Chemical Thermodynamics for 1st year students

chemical thermo

1. Chemistry CHEM1004 Chemical Thermodynamics Suresh Krishnasamy School of Agriculture and Food Science University of Queensland

2. At the end of this lesson, you would be able to: Explain how chemical and physical processes may lead to changes in both heat and work {Part 1} Predict the spontaneity of reactions using enthalpy, entropy and temperature. {Part 2} Solve chemical thermodynamic problems using thermodynamic data and mathematical equations. Learning Outcomes 2 Inspiration for this lecture comes from chapter 8 of the recommended textbook.

3. Part 1 Internal Energy, Heat and Work Done 3

4. Laws of Thermodynamics https://sciencenotes.org/laws-of-thermodynamics/

5. The First Law of Thermodynamics, also known as the law of conservation of energy , states that energy can neither be created nor destroyed but may change from one form to another. BUT the energy in a closed system remains constant. How does that make sense? 1 st Law of Thermodynamics 5 https://qsstudy.com/significance-first-law-thermodynamics/

6. Internal energy is the energy due to the random motion of atoms, molecules, or particles in a system. It is measured at the microscopic scale and includes all macroscopic energies that the system possesses like: Kinetic energy due to translational, vibrational, and rotational motion Potential energy due to intermolecular forces The internal energy is a state function , not a path function.  C hange in internal energy depends on the initial and final states and not on the process taken to make the change. The symbol for internal energy is U and its unit is J and is calculated per mole of a substance. For most practical purposes, the change in internal energy ΔU is usually calculated. Internal Energy 6 http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html  

7. 1 st Law Mathematics!! 7   q = heat w = work done by gas   Heat (J) Mass (g) Temperature Change ( o C ) Specific Heat Capacity (Jg -1o C -1 )   Work Done by Gas (J) Volume Change (m 3 ) Pressure (Pa) https://www.chemistrylearner.com/internal-energy.html

8. Heat transfer is the transfer of energy from one object to another. NOT the same as temperature though more often than not increase in heat causes increase in temperature. Heat is a form of energy, while temperature is how hot or cold something is. Temperature as the average kinetic energy of a substance. Heat 8 https://www.britannica.com/science/heat-transfer

9. For a given mass, some substances require more heat than others to undergo the same change in temperature. Heat capacity is the amount of heat required to raise the temperature of the substance by 1°C (1K). Specific heat capacity is the amount of heat required to raise the temperature of 1g of the substance by 1°C (1K). Regardless of what material is used, it is determined experimentally that the temperature change is directly proportional to the heat absorbed by the material: .   Heat Capacity (C p ) 9   Heat (J) Mass (g) Temperature Change ( o C ) Specific Heat Capacity (Jg -1o C -1 )   Heat (J) Temperature Change ( o C ) Heat Capacity (J o C -1 )

10. A 2.28 g lead weight, initially at 11.1 o C, is submerged in 8.15 g of water at 52.8 o C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium? { C pb = 0.129 J/ g o C , C H 2 O = 4.186 J/ g o C } Example 10

11. A 2.28 g lead weight, initially at 11.1 o C, is submerged in 8.15 g of water at 52.8 o C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium? { C pb = 0.129 J/ g o C , C H 2 O = 4.186 J/ g o C } Example 11 Step 1: List the known quantities. Mass: m pb = 2.28g, m H 2 O = 8.15g Ti : Ti pb = 11.1 o C , Ti H 2 O = 52.8 o C Step 2: Select the right equation.   Step 3: Substitute and Solve.  Final Temperature: 52.4 o C  

12. Imagine an ideal gas confined by a frictionless piston. If P ext = P int , the system is at equilibrium; the piston does not move, and no work is done. If P ext < P int , the gas will expand, performing work on its surroundings  V f > V i . If P ext > P int , the gas will be compressed, and the surroundings will perform work on the gas  V f < V i . Work 12 https://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s22-01-thermodynamics-and-work.html   Work Done by Gas (J) Volume Change (m 3 ) Pressure (Pa)

