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Thermochemistry: Chemical Energy

Thermochemistry: Chemical Energy. Energy and Its Conservation. Energy : The capacity to supply heat or do work. Kinetic Energy ( E K ) : The energy of motion. Potential Energy ( E P ) : Stored energy.

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Thermochemistry: Chemical Energy

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  1. Thermochemistry: Chemical Energy

  2. Energy and Its Conservation Energy: The capacity to supply heat or do work. Kinetic Energy (EK): The energy of motion. Potential Energy (EP): Stored energy. Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form into another. First Law of Thermodynamics ∆E = q + w

  3. Energy and Its Conservation Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object. Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.

  4. Calorimetry The energy released or absorbed during a chemical reaction.

  5. Some definitions Exothermic: energy is given off (temperature increases) Endothermic: energy is required (temperature decreases) Enthalpy, DH: Internal energy in kJ/mol - DH: exothermic + DH: endothermic

  6. Types of Calorimeters Coffee cup Bomb

  7. Coffee Cup Calorimetry Measure the heat flow at constant pressure (∆H).

  8. Bomb Calorimetry Measure the heat flow at constant volume (∆E).

  9. Coffee-cup calorimetry q = -mcDT q: quantity of heat in joules, J m: mass of liquid, g c: specific heat capacity, 4.18J/goC DT: change in temperature in oC, Tfinal – Tinitial q = nΔH or ΔH = q/n

  10. There are two kinds of coffee-cup calorimetry problems. 1st kind: Grams of a chemical are added to a given mass or volume of water. What to do: Grams of water will be plugged into m. DT, Last temperature minus first temperature. Solve for q and change joules to kJ. Grams of the other chemical convert to moles. Solve for DH by putting kJ over moles and divide.

  11. Example #1 In a coffee-cup calorimeter, 5g of sodium are dropped in 300g of water at 20oC the temperature rises to 35oC. Calculate the enthalpy change for this reaction.

  12. Example #2 A 1.5g sample of calcium is dropped into 275mL of water at 20oC in a coffee-cup calorimeter. The temperature rises to 23oC. The density of water is 1g/mL. What is the energy change for this reaction in kJ/mol?

  13. There are two kinds of coffee-cup calorimetry problems. 2nd kind: Two volumes of a solution are given. What to do: Add them and plug into m (the density of most solutions = 1g/mL, so the mL = the grams) DT, Final temperature minus Initial temperature. Solve for q and change joules to kJ. Pick one of the chemicals and multiply (L)(M) to find moles. Solve for DH by putting kJ over moles and divide.

  14. Example #3 450mL of 2M HCl are mixed with 450mL of 2M NaOH in a coffee-cup calorimeter. The temperature rises from 23oC to 25oC. Calculate the enthalpy change for this reaction.

  15. Example #4 In a coffee-cup calorimeter, 275mL of 0.5M Ca(OH)2 react with 275mL of 0.5M H2SO4. The temperature increases from 25oC to 30oC. Calculate the energy in kJ/mol.

  16. Worked Example Calculating the Amount of Heat Released in a Reaction How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a stoichiometric amount of Fe2O3? Strategy According to the balanced equation, 852 kJ of heat is evolved from the reaction of 2 mol of Al. To find out how much heat is evolved from the reaction of 5.00 g of Al, we have to find out how many moles of aluminum are in 5.00 g. Solution The molar mass of Al is 26.98 g/mol, so 5.00 g of Al equals 0.185 mol: Because 2 mol of Al releases 852 kJ of heat, 0.185 mol of Al releases 78.8 kJ of heat: ✔ Ballpark Check Since the molar mass of Al is about 27 g, 5 g of aluminum is roughly 0.2 mol, and the heat evolved is about (852 kJ/2 mol)(0.2 mol), or approximately 85 kJ.

  17. Bomb Calorimetry q = -CDT q: quantity of heat, kJ C: heat capacity, kJ/oC DT: change in temperature (increases positive, decreases negative)

  18. Two kinds of Bomb also: 1st kind: Enthalpy (kJ/mol) is given What to do: Convert grams to moles Multiply the moles by the enthalpy. This answer plugs in for q Find C or DT (whichever it asks for)

  19. Example #5 A 5g sample of hydrogen is burned in a bomb calorimeter. The enthalpy of combustion for hydrogen is -242kJ/mol. What is the heat capacity of the calorimeter if the temperature rises 5oC?

