all of the reactants and products Surroundings: water molecules NOT being used, beaker, air…… etc. ( this is an exothermic process, heat is given off…. what does that mean? )

7 Energy is a property that is determined by specifying the condition or “state” ( temperature, pressure, etc) of a system or substance. (is NOT really solid, liquid or gas….) That’s why its called a STATE FUNCTION (where its value is determined only by its present condition and not on its history) Ohio

8 EXOTHERMIC REACTION Energy is – Given Off! Energy Reaction Coordinate Activation Energy  H < 0

9 ENDOTHERMIC Energy is + Absorbed! Energy Reaction Coordinate Activation Energy  H > 0 Reactants Products

10 Internal Energy of a System Internal energy can be increased by: an increase in temperature a phase change the initiation of a chemical reaction Internal energy can be Decreased by: a decrease in temperature a phase change

11 Total Energy of a System cannot be determined, but the CHANGE may be measured. ∆E = E final - E initial

12 First Energy cannot be created or destroyed but only transferred from one body to another or changed from one form to another. Second Every spontaneous change increases the entropy of the universe/system. Third The entropy of a perfect crystalline substance (no disorder) is zero at 0 K.

13 In a chemical system, the energy exchanged between a system and its surroundings can be accounted for by heat (q) and work (w). ∆E system = q + w ∆E < 0 (negative) means system gave away energy to surroundings ∆E > 0 (positive) means system absorbed energy from surroundings If q < 0 then system gave away heat to surroundings (-) If w < 0 then system did work on surroundings (-) ALL ENERGY MEASURED IN JOULES!

14 2NH 3 (g)  N 2 (g) + 3H 2 (g) More moles of gas are produced than reacted. This system has done work on its surroundings. The system is ‘pushing back’ to make room for the expanding gas.

15 Only pressure/volume work (expansion or contraction of a gas) is of importance in a chemical reaction system, and only when there is an increase or decrease in the amount of of gas present! VERY IMPORTANT AND USEFUL:

16 q+ w=∆E∆E Depends on values of q and w For q: + means the system gained heat - means the system lost heat For w : + means work is done ON system - means system does work ON surroundings

17 More on Work; w ……… The expansion/contraction of a gas against constant external pressure can be expressed: w = -P∆V Useful when volumes of gases are shown in a chemical reaction. **** If there is no change in the total volume of gas before and after a reaction, then there is no work done by OR on the system!**** conversion needed: J/liter atmosphere

18 Remember! To DO work, something MUST MOVE! If a reaction occurs in a rigid container, only the pressure, NOT the volume may change. If no volume change occurs, nothing moved…… NO WORK is done on or by the system! That means that any ∆E change is due ONLY to heat transfers. So…… ∆E = q

19 Example: For the following reactions performed at constant external pressure, is work done on the system by the surroundings, by the system on the surroundings or is the amount of work negligible? A) Sn(s) + 2F 2 (g)  SnF 4 (s) B) AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) C) C(s) + O 2 (g)  CO 2 (g) D) SiI 4 (g) + heat  Si(s) + 2I 2 (g)


21 Most times chemical reactions occur where the external pressure is constant…. most often at atmospheric pressure and in aqueous forms Here, a pressure change or volume change does not happen. So the heat exchange is the important thing to consider……. ∆H = q (at constant pressure) ∆H is ENTHALPY, the exchange of heat between a system and its surroundings at constant pressure Commonly called: heat of reaction

22 The amount of ∆H in any given reaction is directly proportional to the quantity of reactants AND depends on the conditions that the reaction takes place. So… ∆H is associated with a balanced chemical equation and the temperature and pressure are specified and units of enthalpy change is given in Joules/mole

23 A thermochemical equation: 2HgO(s) + heat  2Hg(l) + O 2 (g) ∆H = kJ (endo) K 2 O(s) + CO 2 (g)  K 2 CO 3 (s) ∆H = kJ (exo) HgO(s) + heat  Hg(l) + ½ O 2 (g) ∆H = kJ Also: 2K 2 O(s) + 2CO 2 (g)  2K 2 CO 3 (s) ∆H = kJ What’s going on here with these diff equations?

