• Like

Loading…

Flash Player 9 (or above) is needed to view presentations.
We have detected that you do not have it on your computer. To install it, go here.

Hess Law And Thermodynamics

  • 13,352 views
Uploaded on

ABC Lab 2 intro powerpoint

ABC Lab 2 intro powerpoint

More in: Education
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
  • veery nice ,
    Are you sure you want to
    Your message goes here
No Downloads

Views

Total Views
13,352
On Slideshare
0
From Embeds
0
Number of Embeds
16

Actions

Shares
Downloads
208
Comments
1
Likes
1

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 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 E K of an object is dependent on its mass(m) and velocity (v)
    • E K = 1/2mv 2
    • 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 = mC Δ T)
    • w = work. Force acting over a distance (F x d)
  • 10. E
  • 11. Energy Exchange
    • A system can exchange energy with its surroundings through work or heat
    • Internal energy = heat energy + work done
    • Δ E = 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 = E final – E initial  E= E f – E i
    • 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
    • w system =  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 Calorimetry (Constant-Volume Calorimetry) Calorimetry 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 Calorimetry Calorimetry 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 Calorimetry
    • Atmospheric pressure is constant!
    •  H = q P
    • q system = - q surroundings
    • The surroundings are composed of the water in the calorimeter and the calorimeter.
    • For most calculations, the q calorimeter can be ignored.
    • q system = - q water
    • c system m system  T system = - c water m water  T water
    Calorimetry
  • 26. NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  T water = 16.9 o C – 22.0 o C = -5.1 o C m water = 60.0g c water = 4.184J/g o C m sample = 4.25g Now calculate Δ H in kJ/mol q sample = -q water q sample = -c water m water  T water q sample = -(4.184J/g o C)(60.0g)(-5.1 o C) q sample = 1280.3J Calorimetry
  • 27. Problem: Heats of Chemical Reaction
    • 100 ml solutions of 1.00 M NaCl and 1.00 M AgNO 3 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 AgNO 3 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(NO 3 ) 2 solution at 25.1 ° C are mixed with 50.0 mL of 0.25 M Na 2 SO 4 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 = H products - H reactants
    • If  H >0, then q p >0. The reaction is endothermic
    • If  H <0, then q p <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’ 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:  H 1 =  H 2 +  H 3
  • 35. Calculations via Hess’s Law
    • 1. If a reaction is reversed ,  H is also reversed.
    • N 2 ( g ) + O 2 ( g )  2NO( g )  H = 180 kJ
    • 2NO( g )  N 2 ( g ) + O 2 ( 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 )  N 2 ( g ) + O 2 ( g )]  H = 3 [  180 kJ]
    • 6 NO( g )  3 N 2 ( g ) + 3 O 2 ( g )  H =  540 kJ
  • 36.
    • A consequence of Hess’ Law is that reactions can be added just like algebraic equations.
    • For example:
    Hess’ Law CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( g )  H = -802 kJ 2H 2 O( g )  2H 2 O( l)  H = -88 kJ CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( l )  H = -890 kJ
  • 37. Hess’s Law Hess‘s Law is particularly useful for calculating  f H o which would not be easy to measure experimentally.  H o f for CO cannot be measured as CO 2 is also formed when graphite is burned C(s) + 1/2O 2  CO  H o 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  H o f CO CO + 1/2O 2  CO 2  rxn H o = -283 kJmol -1 C(s) + O 2  CO 2  f H o = -393.5 kJmol -1
  • 38. Rearrange the equations to calculate the for  Hof CO CO + 1/2O 2  CO 2  H o = -283 kJmol -1 C(s) + O 2  CO 2  H o = -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.
  • 40. Hess’ Law: 1 st Example
  • 41. 1 st Example (cont.)
    • If we take the previous two reactions and add them, we get the original reaction of interest:
    • N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ
    • 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ
    • N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ
  • 42. 1 st Example (cont.)
