Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 6. = the capacity to do work or to produce heat Kinetic energy = the energy due to motion depends on mass & velocity Potential Energy = energy.

Similar presentations


Presentation on theme: "Chapter 6. = the capacity to do work or to produce heat Kinetic energy = the energy due to motion depends on mass & velocity Potential Energy = energy."— Presentation transcript:

1 Chapter 6

2 = the capacity to do work or to produce heat Kinetic energy = the energy due to motion depends on mass & velocity Potential Energy = energy due to the position or composition.

3  Temperature = reflects the random motions of particles in a substance. The more motion the higher the temperature.  Heat = Involves the transfer of energy between two objects due to a temperature difference.

4  The portions of universe that is identified  Two Types: Open system: the designated part is open to the atmosphere. Closed system: the designated part is closed to the atmosphere.

5 2 ways 1. work = force x distance 2. heat

6  A property that is related only to the current conditions—There is no consideration as to how it got to the current situation.  Examples: pressure, volume, temperature, energy and enthalpy

7  System-reactants and products  Surrounding-everything else

8

9

10

11  Also known as the Law of Conservation of Energy  Energy cannot be created nor destroyed but may be conserved.  Concept describes the universe not a system.

12  Thus, energy can be lost or gained by a system.  Energy in the universe is constant.  Thermodynamics = the study of energy and its conversions

13 = sum of the kinetic and potential energies of all “particles” in the system. ΔE = q + w q = heat W = work

14  Consists of two parts: a. Number = gives the magnitude of the change b. sign 1. (+) = endothermic 2. (-) = exothermic

15  The energy is exchanged with the environment in terms of heat or work.  ΔE = q + w  q = (+) means that heat is added to the system  q = (-) means that heat is subtracted from the system

16  Negative work = energy flows out of the system so the system does work on the surroundings --exothermic  Positive work = energy flows into the system so the surrounding do work on the system --endothermic  When the systems are under relatively standard conditions the effects of work is ignored.

17

18 ΔH = ΔE + PΔV  ΔE = change in internal energy  P = pressure of the system  ΔV = change in volume of the system  ΔH = is equal to the energy flow as

19 ΔH = ΔH products – ΔH reactants  -ΔH = exothermic  +ΔH = endothermic

20

21  Loss or gain of heat by a system is enthalpy. (ΔH)  State Function  ΔH = H f – H i = q p  q p is heat associated with constant pressure

22  Positive value of ΔH means that the system has gained heat from the surrounding. (endo)  Negative value of ΔH means that the system has lost heat to the surroundings. (exo)

23  Heat Capacity is the amount of heat required to raise the temperature of a substance 1°C.  Molar Heat Capacity is the heat capacity of one mole of the substance.  Specific Heat Capacity is the heat capacity of gram values of a substance.  The specific heat of a substance is the amount of heat required to raise 1 gram of the substance 1°C.

24  q = m x c x ΔT  q = heat  M = mass in grams  c = specific heat in J/g°C  ΔT is the difference between final and initial temperature in°C

25  A 2.50 kg piece of copper metal is heated from 25°C to 225°C. How much heat kJ, is absorbed by the copper. The specific heat is 0.384 J/g°C for copper.  q = 192 kJ

26  The enthalpy of a reaction is equal to the sum of the enthalpies for each step.  Allows us to calculate the enthalpy of the reaction by using information about each reactant.

27  ΔH°  Enthalpy for a reaction when all reactants and products are in their standard state.  Standard state is 25°C and 1 atm  ΔΔΔ

28  ΔH f  Represents the enthalpy change that occurs when a compound is formed from its constituent elements.

29  ΔH° f  References to one mole of a compound formed from its constituent elements in their standard state.

30

31

32

33

34

35

36

37 Relationships of Energies of Reactants, Products and Reactions Chapter 16

38  Occurs without outside intervention  Can occur fast or slow  Ex. carbon to diamond

39  = disorder  The driving force for a spontaneous process is an increase in entropy  Has to do with the probability everything is in order

40  Higher the positional probability the larger the entropy, +S  Increases going from a solid to a liquid, to a gas  Increases the larger the volume you have

41  In any spontaneous process there is always an increase in the entropy of the universe  It occurs in one direction. ∆S univ = ∆S sys + ∆S Surr +∆S univ = process is spontaneous in direction written -∆S univ = spontaneous reverse direction

42 1) The sign ∆S surr depends on the direction of the heat flow - ∆S surr = endothermic + ∆S surr = exothermic

43 2) The magnitude of ∆S depends on temperature. -The impact of the transfer of energy as heat to and from the surroundings has greater impact at lower temperatures.

44  ∆H = heat flow = change in enthalpy -∆H sys = endothermic +∆H sys = exothermic

45 ∆S surr = - ∆H / T *the minus sign changes the point of view from the system to the surroundings - For constant pressure and temperature

46 ∆G = ∆H -T∆S H = enthalpy T = temperature in Kelvin (constant) S = entropy

47  A reaction is spontaneous if ∆G is negative and carried out under constant pressure and temperature.

48  The change in positional entropy is dominated by the relative #’s of molecules of gaseous products and reactants

49 1) N 2 + 3H 2  2NH 3 2) H 2  2H 3) 4NH 3 + 5O 2  4NO + 6H 2 O

50  The entropy of a perfect solids at 0K is zero.  An increase in motion is associated with higher entropy value.

51  When a solid melts  When a solid dissolves  When a solid or liquid becomes a gas  When a gaseous chemical reaction produces more molecules  When the temperature increases

52  Given in appendix 4  ∆S o reaction = Σnp S o – Σnr S o

53  We cannot measure ∆G directly we must use other measured quantities.  The more negative the value of ∆G the further a reaction will go to the right to reach equilibrium

54 1) Using the formula ∆G o = ∆H o -T∆S 2) By taking advantage of the fact that ∆G is a state function and solving like Hess’s law

55 3) By using ∆G o f, standard free energy of formation ∆G o = Σnp G f o (products)– Σnr G f o (reactants)

56  Enthalpy is not affected by pressure  Entropy is dependent on pressure because entropy is dependent on volume. S large volume > S small volume so S low pressure > S high pressure

57  ∆G = ∆G o + RT lnQ  ∆G = free energy change for rxn for specified pressure  ∆G o = free energy change at standard pressure  R = 8.31 J/K * mol  T = temp in Kelvin  Q = reaction quotient

58  The equilibrium point occurs at the lowest value of free energy available to the reaction system

59  ∆G O = 0 the system is at equilibrium, K = 1, pressure = 1 atm  ∆G o 1, pressures of products > 1 atm, pressure of reactants < 1  ∆G 0 > 0 not at equilibrium, K <1, shift left because reactants have less energy

60  Free energy can tell us how much work can be done with a given process.

61 - ∆G means the amount of free energy available \ to do use work + ∆G is the minimium amount of work that must be expended to make the reaction occur.


Download ppt "Chapter 6. = the capacity to do work or to produce heat Kinetic energy = the energy due to motion depends on mass & velocity Potential Energy = energy."

Similar presentations


Ads by Google