13. The lung volume of a 70 kg man at rest changed from 2200mL to 2700mL when he inhaled, while his lungs maintained a pressure of approximately 1.0atm. How much work was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work did he require to take a breath while exercising? Example 13

14. The lung volume of a 70 kg man at rest changed from 2200mL to 2700mL when he inhaled, while his lungs maintained a pressure of approximately 1.0atm. How much work was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work did he require to take a breath while exercising? Example 14 Step 1: List the known quantities. P ext = 1.0 atm = 10 5 Pa ΔV rest = 2700mL – 2200mL = 500mL = 0.0005m 3 ΔV exercise = 5200mL – 2200mL = 3000mL = 0.003m 3 Step 2: Select the right equation.   Step 3: Substitute and Solve.  w rest = -50 J   Step 3: Substitute and Solve.  w exercise = -300 J   negative work done  work done by the gas

15. Heat If heat enters the gas, q will be positive. If heat exits the gas, q will be negative. Work If the gas is compressed, the work done on the gas, w will be positive. If the gas expands, the work done by the gas w will be negative. Heat and Work 15 https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/a/what-is-the-first-law-of-thermodynamics    

16. (Increase in Internal Energy / Temperature) (Heat enter the system) (work done on gas / Volume decreased) (Decrease in Internal Energy / Temperature) (Heat exits system) (work done by gas / Volume increased) (no change in Internal Energy / Temperature) [Constant Temperature] (no heat exchanged) (no work done) [Constant Volume] Overview 16 Isothermal  Constant Temperature Isobaric  Constant Pressure Adiabatic Isochoric  Constant Volume

17. A balloon filled with 39.1 mole of helium has a volume of 876L at 0 o C and 1atm. The temperature of the balloon increased to 38 o C as it expands to a volume of 998L at constant pressure. Calculate q, w and for the helium in the balloon. { C He = 20.8 J/ mol o C }   Example 17

18. A balloon filled with 39.1 mole of helium has a volume of 876L at 0 o C and 1atm. The temperature of the balloon increased to 38 o C as it expands to a volume of 998L at constant pressure. Calculate q, w and for the helium in the balloon. { C He = 20.8 J/ mol o C }   Example 18 Step 1: List the known quantities. P ext = 1.0 atm = 10 5 Pa ΔV= 998L – 876L = 122L = 0.000122m 3 ΔT= 38 o C - 0 o C = 38 o C Step 2: Select the right equation.    Step 3: Substitute and Solve. KJ   KJ KJ KJ   Step 3: Substitute and Solve. KJ   Step 3: Substitute and Solve. J  

19. Part 2 Enthalpy, Entropy and Gibbs Free Energy 19

20. Enthalpy

21. Enthalpy, H, of a substance is its total energy content held at constant pressure. Note: Chemical reactions done in the laboratory are constant pressure processes. H cannot be measured directly since it depends on number and nature of particles kinetic energy of the particles space occupied by the particles It is, however, possible to measure enthalpy change ∆H during a chemical reaction. In a reaction, ∆H = Hproducts - Hreactants Enthalpy (H) & Enthalpy Change (∆H) 21

22. In an exothermic reaction, the reactants have higher energy compared to the products. As the reaction proceeds, energy is released into the surroundings. Lower energy can be thought of as providing a greater degree of stability to a chemical system. Since the energy of the system decreases during an exothermic reaction, the products of the system are more stable than the reactants. We can say that an exothermic reaction is an energetically favorable reaction. Let’s Talk About ∆H 22 https://iteachly.com/endothermic-and-exothermic-reactions-lab/ ∆H<0 ∆H>0

23. Enthalpy change when molar quantities of reactants as stated in the chemical equation react under the standard conditions of 298K and 10 5 Pa pressure . The physical states of reactants and products affect the enthalpy change of a reaction. Hence physical states and + ve and – ve sign of  H  MUST always BE STATED in all thermochemical equations. Standard Enthalpy Change of Reaction ∆H θ 23 4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O (l)  H  = -1260 kJ mol -1 NH 3 (g) + ¾ O 2 (g)  ½ N 2 (g) + 3/2 H 2 O (l)  H  = -315 kJ mol -1 2N 2 (g) + 6H 2 O (l)  4NH 3 (g) + 3O 2 (g)  H  = +1260 kJ mol -1 4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O (g)  H  = -1216 kJ mol -1