24 One special set of conditions used with chemical equations are the Standard State Conditions: Standard temperature = 25 o C Standard pressure = 1 atm Standard concentration = 1 molar When these conditions are met, then the ∆H measure is called Standard Enthalpy of Reaction: ∆H o rxn (has an o by the letter)

25 Standard Enthalpy of Reaction: ∆H o rxn The amount of energy associated with with reactions at standard states has been calculated for many reactions and can be found on lists of ∆H o rxn If the sign of the kJ/mol is negative then the reaction is exothermic. (some equation data can be made available from the q equation at standard conditions)

26 If there is no list available to you, then these numbers can be calculated in a variety of ways. First, let’s look at heat used/given off by the formation of compounds from their elements.

27 Standard Enthalpy of Formation (∆H° f ) change in enthalpy (heat) that accompanies the formation of 1 mole of that substance from its elements, with all substances in their standard states First Way:

28 The standard enthalpy of formation for ethanol (C 2 H 5 OH): 2C (s) + 3H 2 (g) + 1/2O 2 (g)  C 2 H 5 OH (l)  H o f = kJ/mol only elements as reactants only 1 mole of product made

29 Remember: the standard enthalpy of the formation of element has got to be zero. Why is that?

30 H = H products - H reactants Another way of calculating  H is to use combinations of  H o f of multiple compounds Fe 2 O CO  2Fe + 3CO 2 Calculate the standard enthalpy of above reaction using this  H o f data: CO = kJ/mol CO 2 = kJ/mol Fe 2 O 3 = kJ/mol Second Way:

31  H =  H products -  H reactants  H rxn = [ 3(-393.7)] – [3(-110.4) + (-822.2)] Fe 2 O CO  2Fe + 3CO 2 No #’s needed for the 2Fe!

32 What if asked to find  H o f for SrCO 3 ? But there isn’t a listing for enthalpy of formation for SrCO 3 from its elements? According to Hess’s law, we can do some creative equation manipulation! Third Way:

33 Add these equations together and get: Sr + 3/2 O 2 + C  SrCO 3

34 The enthalpy change for any reaction depends on the products and reactants and is independent of the pathway or the number of steps between the reactant and product.

35 Find other reactions that are easier to perform and when summed up, equal the original desired reaction. According to Hess's Law, the sum of the heat flows for these reactions is equal to the heat flow for the original reaction.

36 Find the standard enthalpy of formation of N 2 H 4 N 2 + 2O 2  2NO 2  H o rxn = N 2 H 4 + 3O 2  2NO 2 + 2H 2 O  H o rxn = H 2 + O 2  2H 2 O  H o rxn = We need: N 2 + 2H 2  N 2 H 4 (coefficient of 1 before N 2 H 4 )

37 N 2 + 2O 2  2NO 2  H o rxn = H 2 + O 2  2H 2 O  H o rxn = NO 2 + 2H 2 O  N 2 H 4 + 3O 2  H o rxn = Needed to flip one equation, change the sign N 2 + 2H 2  N 2 H 4

38 Another kind of equation that can be used is the heat of combustion:  H o c Combustion is the adding of oxygen to a compound or element. When using a hydrocarbon, usually CO 2 and H 2 O are formed. Combustion of CO: CO + 1/2O 2  CO 2 Combustion of Al : 2Al + 3/2O 2  Al 2 O 3 You may have to create this equation if given the kJ/mol of heat. Fourth Way:

39 Calculate the standard enthalpy of formation of ethane, C 2 H 6, given the following combustion data; C = -393, H 2 = -286 and C 2 H 6 = -1560kJ/mol 1. Must write the equations for combustion for each of these items 2. Arrange those elements to give you: 2C + 3H 2  C 2 H 6 3. Add up the correct kJ’s

