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Hess Law And Thermodynamics

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    • 1. Thermodynamics and Hess’s Law
    • 2. Energy is... • The ability to do work. • Conserved. • Work is done when energy is transferred – At macroscopic level- work is done as force acting over a distance to move an object – At microscopic level- work is done on the particles as HEAT • a state function. – independent of the path, or how you get from point A to B.
    • 3. Energy • Kinetic Energy- MOTION » Thermal - Heat » Mechanical » Electrical » Sound • Potential Energy - POSITION » Chemical » Gravitational » Electrostatic
    • 4. K.E. kinetic energy- motion • Magnitude of K.E. or EKof an object is dependent on its mass(m) and velocity (v) • EK = 1/2mv2 • Mass and velocity determine how much work an object can do
    • 5. Potential Energy • Potential energy: stored energy as a result of the energy an object possesses by its position relative to other objects and some force acting between • Charged particles have P.E. due to attractive and repulsive forces (electrostatic forces) between them
    • 6. The Universe • is divided into two halves. • the system and the surroundings. • The system is the part you are concerned with. • The surroundings are the rest. q into system = -q from surroundings
    • 7. State Function • A system has a fixed value of energy for a given set of conditions • Value independent of pathway • Value depends only on present condition not how it got there • Example – PE of bowling ball at top of stadium
    • 8. Energy is transferred as ... • Work is a force acting over a distance. • Heat is energy transferred between objects because of temperature difference.
    • 9. First Law of Thermodynamics Energy of the Universe is Constant E = q + w q = heat. Transferred between two bodies due to difference in temperature (q = mCΔT) w = work. Force acting over a distance (F x d)
    • 10. 10 E
    • 11. Energy Exchange • A system can exchange energy with its surroundings through work or heat • Internal energy = heat energy + work done ΔE = q + w • q is pos. gains heat from surroundings q>0 • w pos. work done on system by surroundings w>0 • q- neg. value heat lost to surroundings q<0 • w neg. work done on surroundings w<0
    • 12. Change internal energy = difference between internal energy of system at the completion of a process and that of the beginning ∆E = Efinal – Einitial ∆E= Ef – Ei • However, it is impossible to know the internal energy at the beginning and the end • So, we may only determine the change in internal energy by measuring energy transferred as heat and work
    • 13. Exothermic versus endothermic.
    • 14. ‘Heat’ & ‘Work’ • Heat stimulates disorderly motion of surroundings • Work of expansion stimulates organized motion of surroundings
    • 15. Energy transfer as work/heat
    • 16. Work • Gas pushes back the surrounding atmosphere • Does work on its surroundings • Exerts a force over a distance to push up the piston h
    • 17. Expansion work • Work = force x distance • Pressure = Force /Area • Δ Volume = Area x height F
    • 18. work = force × distance (h) since pressure = force / area, and volume = area x distance (h) work = pressure × volume Force (area x height) area wsystem = −P∆V Work
    • 19. Indicator Diagram
    • 20. The piston, moving a distance ∆h against a pressure P, does work on the surroundings. Since the volume of a cylinder is the area of the base times its height, the change in volume of the gas is given by ∆h x A = V.
    • 21. Bomb CalorimetryBomb Calorimetry (Constant-Volume Calorimetry)(Constant-Volume Calorimetry) CalorimetryCalorimetry No Δ in volume No work done Energy transfer only as heat E (internal energy) can be measured in bomb calorimeter, since no energy is transferred as work. Measuring the heat transfer alone is enough.
    • 22. Constant Volume Calorimetry • Constant Volume • NO MOVEMENT – NO DISTANCE • NO WORK • So the ONLY energy transfer is as heat. • Thus measuring the heat transfer is a measure of change in internal energy
    • 23. Constant-Pressure CalorimetryConstant-Pressure Calorimetry CalorimetryCalorimetry Enthalpy is measured at constant pressure. The work done by the system will be ignored.
    • 24. Constant Pressure Calorimetry • The change in volume is ignored • Thus the work done is ignored • Only heat transfer will be measured • The change in enthalpy is defined as the heat transferred at constant pressure- thus ignores work done
    • 25. Constant-Pressure CalorimetryConstant-Pressure Calorimetry Atmospheric pressure is constant! ∆H = qP qsystem = -qsurroundings - The surroundings are composed of the water in the calorimeter and the calorimeter. - For most calculations, the qcalorimeter can be ignored. qsystem = - qwater csystemmsystem ∆Tsystem = - cwatermwater ∆Twater CalorimetryCalorimetry
    • 26. NH4NO3(s)  NH4 + (aq) + NO3 - (aq) ∆Twater = 16.9o C – 22.0o C = -5.1o C mwater = 60.0g cwater = 4.184J/go C msample = 4.25g Now calculate ΔH in kJ/mol qsample = -qwater qsample = -cwatermwater ∆Twater qsample = -(4.184J/go C)(60.0g)(-5.1o C) qsample = 1280.3J CalorimetryCalorimetry
    • 27. Problem: Heats of Chemical Reaction 100 ml solutions of 1.00 M NaCl and 1.00 M AgNO3 at 22.4 o C are mixed in coffee cup calorimeter and the resulting temperature rises to 30.2 o C. What is the heat per mole of product? Assume the solution density and specific heat are the same as pure water.