24. Method: Reaction is carried out in a calorimeter The initial and final temperatures are measured using a thermometer q = m solution c solution ∆T + m cal c cal ∆T ≈ m solution c solution ∆T Assumptions [made when information not provided]: Heat capacities of containers are small (compared to that of aqueous solutions) and assumed to be zero Density of aqueous solution assumed as 1.00gcm -3 Specific heat capacity of aqueous solutions assumed as 4.18 J K -1 g -1 . Measurement of Enthalpy Change of Reaction q ∆H J or KJ KJ/mol

25. 50.0 mL of water at 40.5°C is added to a calorimeter containing 50.0 mL of water at 17.4°C. After waiting for the system to equilibrate, the final temperature reached is 28.3°C. Calculate the heat capacity of the calorimeter. (c H2O = 4.18 J K -1 g -1 ) Example 25

26. 50.0 mL of water at 40.5°C is added to a calorimeter containing 50.0 mL of water at 17.4°C. After waiting for the system to equilibrate, the final temperature reached is 28.3°C. Calculate the heat capacity of the calorimeter. (c H2O = 4.18 J K -1 g -1 ) Example 26 Step 1: List the known quantities. Mass: m hot = 50.0g, m cold = 50.0g ∆T: ∆T hot = -12.2 o C, ∆ T cold = 10.9 o C, ∆ T cal = 10.9 o C Step 2: Select the right equation. or   Step 3: Substitute and Solve.  = 24.9 JK -1  

27. Enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under the standard conditions of 298K and 10 5 Pa pressure. By definition, the standard enthalpy of an element is zero. Standard Enthalpy Change of Formation, ∆H f θ 27  H r  =  m  H f (products) -  n  H f (reactants) H 2 (g) + ½O 2 (g)  H 2 O(l)  H f  (H 2 O) = -286 kJ mol -1 C(s) + 2H 2 (g)  CH 4 (g)  H f  (CH 4 ) = -75 kJ mol -1 Na(s)  Na(s)  H f  (Na) = 0 kJ mol -1

28. Enthalpy change when 1 mole of a substance is completely burnt from its constituent elements in their standard states under the standard conditions of 298K and 10 5 Pa pressure. Standard Enthalpy Change of Combustion, ∆ H c θ 28  H r  =  m  H f (reactants) -  n  H f (products) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H c  = -890 kJ mol -1 H 2 (g) + ½O 2 (g)  H 2 O(l)  H c  = -286 kJ mol -1 =  H f  ( H 2 O) C(s) + O 2 (g)  CO 2 (g)  H c  (graphite) = -393 kJ mol -1

29. Bond energy (or bond enthalpy) is the energy absorbed when a mole of covalent bond between atoms in a molecule in the gaseous state is broken to form two moles of free gaseous atoms . Enthalpy change when 1 one mole of gaseous atoms is formed from its element under the standard conditions of 298K and 10 5 Pa pressure. Bond Energy and Standard Enthalpy Change of Atomisation, ∆ H atm θ 29  H  reaction =  BE (reactants) -  BE (products) 2H 2 (g)  4H(g)  H = 4  H  at­ = 2 BE (H-H) = 2(+436) = +872 kJ mol -1 Bond Bond energy (kJ mol -1 ) C ––– H 410 Cl ––– Cl 244 C ––– Cl 340 H ––– Cl 431 Na(s)  Na(g)  H  atm = +108 kJ mol -1 ½C l 2 (g)  C l (g)  H  at = +122 kJ mol -1 [= ½ BE(C l - C l )]

30. Enthalpy change of a chemical reaction is the same regardless of the route taken, provided initial and final states are the same in each route. Law used to find enthalpy changes which cannot be measured directly in laboratory. e.g.,  H f  (CH 4 ) Hess’ Law 30 Route 1 : D H 1 A B Route 2 C D D H 2 D H 4 D H 3 By Hess ’ Law,  H 1 =  H 2 +  H 3 +  H 4