40 2(C + O 2  CO 2) 2( -393 kJ) 3( 1/2 O 2 + H 2  H 2 O) 3(-286 kJ) (Flip!) C 2 H 6 + 7/2 O 2  3H 2 O + 2CO kJ 2C + 3H 2  C 2 H 6

41 Still another way to describe energy changes in compound includes the energy changes from breaking and forming bonds  H rxn = ΣH bonds broken - ΣH bonds formed (pgs in text) add up the bonds broken and formed in a given equation and that can equal the heat of reaction. Fifth Way:

42 Bond Energy (kJ/mol) Bond Energy Bond Energy H--H436H-C412 C--C348H-N391 N--N170H-O463 O--O145H-F565 F--F154H-Cl427 Cl-Cl239H-Br366 Br-Br193H-I299 I--I151C-C 348 C--C348C=C612 C--N305 CCCC 837 C--O360 C--S272 O--O145 C--F485O=O498 C--Cl339 C--Br276 N--N170 C--I238 NNNN 945 C=O 743

43 CH 3 CH=CH 2 + H 2  CH 3 CH 2 CH 3 (Broken – Formed) BrokenFormed C=C H-H C-H 2(412) 700 kJ 824 kJ  H rxn = 700 – 824 = -124 kJ/mol NOT the same as  H =  H products -  H reactants


45 Spontaneous Processes and Entropy Why does a chemical reaction go naturally in one particular direction? A spontaneous process is one that, once initiated, continues to proceed in the forward direction with no further input of energy. Using certain characteristics of a reaction, we can describe the spontaneity of the reaction.


47 What are the forces that could lend a reaction to become spontaneous or nonspontaneous?

48 Movement from order to disorder [Many times overcoming a high positive activation energy (high +  H)]

49 Degree of disorder of a system = S Called Entropy Increase in disorder (entropy) will result in + quantity. Decrease will be – quantity. Spontaneous reactions usually have + value for Entropy

50 Entropy can take many forms: In terms of Solid, Liquid and Gas Phases, the phases that provide more random motion would have greater entropy. Within a gas sample, the sample that has the higher temperature or volume would provide more random motion and have greater entropy. Substances that are made of larger particles compared to substances of smaller particles, have greater entropy. Mixtures of gases, solids and solutions are usually viewed has having greater entropy.

51 Entropy is a STATE function……. Means it depends on the quantity and conditions of the substance. More substance reacting = more entropy

52 H 2 O(l)  H 2 O(g) Is liquid water more or less ordered than gaseous water? ∆S rxn = S products – S reactants There are starting points of Entropy for substances. Measured at the standard state conditions. These are listed in appendix of the text. (  S o ) ∆S rxn = ( ) J/mol = 119 J/mol + Entropy means greater disorder! Calculating Entropy

53 Consider this spontaneous reaction: N 2 + 3H 2  2NH 3 Why is this Spontaneous, Is the product more or less ordered?

54 N 2 + 3H 2  2NH 3 Using :  S rxn = S products – S reactants  S rxn = 2(193) – [ (131)] = -199 J/mol Negative sign says the product is more ordered. Entropy is not the reason this reaction occurs. so what DID cause this reaction to be spontaneous?

55 To determine the spontaneity of a process, more than a few things need consideration. Second (Law of Thermodynamics) Every spontaneous change increases the total entropy of a system and its surroundings. For a chemical process to be spontaneous, it must increase the entropy of the universe!  S univ =  S reaction +  S surr (What is meant by the term….. Universe, here?) +  S univ = Spontaneous -  S univ = NOT Spontaneous

56 The entropy of the surroundings can be calculated:  S surr = -H-H T We know how to calculate the Entropy of the reaction…. (Sometimes q)

57  S univ =  S rxn +  S surr  S univ = [  S products -  S reactants ] + -  H T Putting these two ideas together….. (let’s see this in action…….)

