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Topic # 9 Thermochemistry.

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1 Topic # 9 Thermochemistry

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

6 System: all of the reactants and products
Concentrated Sulfuric acid is added to distilled water in a beaker. H2SO4(l) + H2O(l)  HSO4-(aq) + H3O+ (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! Activation Energy Energy
Reaction Coordinate Activation Energy H < 0 Energy is – Given Off!

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

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

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

12 The Laws of Thermodynamics
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 ∆Esystem = q + w In a chemical system, the energy exchanged between a
system and its surroundings can be accounted for by heat (q) and work (w). ∆Esystem = 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 2NH3(g)  N2(g) + 3H2(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.

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!

16 For q: + means the system gained heat - means the system lost heat
w = ∆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 Useful when volumes of gases are shown in a
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. conversion needed: J/liter atmosphere **** 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!****

18 ∆E = q 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 B) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
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) + 2F2(g) SnF4 (s) B) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) C) C(s) + O2(g)  CO2(g) D) SiI4(g) + heat  Si(s) + 2I2(g)


21 ∆H = q (at constant pressure)
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) + O2(g) ∆H = kJ (endo) K2O(s) + CO2(g)  K2CO3(s) ∆H = kJ (exo) Also: HgO(s) + heat  Hg(l) + ½ O2(g) ∆H = kJ 2K2O(s) + 2CO2(g)  2K2CO3(s) ∆H = kJ What’s going on here with these diff equations?

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

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

26 First, let’s look at heat used/given off
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 )
First Way: 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

28 The standard enthalpy of formation
for ethanol (C2H5OH): 2C(s) + 3H2(g) + 1/2O2(g)  C2H5OH(l) Hof = 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 = Hproducts - Hreactants
Second Way: Another way of calculating H is to use combinations of Hof of multiple compounds H = Hproducts - Hreactants Fe2O CO  2Fe + 3CO2 Calculate the standard enthalpy of above reaction using this Hof data: CO = kJ/mol CO2 = kJ/mol Fe2O3 = kJ/mol

31 Fe2O3 + 3 CO  2Fe + 3CO2 H = Hproducts - Hreactants
Hrxn = [ 3(-393.7)] – [3(-110.4) + (-822.2)] No #’s needed for the 2Fe!

32 According to Hess’s law, we can do some
Third Way: What if asked to find Hof for SrCO3? But there isn’t a listing for enthalpy of formation for SrCO3 from its elements? According to Hess’s law, we can do some creative equation manipulation!

33 Add these equations together and get:
Sr + 3/2 O2 + C  SrCO3

34 Hess's Law 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 N2H4 N2 + 2O2  2NO Horxn= +67.4 N2H4 + 3O2  2NO2 + 2H2O Horxn= 2H2 + O2  2H2O Horxn = We need: N2 + 2H2  N2H4 (coefficient of 1 before N2H4)

37 N2 + 2H2  N2H4 N2 + 2O2  2NO2 Horxn= +67.4
2H2 + O2  2H2O Horxn = 2NO2 + 2H2O N2H4 + 3O2 Horxn= Needed to flip one equation, change the sign N2 + 2H2  N2H4

38 Another kind of equation that can be used
Fourth Way: Another kind of equation that can be used is the heat of combustion: Hoc Combustion is the adding of oxygen to a compound or element. When using a hydrocarbon, usually CO2 and H2O are formed. Combustion of CO: CO + 1/2O2  CO2 Combustion of Al : 2Al + 3/2O2  Al2O3 You may have to create this equation if given the kJ/mol of heat.

39 Must write the equations for combustion for each of these items
Calculate the standard enthalpy of formation of ethane, C2H6, given the following combustion data; C = -393, H2 = -286 and C2H6 = -1560kJ/mol Must write the equations for combustion for each of these items 2. Arrange those elements to give you: 2C + 3H2  C2H6 3. Add up the correct kJ’s

40 2(C + O2  CO2) ( -393 kJ) 3(1/2O H2  H2O) (-286 kJ) (Flip!)C2H6 + 7/2O2  3H2O + 2CO kJ 2C + 3H2  C2H6