    • 28. Problem: Heats of Chemical Reaction • 100 ml solutions of 1.00 M NaCl and 1.00 M AgNO3 at 22.4 o C are mixed in coffee cup calorimeter and the resulting temperature rises to 30.2 o C. What is the heat per mole of product? Assume the solution density and specific heat are the same as pure water. • Write balanced chemical reaction: • Net ionic: Ag+(aq) + Cl-(aq) AgCl(s)→ • Determine heat of reaction: • qrxn= -qcal = -m×c× T∆ • m = 200 ml × 1.0g/ml = 200g • c = cH2O = 4.18 J/g-oC = -200g × 4.18 J/g-o C × (30.2-22.4) • = -6,520 J • Determine heat per mole of product: • stoichiometric reactants, 0.1 mol in 100 ml • qrxn/mol = -6.52 kJ/0.1 mol • = -65.2 kJ/mol
    • 29. Practice calorimetry calculation When 50.0 mL of 0.25 M Ba(NO3)2 solution at 25.1° C are mixed with 50.0 mL of 0.25 M Na2SO4 at 25.1° C in a calorimeter, the white solid barium sulfate forms and the temperature increases to 26.6° C. Assuming the calorimeter absorbs only minimal heat and that the specific heat and density of the solution are equivalent to water, calculate the enthalpy per mole of barium sulfate. • Write the balanced and net ionic equation. • Show the work to solve for the q rxn • Show the work to find the molar enthalpy
    • 30. Definition of Enthalpy • Thermodynamic Definition of Enthalpy (H): H = E + PV E = energy of the system P = pressure of the system V = volume of the system
    • 31. Changes in Enthalpy • Consider the following expression for a chemical process: ∆H = Hproducts - Hreactants If ∆H >0, then qp >0. The reaction is endothermic If ∆H <0, then qp <0. The reaction is exothermic
    • 32. Thermodynamic State Functions • Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: ∆E and ∆H) • Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.
    • 33. Hess’ Law Defined • Enthalpy is a state function. As such, ∆H for going from some initial state to some final state is pathway independent. • Hess’ Law: ∆H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate ∆H for a reaction.
    • 34. Hess’ LawHess’ Law : if a reaction is carried out in a number of steps, ∆H for the overall reaction is the sum of ∆H for each individual step. Note that: ∆H1 = ∆H2 + ∆H3
    • 35. Calculations via Hess’s Law 1. If a reaction is reversed, ∆H is also reversed. N2(g) + O2(g) → 2NO(g) ∆H = 180 kJ 2NO(g) → N2(g) + O2(g) ∆H = −180 kJ 2. If the coefficients of a reaction are multiplied by an integer, ∆H is multiplied by that same integer. 3[2NO(g) → N2(g) + O2(g)] ∆H = 3[−180 kJ] 6NO(g) → 3N2(g) + 3O2(g) ∆H = −540 kJ
    • 36. • A consequence of Hess’ Law is that reactions can be added just like algebraic equations. • For example: Hess’ LawHess’ Law CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = -802 kJ 2H2O(g) → 2H2O(l) ∆H = -88 kJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = -890 kJ
    • 37. Hess’s Law Hess‘s Law is particularly useful for calculating ∆fHo which would not be easy to measure experimentally. ∆Ho f for CO cannot be measured as CO2 is also formed when graphite is burned C(s) + 1/2O2  CO ∆Ho f = x Standard heats of formation are the enthalpy change associated with forming one mole of the compound from its elements in their standard states. Rearrange the equations to calculate the for ∆Ho f CO CO + 1/2O2  CO2 ∆rxnHo = -283 kJmol-1 C(s) + O2  CO2 ∆fHo = -393.5 kJmol-1
    • 38. Rearrange the equations to calculate the for ∆Hof CO CO + 1/2O2  CO2 ∆Ho = -283 kJmol-1 C(s) + O2  CO2 ∆Ho = -393.5 kJmol-1
    • 39. Using Hess’ Law • When calculating ∆H for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine ∆H for our “single step” reaction. 2NO2 (g) N2 (g) + 2O2 (g) q 2NO2 (g)N2 (g) + 2O2 (g)
    • 40. Hess’ Law: 1st Example
    • 41. 1st Example (cont.) • If we take the previous two reactions and add them, we get the original reaction of interest: N2 (g) + O2 (g) 2NO(g) ∆H = 180 kJ 2NO (g) + O2 (g) 2NO2(g) ∆H = -112 kJ N2 (g) + 2O2 (g) 2NO2(g) ∆H = 68 kJ
    • 42. 1st Example (cont.) • Our reaction of interest is: N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ • This reaction can also be carried out in two steps: N2 (g) + O2 (g) 2NO(g) ∆H = 180 kJ 2NO (g) + O2 (g) 2NO2(g) ∆H = -112 kJ
    • 43. 1st Example (cont.) • Note the important things about this example, the sum of ∆H for the two reaction steps is equal to the ∆H for the reaction of interest. • We can combine reactions of known ∆H to determine the ∆H for the “combined” reaction.