  20. Example #6 30g of magnesium are burned in a bomb calorimeter that has a heat capacity of 18kJ/oC. If the enthalpy of combustion for magnesium is -602kJ/mol, how much will the temperature rise?

  21. Still Bomb 2nd kind: Enthalpy is asked for (or energy change in kJ/mol) What to do: Convert grams to moles. Plug in C and DT and solve for q. Solve for DH by putting kJ over moles and divide.

  22. Example #7 If 10g of Ca are burned in a bomb calorimeter that has a heat capacity of 12.8kJ/oC, the temperature increases by 12.38oC. What is the enthalpy of combustion of calcium?

  23. Example #8 An 18g sample of sulfur is burned in a bomb calorimeter. The heat capacity of the calorimeter is 8.9kJ/oC and the temperature increases 18.74oC. What is the enthalpy of combustion for sulfur?

  24. What is “Thermodynamics”? The study of energy and its interconversions.

  25. Some terms: Enthalpy: Internal energy (energy stored in the bonds) DH kJ/mol (+ endothermic, - exothermic) Entropy: disorder of a substance DS J/mol.K gases  aqueous  liquids  solids most disordered to least disordered Gibbs Free Energy: stored energy less the degree of disorder DG kJ/mol

  26. How to calculate DH, DS & DG DHo = SDHfo for products – SDHfo for reactants DSo = S So of products – SSo for reactants DGo = SDGfo of products – SDGfo for reactants O superscripted means standard states (1atm, 273K, etc)

  27. Calculating Enthalpy Changes Using Standard Heats of Formation C(s) + 2 H2(g) CH4(g) Standard Heat of Formation (∆H°f): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. Standard states ∆H°f = -74.8 kJ 1 mol of 1 substance

  28. Calculating Enthalpy Changes Using Standard Heats of Formation a A + b B c C + d D ∆H° = ∆H°f (Products) - ∆H°f (Reactants) ∆H°= [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)] Products Reactants

  29. Calculating Enthalpy Changes Using Standard Heats of Formation 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g) Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. ∆H° = ? H°= [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))] ∆H°= [(1 mol)(-1273.3 kJ/mol)] - [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] = 2802.5 kJ

  30. How to calculate DH, DS & DG DSoand DGocan be calculated the same way as DHo DSo = S So of products – SSo for reactants DGo = SDGfo of products – SDGfo for reactants O superscripted means standard states (1atm, 273K, etc)

  31. Example #1 Calculate DHo, DSo and DGo for the reaction below. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

  32. Predicting Entropy To decide which substance has more entropy, first look at its state (solid, liquid, etc.). The more disordered state will have more disorder. If they have the same state, the larger molecule will have more entropy. Both gaseous, higher temperature, less pressure, more disorder.

  33. Example #2 Which in each pair has more entropy? A. H2O(g) H2O(l) B. C6H12O6(s) C12H22O11(s) C. HCl(aq) HCl(g)

  34. Predicting DS change in a reaction. Look at both sides of the reaction, identify the state of the chemicals, how many moles of chemicals are present (more particles, more disorder)and how big the particles are. These are in order of importance. If the right has more disorder, DS goes up. If the right has less disorder, DS goes down.

  35. Example #3 Predict the sign for DS for each reaction. A. 12CO2(g) + 11H2O(l) C12H22O11(s) + 12O2(g) B. 2H2O(g) 2H2(g) + O2(g)

  36. Worked Example Predicting the Sign of ΔS for a Reaction Predict whether ΔS° is likely to be positive or negative for each of the following reactions: a. b. Strategy Look at each reaction, and try to decide whether molecular randomness increases or decreases. Reactions that increase the number of gaseous molecules generally have a positive ΔS, while reactions that decrease the number of gaseous molecules have a negative ΔS. Solution a. The amount of molecular randomness in the system decreases when 2 mol of gaseous reactants combine to give 1 mol of liquid product, so the reaction has a negative ΔS° . b. The amount of molecular randomness in the system increases when 9 mol of gaseous reactants give 10 mol of gaseous products, so the reaction has a positive ΔS° .