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

42 Bond Energy (kJ/mol) C=O 743 H--H 436 H-C 412 C--C 348 H-N 391 N--N
170 H-O 463 O--O 145 H-F 565 F--F 154 H-Cl 427 Cl-Cl 239 H-Br 366 Br-Br 193 H-I 299 I--I 151 C-C C=C 612 C--N 305 CºC 837 C--O 360 C--S 272 C--F 485 O=O 498 C--Cl 339 C--Br 276 C--I 238 NºN 945 C=O

43 (Broken – Formed) CH3CH=CH2 + H2  CH3CH2CH3 Broken Formed C=C H-H 2 C-H 2(412) 824 kJ 700 kJ Hrxn = 700 – 824 = kJ/mol NOT the same as H = Hproducts - Hreactants


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 Entropy measure of disorder 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 ∆Srxn = Sproducts – Sreactants
Calculating Entropy There are starting points of Entropy for substances. Measured at the standard state conditions. These are listed in appendix of the text. (So ) H2O(l)  H2O(g) Is liquid water more or less ordered than gaseous water? ∆Srxn = Sproducts – Sreactants ∆Srxn = ( ) J/mol = 119 J/mol + Entropy means greater disorder!

53 Consider this spontaneous reaction:
N H2  2NH3 Why is this Spontaneous, Is the product more or less ordered?

54 N2 + 3H2  2NH3 Using : Srxn = Sproducts – Sreactants
= 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 Suniv = Sreaction + Ssurr
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! Suniv = Sreaction + Ssurr + Suniv = Spontaneous Suniv = NOT Spontaneous (What is meant by the term….. Universe, here?)

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

57 Suniv = Srxn + Ssurr Putting these two ideas together…..
Suniv = [Sproducts -Sreactants] + -H T (let’s see this in action…….)

58 Consider this spontaneous situation:
2H2(g) + O2(g)  2H2O(l) kJ/mole ∆So for the following are: H2(g) = 131 J/mol O2(g)= 205J/mol H2O(l) = 70 J/mol S = Sproducts – Sreactants ∆Sorxn = [2(70)] – [2(131)+ (205)] ∆Sorxn = 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? -Hrxn Ssurr = -(-572 kJ/mol) 298K = T This equals 1919 Joules/mole (converted to Joules) Need to calculate the entropy for the universe…….. (OH! A lot bigger than -327!)

60 Suniv = Srxn + Ssurr Suniv = -327 J/mole + 1919 J/mole
Answer: J/mole + answer means that entropy for the overall process, including system AND surroundings DID, in fact, increase! Suniv > so IS spontaneous

61 ∆Gorxn ∆Gorxn = ∆Horxn – (T)∆Sorxn
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. ∆Gorxn ∆Gorxn = ∆Horxn – (T)∆Sorxn ∆Gorxn < 0 FOR spontaneous reactions ( WORK? What work?)

62 2H2(g) + O2(g)  2H2O(l) + 572 kJ/mole
(where’s the work?) 2H2(g) + O2(g)  2H2O(l) kJ/mole 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 ∆Gorxn = ∆Horxn – (T)∆Sorxn
2H2(g) + O2(g)  2H2O(l) kJ/mole ∆Hosurr > (0.327kJ/mol K given to surroundings) * (298K) = 97kJ/mole available to do work on surroundings ∆Gorxn = ∆Horxn – (T)∆Sorxn ∆Gorxn = needs to be negative to have a spontaneous reaction ∆Gorxn = kJ/mol – [ (298K)(-0.327kJ/mole)] ∆Gorxn = kJ/mol Number is – so reaction is spontaneous

64 ∆Gorxn = ∆Horxn – (T)∆Sorxn
For any reaction to be spontaneous the sole condition is that Gibbs Free Energy needs to be negative! ∆Gorxn = ∆Horxn – (T)∆Sorxn

65 Go is negative for any reaction for which Ho is
Go is negative for any reaction for which  Ho is negative and  So is positive.  Go is negative for any reaction where both the enthalpy and entropy terms are negative.

66 NO Yes Keys to determining why a reaction happens H Sreaction G +
at high temp + at low temp - - at low temp + at high temp 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 Formation of LiF Li(s) + ½ F2(g) (pgs. 363 – 367 of text)
Li+ (g) + F (g) Li+ (g) + ½ F2(g) 77kJ -328 kJ Li+ (g) + F- (g) 520kJ Li (g) + ½ F2(g) 161kJ Li(s) + ½ F2(g) -1047 kJ Overall -617 kJ LiF


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