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Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
Hess Law And Thermodynamics
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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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