58 Consider this spontaneous situation: 2H 2 (g) + O 2 (g)  2H 2 O(l) kJ/mole ∆S o for the following are: H 2 (g) = 131 J/mol O 2 (g)= 205J/mol H 2 O(l) = 70 J/mol  S = S products – S reactants ∆S o rxn = [2(70)] – [2(131)+ (205)] ∆S o rxn = -327 J/mole ( lower entropy) ( Seems correct, water is more ordered than the two gases)

59 Why is this reaction spontaneous? Disorder is not increased! What’s the force, here?  S surr = -  H rxn T = -(-572 kJ/mol) 298K This equals 1919 Joules/mole (converted to Joules) (OH! A lot bigger than -327!) Need to calculate the entropy for the universe……..

60  S univ =  S rxn +  S surr  S univ = -327 J/mole J/mole Answer: J/mole + answer means that entropy for the overall process, including system AND surroundings DID, in fact, increase!  S univ > 0 so IS spontaneous

61 Further study of these calculations and processes lead to a culmination formula that connects ∆H, heat or enthalpy and ∆S, entropy or disorder: This is another way to calculate/prove that a reaction is spontaneous…… Gibbs Free Energy: The amount of energy that is available to do work on the UNIVERSE in a chemical reaction. ∆G o rxn ( WORK? What work?) ∆G o rxn = ∆H o rxn – (T)∆S o rxn ∆G o rxn < 0 FOR spontaneous reactions

62 2H 2 (g) + O 2 (g)  2H 2 O(l) kJ/mole (where’s the work?) Remember negligible work? There is not much energy available to do work on this system; three moles of gas on one side, NONE on the other. Negligible to zero work done on the system. Most of the energy here is given off as enthalpy, not work. But according to the Second Law, for this reaction to proceed as spontaneous, enough energy as work must be given away to the surroundings so that the increase in entropy of the surroundings offsets the decrease in the entropy of the system. ( how much energy is that?)

63 ∆H o surr > (0.327kJ/mol K given to surroundings ) * (298K) = 97kJ/mole available to do work on surroundings 2H 2 (g) + O 2 (g)  2H 2 O(l) kJ/mole ∆G o rxn = ∆H o rxn – (T)∆S o rxn ∆G o rxn = needs to be negative to have a spontaneous reaction ∆G o rxn = -572 kJ/mol – [ (298K)(-0.327kJ/mole)] ∆G o rxn = -475 kJ/mol Number is – so reaction is spontaneous

64 ∆G o rxn = ∆H o rxn – (T)∆S o rxn For any reaction to be spontaneous the sole condition is that Gibbs Free Energy needs to be negative!

65  G o is negative for any reaction for which  H o is negative and  S o is positive.  G o is negative for any reaction where both the enthalpy and entropy terms are negative.

66 HH  S reaction  G at high temp + at low temp at low temp + at high temp Keys to determining why a reaction happens NO Yes

67 Our description of why a reaction occurs sometimes needs to include HOW need to describe a step-wise progression on exactly how a compound is formed. needs to include all energies along the way The Born Haber Cycle (normally associated with an ionically bonded compound)

68 Include these energy changes as needed: Standard enthalpy of formation Standard enthalpy of vaporization (substance into gas) First Ionization energy ( gas losing an electron) Enthalpy of bond dissociation (molecules to atoms) First electron affinity (gas gaining an electron) Enthalpy of Lattice (energy associated with one mole of solid is formed from its gaseous ions)

69 (pgs. 363 – 367 of text) Formation of LiF Li(s) + ½ F 2 (g) 161kJ Li (g) + ½ F 2 (g) 520kJ Li + (g) + ½ F 2 (g)77kJ Li + (g) + F (g) Li+ (g) + F - (g) -328 kJ LiF kJ Overall -617 kJ


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