31. Given the following standard enthalpy changes of formation in kJ mol -1 Calculate the standard enthalpy change for following reaction: CH 4 (g) + 4F 2 (g)  CF 4 (g) + 4HF(g) Example 31 CH 4 (g) -76 CF 4 (g) -925 HF(g) -271

32. Given the following standard enthalpy changes of formation in kJ mol -1 Calculate the standard enthalpy change for following reaction: CH 4 (g) + 4F 2 (g)  CF 4 (g) + 4HF(g) Example 32 CH 4 (g) -76 CF 4 (g) -925 HF(g) -271 1 2

33. Given the following standard enthalpy changes of formation in kJ mol -1 Calculate the standard enthalpy change for following reaction: CH 4 (g) + 4F 2 (g)  CF 4 (g) + 4HF(g) Example 33 CH 4 (g) -76 CF 4 (g) -925 HF(g) -271  H  reaction =  m  H f  (products) -  n  H f  (reactants) =  H f  (CF 4 ) + 4  H f  (HF) -  H f  (CH 4 ) = [(-925) + 4(-271)] - [(-76) + (0) = [-2009] - [-76] = - 1933 kJ mol -1 By Hess’ Law,  H  reaction = -  H f  (CH 4 ) +  H f  (CF 4 ) + 4  H f  (HF)  H  reaction = - (-76) + -925 + 4(-271)  H  reaction = -1933 kJ mol -1 1 2

34. Entropy 34

35. The second law of thermodynamics started from the idea that heat can flow spontaneously from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object. The second law of thermodynamics states that as energy is transferred or transformed, more and more of it is wasted.  The “wasted” energy goes towards Entropy The 2 nd Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state. 2 nd Law of Thermodynamics 35 Entropy (S) is a measure of disorder of a system. √ ? https://lms.su.edu.pk/lesson/826/week-5-thermal-equilibrium-definition-and-formulation-of-the-first-law-of-thermodynamics

36. Increasing Entropy 36 https://courses.lumenlearning.com/suny-chem-atoms-first/chapter/entropy/   Increasing Temperature

37. Increasing Entropy 37

38. Order in Our Natural World 38 https://www.wsj.com/articles/the-natural-order-of-things-1460126135 https://www.chemistryworld.com/news/snowflake-symmetry-mirrors-ice-crystals-molecular-structure/3007369.article As open systems, organisms can increase their order as long as the order of their surrounding decreases, with an overall increase in entropy in the universe. https://scied.ucar.edu/learning-zone/how-climate-works/tree-rings-and-climate https://en.wikipedia.org/wiki/File:Metabolic_Metro_Map.svg

39. Gibbs Free Energy and Spontaneity 39

40. The Gibbs free energy (G) of a system is a measure of the amount of usable energy in that system. The change in Gibbs free energy (ΔG) during a reaction provides useful information about the reaction's energetics and spontaneity. When we work with Gibbs free energy, we have to make some assumptions, such as constant temperature and pressure; however, these conditions hold roughly true for cells and other living systems. Bringing Them Together - Gibbs Free Energy 40   Enthalpy (KJ/mol) Entropy (J/K) Temperature in Kelvin °C + 273 = 273K

41. (endothermic) (exothermic) (Increase in Entropy) at low T Process is spontaneous at any T (Decrease in Entropy) Process is NOT spontaneous at any T at high T at low T Exergonic and Endergonic Reactions 41   Endergonic reactions are NOT spontaneous   Exergonic reactions are spontaneous  

42. At Equilibrium 42 https://www.scienceabc.com/nature/universe/what-is-the-big-freeze-hypothesis.html at T > 0 o C   at T < 0 o C   at T < 0 o C   at T > 0 o C   at T = 0 o C   at T = 0 o C  

43. Would you expect each of the following reactions to be spontaneous at low temperatures, high temperatures, all temperatures, or not at all? Explain. 2 N 2 O( g ) → 2 N 2 ( g ) + O 2 ( g ) Δ H° = –164.1 kJ H 2 O( g ) + ½O 2 ( g ) → H 2 O 2 ( g ) Δ H° = +105.5 kJ CH 4 ( g ) + O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( g ) Δ H° = –802.3 kJ Example 43