  37. Spontaneity A spontaneous reaction occurs without the need of outside energy. Do not confuse spontaneous with instantaneous. Spontaneous happens on its own, instantaneous happens right away.

  38. How can you tell? DS + tends to be spontaneous DH – tends to be spontaneous The only way to know for sure is if DG is -, then it is guaranteed to be spontaneous. DGo = DHo - T DSo DG = DH - T DS

  39. Example #4 Predicting the signs of ΔH, ΔS and ΔG A. When solid ammonium chloride is dissolved in water, it forms an aqueous solution and the temperature drops. B. When hydrogen gas is mixed with oxygen gas and a spark ignited, a fire ball to the ceiling occurs. 2H2(g) + O2(g) 2H2O(g)

  40. Worked Conceptual ExamplePredicting the Signs of ΔH, ΔS, and ΔG for a Reaction What are the signs of ΔH, ΔS, and ΔG for the following nonspontaneous transformation? Strategy First, decide what kind of process is represented in the drawing. Then decide whether the process increases or decreases the entropy (molecular randomness) of the system and whether it is exothermic or endothermic. Solution The drawing shows ordered particles in a solid subliming to give a gas. Formation of a gas from a solid increases molecular randomness, so ΔS is positive. Furthermore, because we’re told that the process is nonspontaneous, ΔG is also positive. Because the process is favored by ΔS (positive) yet still nonspontaneous, ΔH must be unfavorable (positive). This makes sense, because conversion of a solid to a liquid or gas requires energy and is always endothermic.

  41. Calculating DG using DH & DS Be sure to change DS into kJ from J. Change temperature to Kelvin by adding 273 to celsius. Plug in and solve for the unknown.

  42. Example #5 For the reaction: H2O(g) H2O(l) DHo = -44kJ/mol DSo = 119J/mol.K Calculate DGo at 10oC, 0oC and -10oC.

  43. More info about DG If DG is negative the reaction is spontaneous (runs forwards) If DG is positive it is not spontaneous forwards (can be spontaneous backwards) If DG = 0 it is at equilibrium (no net forward or backward movement)

  44. Temperature and spontaneity You can change the temperature of reactions and they will become spontaneous at their new temperature. Phase changes: The boiling point and freezing (melting) points are temperatures where DG =0.

  45. Free Energy Reversible

  46. Example #6 For the reaction: H2O(g) H2O(l) DHo = -44kJ/mol DSo = 119J/mol.K At what temperature does this reaction become spontaneous?

  47. Example #7 For the phase change: Br2(l) Br2(g) DHo = 31kJ/mol and DSo = 93J/mol.K What is the normal boiling point of bromine?

  48. Worked Example Using the Free-Energy Equation to Calculate Equilibrium Temperature Lime (CaO) is produced by heating limestone (CaCO3) to drive off CO2 gas, a reaction used to make Portland cement. Is the reaction spontaneous under standard conditions at 25 ° °°C? Calculate the temperature at which the reaction becomes spontaneous. Strategy The spontaneity of the reaction at a given temperature can be found by determining whether ΔGis positive or negative at that temperature. The changeover point between spontaneous and nonspontaneous can be found by setting ΔG= 0 and solving for T. Solution At 25 °C(298 K), we have Because ΔGis positive at this temperature, the reaction is nonspontaneous. The changeover point between spontaneous and nonspontaneous is approximately The reaction becomes spontaneous above approximately 1120 K (847 ° °°C).

  49. Calculating Enthalpy Changes Using Hess’s Law N2H4(g) + H2(g) 2 H2(g) + N2(g) 3 H2(g) + N2(g) 3 H2(g) + N2(g) N2H4(g) 2 NH3(g) 2 NH3(g) 2 NH3(g) Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Haber Process: ∆H°= -92.2 kJ Multiple-Step Process ∆H°1 = ? ∆H°2 = -187.6 kJ ∆H°reaction = -92.2 kJ

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