Topic # 9 Thermochemistry

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

9. Activation Energy Energy H > 0 Reactants Reaction Coordinate 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 a phase change

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

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. The Laws of Thermodynamics

13. 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.

15. VERY IMPORTANT AND USEFUL: 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 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. conversion needed: 101.325 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. 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) + 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. 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 = 181.67 kJ (endo) K2O(s) + CO2(g)  K2CO3(s) ∆H = -391.1 kJ (exo) Also: HgO(s) + heat Hg(l) + ½ O2(g) ∆H = 90.84 kJ 2K2O(s) + 2CO2(g)  2K2CO3(s) ∆H = -782.2kJ 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 = 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. 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. 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 = -277.7kJ/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 Fe2O3 + 3 CO  2Fe + 3CO2 Calculate the standard enthalpy of above reaction using this Hof data: CO = -110.4kJ/mol CO2 = -393.7 kJ/mol Fe2O3 = -822.2 kJ/mol

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

32. 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 2NO2Horxn= +67.4 N2H4 + 3O2  2NO2 + 2H2O Horxn= -554.4 2H2 + O2  2H2O Horxn = -571.5 We need: N2 + 2H2 N2H4 (coefficient of 1 before N2H4)

37. N2 + 2O2 2NO2Horxn= +67.4 2H2 + O2  2H2OHorxn = -571.5 2NO2 + 2H2O N2H4 + 3O2Horxn= +554.4 Needed to flip one equation, change the sign N2 + 2H2 N2H4

38. 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. 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) 2( -393 kJ) 3(1/2O2 + H2 H2O) 3(-286 kJ) (Flip!)C2H6 + 7/2O2 3H2O + 2CO2 -1560 kJ 2C + 3H2 C2H6

41. 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.373-375 in text)

42. C=O 743

43. (Broken – Formed) CH3CH=CH2 + H2 CH3CH2CH3 Broken Formed C=C 612-348 H-H 436 2 C-H 2(412) 824 kJ 700 kJ Hrxn = 700 – 824 = -124 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 processis 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.