44. Would you expect each of the following reactions to be spontaneous at low temperatures, high temperatures, all temperatures, or not at all? Explain. 2 N 2 O( g ) → 2 N 2 ( g ) + O 2 ( g ) Δ H° = –164.1 kJ H 2 O( g ) + ½O 2 ( g ) → H 2 O 2 ( g ) Δ H° = +105.5 kJ CH 4 ( g ) + O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( g ) Δ H° = –802.3 kJ Example 44 ΔS° is predicted to be positive since there are more moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TΔS° is small) and the reaction will be spontaneous (ΔH° < 0) while at high temperature the entropy dominates and the reaction is also spontaneous (TΔS° becomes large and positive). Therefore, the reaction is expected to be spontaneous over all temperatures. ΔS° will be negative (1.5 moles gases in reactants but only 1 mole of gases in the product) and ΔH° is positive so ΔG° will be positive at all temperatures. Thus, the reaction is never spontaneous. ΔS° will be positive (2 moles gases in reactants but 3 moles of gases in the product) and ΔH° is negative so ΔG° will be negative at all temperatures. Thus, the reaction is spontaneous at all temperatures.

45. The reaction of aqueous solutions of barium nitrate with sodium iodide is described by the following equation: Ba(NO 3 ) 2 ( aq ) + 2NaI( aq ) → BaI 2 ( aq ) + 2NaNO 3 ( aq ) You want to determine the entropy of formation for BaI 2 , but that information is not listed in your tables. However, you have been able to obtain the following information: You know that ΔG° for the reaction at 25°C is 22.64 kJ/mol. What is ΔH° for this reaction? What is S° for BaI 2 ? Example 45 Ba(NO 3 ) 2 NaI BaI 2 NaNO 3 Δ H ∘ f  (kJ/mol) −952.36 −295.31 −605.4 −447.5 Δ S ∘ f   [J/( mol·K )] 302.5 170.3 ? 205.4

46. Example 46 The reaction of aqueous solutions of barium nitrate with sodium iodide is described by the following equation: Ba(NO 3 ) 2 ( aq ) + 2NaI( aq ) → BaI 2 ( aq ) + 2NaNO 3 ( aq ) You want to determine the entropy of formation for BaI 2 , but that information is not listed in your tables. However, you have been able to obtain the following information: You know that ΔG° for the reaction at 25°C is 22.64 kJ/mol. What is ΔH° for this reaction? What is S° for BaI 2 ? Ba(NO 3 ) 2 NaI BaI 2 NaNO 3 Δ H ∘ f  (kJ/mol) −952.36 −295.31 −605.4 −447.5 Δ S ∘ f   [J/( mol·K )] 302.5 170.3 ? 205.4 Ba(NO 3 ) 2 ( aq ) + 2NaI( aq ) → BaI 2 ( aq ) + 2NaNO 3 ( aq ) Ba(s) + N 2 (g) + 3O 2 (g) + 2Na(s) + I 2 (l) −952.36 2(−295.31) −605.4 2(−447.5) Δ H ∘ r Δ H ∘ f = - (-952.36) - 2(-295.31) + (-605.4) + 2(-447.5) Δ H ∘ f = +42.6 KJ/mol   Ba(NO 3 ) 2 ( aq ) + 2NaI( aq ) → BaI 2 ( aq ) + 2NaNO 3 ( aq ) Ba(s) + N 2 (g) + 3O 2 (g) + 2Na(s) + I 2 (l) 302.5 2(170.3) 2(205.4) Δ S ∘ r -67 = - (302.5) - 2(170.3) + Δ S ∘ BaI2 + 2(205.4) Δ S ∘ BaI2 = +165.3 J/( mol·K ) Δ S ∘ BaI2

47. Thank you Suresh Krishnasamy | Associate Lecturer School of Agriculture and Food Sciences [email protected] 07 5460 1186