    • 44. Hess’ Law: Details • One can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of ∆H changes. N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ 2NO2(g) N2(g) + 2O2(g) ∆H = -68 kJ
    • 45. Details (cont.) • The magnitude of ∆H is directly proportional to the quantities involved (it is an “extensive” quantity). • As such, if the coefficients of a reaction are multiplied by a constant, the value of ∆H is also multiplied by the same integer. N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ 2N2(g) + 4O2(g) 4NO2(g) ∆H = 136 kJ
    • 46. Using Hess’ Law • When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest. • 2nd Example: What is ∆H for the following reaction? 3C (gr) + 4H2 (g)  C3H8 (g) C(gr) indicates that it is carbon in the form of graphic rather than diamond.
    • 47. 2nd Example (cont.) 3C (gr) + 4H2 (g) C3H8 (g) ∆H = ? • You’re given the following reactions: C (gr) + O2 (g) CO2 (g) ∆H = -394 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) ∆H = -2220 kJ H2 (g) + 1/2O2 (g) H2O (l) ∆H = -286 kJ
    • 48. Example (cont.) • Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. 3[C (gr) + O2 (g) CO2 (g)] 3[ ∆H = -394 kJ] 3C (gr) + 3O2 (g) 3CO2 (g) ∆H = -1182 kJ Given: After manipulating:
    • 49. Example (cont.) • Step 2. To get C3H8 on the product side of the reaction, we need to reverse reaction 2. 3CO2 (g) +4H2O (l) C3H8 (g)+5O2 (g) -[∆H = - 2220 kJ] C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) ∆H = -2220 kJ Given: After manipulating:reverse
    • 50. Example (cont.) • Step 3: Add two “new” reactions together to see what is left: 3C (gr) + 3O2 (g) 3CO2 (g) ∆H = -1182 kJ 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) ∆H = +2220 kJ 2 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ
    • 51. Example (cont.) • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ H2 (g) + 1/2O2 (g) H2O (l) ∆H = -286 kJ 3C (gr) + 4H2 (g) C3H8 (g) Need to multiply second reaction by 4
    • 52. Example (cont.) • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) ∆H = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g)
    • 53. Example (cont.) • Step 4 (cont.): 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) ∆H = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g) ∆H = -106 kJ
    • 54. Calculate the change in enthalpy.
    • 55. Calculate the change in enthalpy for the reaction.
    • 56. Practice Calculate ∆H for : • 2C(s) + H2(g)  C2H2(g) given the following: • Correct moles and divide so function of coefficients moles for wanted product is one • C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l) ∆H = -1299.6kJ • C(s) + O2(g)  CO2(g) ∆H = -393.5kJ • H2(g) + 1/2O2(g)  H2O(l) ∆H = -285.9kJ • Rearrange the equations so similar to algebra:
    • 57. Practice Calculate ∆H for : • NO(g) + O(g)  NO2(g) given: • NO(g) + O3(g)  NO2(g) + O2(g) ∆H = -198.9kJ • O3(g)  3/2O2(g) ∆H = -142.3kJ • O2(g)  2O(g) ∆H = 495.0kJ
    • 58. Another Example • Calculate ∆H for the following reaction: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) ∆H = -176 kJ N2 (g) + 3H2 (g) 2NH3 (g) ∆H = -92 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) ∆H = -629 kJ
    • 59. Changes in Enthalpy • Consider the following expression for a chemical process: ∆H = Hproducts - Hreactants If ∆H >0, then qp >0. The reaction is endothermic If ∆H <0, then qp <0. The reaction is exothermic
    • 60. Standard Heat of Formation (ΔHf ° ) • Standard heat of formation is the enthalpy change associated with forming one mole of compound from its elements in their standard states • Standard heats of formations are given in a table of thermodynamic values • Finding the sum of the heat formations for products minus reactants is an application of Hess’s Law ΔH° = Σ ΔHf ° products- Σ ΔHf ° reactants
    • 61. Standard States Compound For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.
    • 62. Changes in Enthalpy • Consider the following expression for a chemical process: ∆H = Hproducts - Hreactants If ∆H >0, then qp >0. The reaction is endothermic If ∆H <0, then qp <0. The reaction is exothermic
    • 63. Using the equation ΔH° = Σ ΔHf ° products - Σ ΔHf ° reactants We will calculate the theoretical ΔH for the reaction of hydrochloric acid and sodium hydroxide.
    • 64. Experimental ΔH Heat of reaction at constant pressure! Use a “coffee-cup” calorimeter to measure it Practice: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee-cup calorimeter, the temperature increases from 21o C to 27.5o C. What is the enthalpy change, if the density and specific heat of the solution are assumed to be the same as water?

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