    • Our reaction of interest is:
    • N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ
    • This reaction can also be carried out in two steps: N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ
  • 43. 1 st 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.
    N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ 2NO 2 (g) N 2 (g) + 2O 2 (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.
    N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ  N 2 (g) + 4O 2 (g) 4NO 2 (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.
    • 2 nd Example:
    • What is  H for the following reaction?
    • 3C (gr) + 4H 2 (g)  C 3 H 8 (g)
    • C(gr) indicates that it is carbon in the form of graphic rather than diamond.
  • 47. 2 nd Example (cont.)
    • 3C (gr) + 4H 2 (g) C 3 H 8 (g)  H = ?
    • You’re given the following reactions: C (gr) + O 2 (g) CO 2 (g)  H = -394 kJ C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l)  H = -2220 kJ H 2 (g) + 1/2O 2 (g) H 2 O (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) + O 2 (g) CO 2 (g) ] 3 [  H = -394 kJ ] 3C (gr) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ Given: After manipulating:
  • 49. Example (cont.)
    • Step 2. To get C 3 H 8 on the product side of the reaction, we need to reverse reaction 2.
    3CO 2 (g) +4H 2 O (l)  C 3 H 8 (g)+5O 2 (g) -[  H = - 2220 kJ ] C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (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) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ 3CO 2 (g) + 4H 2 O (l) C 3 H 8 (g) + 5O 2 (g)  H = +2220 kJ 2 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ
  • 51. Example (cont.)
    • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
    • 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ
    • H 2 (g) + 1/2O 2 (g) H 2 O (l)  H = -286 kJ
    3C (gr) + 4H 2 (g) C 3 H 8 (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) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ
    • 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l)  H = -1144 kJ
    3C (gr) + 4H 2 (g) C 3 H 8 (g)
  • 53. Example (cont.)
    • Step 4 (cont.):
    • 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ
    • 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l)  H = -1144 kJ
    3C (gr) + 4H 2 (g) C 3 H 8 (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) + H 2(g)  C 2 H 2(g) given the following:
    • Correct moles and divide so function of coefficients moles for wanted product is one
    • C 2 H 2(g) + 5/2O 2(g)  2CO 2(g) + H 2 O (l)   H = -1299.6kJ
    • C (s) + O 2(g)  CO 2(g)  H = -393.5kJ
    • H 2(g) + 1/2O 2(g)  H 2 O (l)  H = -285.9kJ
    • Rearrange the equations so similar to algebra:
  • 57. Practice Calculate  H for :
    • NO (g) + O (g)  NO 2(g) given:
    • NO (g) + O 3(g)  NO 2(g) + O 2(g)  H = -198.9kJ
    • O 3(g)  3/2O 2(g)  H = -142.3kJ
    • O 2(g)  2O (g)  H = 495.0kJ
  • 58. Another Example
    • Calculate  H for the following reaction:
    • H 2 (g) + Cl 2 (g) 2HCl(g)
    • Given the following:
    • NH 3 (g) + HCl (g) NH 4 Cl(s)  H = -176 kJ
    • N 2 (g) + 3H 2 (g) 2NH 3 (g)  H = -92 kJ
    • N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 4 Cl(s)  H = -629 kJ
  • 59. Changes in Enthalpy
    • Consider the following expression for a chemical process:
    •  H = H products - H reactants
    • If  H >0, then q p >0. The reaction is endothermic
    • If  H <0, then q p <0. The reaction is exothermic
  • 60. Standard Heat of Formation ( Δ H f ° )
    • 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 Δ H ° = Σ Δ H f ° products - Σ Δ H f ° 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 [N 2 ( 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 = H products - H reactants
    • If  H >0, then q p >0. The reaction is endothermic
    • If  H <0, then q p <0. The reaction is exothermic
  • 63. Using the equation Δ H ° = Σ Δ H f ° products - Σ Δ H f ° reactants
    • We will calculate the theoretical Δ H
    • for the reaction
    • of hydrochloric acid and sodium hydroxide.
  • 64. Experimental Δ H
    • 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 21 o C to 27.5 o C. What is the enthalpy change, if the density and specific heat of the solution are assumed to be the same as water?