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Chem 2 Heider > Ch 17 - Thermodynamics > Flashcards

Ch 17 - Thermodynamics Flashcards

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1
Q

entropy

A

the dispersion(spreading out) of energy

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2
Q

in our universe entropy always

A

increases

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3
Q

heat tax

A

an unavoidable cut of energy is taken from ever transaction by nature
- can not be 100% efficient

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4
Q

in general, the most efficient use of energy occurs with

A

the smallest number of transactions

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5
Q

there is always a loss of energy as

A

heat

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6
Q

fundamental goal of thermodynamics is to

A

predict spontaneity

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7
Q

spontaneous process

A

a process that occurs without ongoing outside intervention(such as the performance of work by some external force)

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8
Q

spontaneity of a process does not depend on

A

the speed of a reaction

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9
Q

Entropy(S)

A

a thermodynamic function that increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state

	- S = k(lnW)
	- K = 1.38*10^-23J/K
	- w = number of possible microstates
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10
Q

the more microstates present the greater entropy

A

more energetically equivalent ways to arrange the components of the system

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11
Q

the state with the highest entropy also has the greatest

A

dispersal of energy

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12
Q

second law of thermodynamics

A

for any spontaneous process, the entropy of the universe increases(deltaSuniverse > 0)

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13
Q

the criterion for spontaneity is the

A

entropy of the universe

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14
Q

the change in entropy is the entropy of the final state minus the initial

A

deltaS = Sfinal – Sinitial

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15
Q

solids -> gas

A

entropy increases as state becomes more fluid

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16
Q

even though the entropy of water decreases during freezing the entropy of the universe

A

increases

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17
Q

deltaSuniv =

A

deltaSsys+ deltaSsurr

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18
Q

the entropy of a system can decrease(deltaSsys < 0) as long as the entropy of the surroundings increases by a greater amount

A

deltaSsurr > -deltaSsys

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19
Q

according to 2nd law of thermodynamics the entropy of the universe must increase for a process to be spontaneous

A
  • deltaSuniv > 0

- overall net entropy of the universe has a net increase

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20
Q

the release of heat energy by the system disperses that energy into the surroundings,

A

increasing the entropy of the surroundings

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21
Q

water freezing becomes more ordered but is spontaneous

A

the heat given off to the surroundings increases the entropy to a sufficient degree to overcome the entropy decrease in the water

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22
Q

An exothermic process increases

A

the entropy of the surroundings

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23
Q

an endothermic process decreases

A

the entropy of the surroundings

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24
Q

a process that emits heat into the surroundings(Qsys negative) increases

A

the entropy of the surroundings(+ deltaSsurr)

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25
Q

a process that absorbs heat from the surroundings(Qsys positive) decreases

A

the entropy of the surroundings(- deltaSsurr)

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26
Q

the magnitude of the change in entropy of the surroundings is

A

proportional to the magnitude of Qsys

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27
Q

in general the higher the temperature the lower the

A

magnitude of deltaSsurr(inverse relationship)

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28
Q

-Qsys = Qsurr as long as there is

A

constant temperature and pressure

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29
Q

as temperature increases in exothermic process the spontaneity tends to

A

decrease and being exothermic becomes less of a determining factor for spontaneity as temperature increases

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30
Q

at constant pressure Qsys = deltaHsys

A

deltaSsurr = (-deltaHsys)/T

- constant P,T

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31
Q

Gibbs free energy(G)

A
  • G = H – TS
    - H = enthalpy
    - T = temp in K
    - S = entropy
    - deltaG = deltaH – T(deltaS)
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32
Q

the change in Gibbs free energy for a process occurring at constant temperature and pressure is proportional to

A

the negative deltaSuniv

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33
Q

deltaG is also a criterion for spontaneity

A

although opposite in sign

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34
Q

Gibbs free energy is also called chemical potential because

A

it is analogous to mechanical potential

- chemical systems tend toward lower Gibbs free energy(lower chemical potential)

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35
Q

deltaG is proportional to the

A

negative of deltaSuniv

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36
Q

a decrease in Gibbs free energy(deltaG<0) corresponds to

A

a spontaneous process

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37
Q

an increase in Gibbs free energy(deltaG>0) corresponds to

A

a nonspontaneous process

38
Q

to calculate Gibbs free energy you only need to determine the entropy of the system(deltaSsys) and the change in enthalpy of the system(deltaHsys)

A

combined you can calculate deltaG = deltaH – T(deltaSsys)

39
Q

delta H(-), delta S(+)

A
  • Low temp(delta G < 0)
    - High Temp(delta G < 0)
    - 2N2O(g) -> 2N2(g) + O2(g)
40
Q

delta H(+), delta S(-)

A
  • Low temp(delta G >0) = Nonspontaneous(delta G > 0)
    - High Temp(delta G >0) = Nonspontaneous(delta G > 0)
    - 3O2(g) -> 2O3(g)
41
Q

delta H(-), delta S(-)

A
  • Low temp(delta G < 0)
    - High Temp(delta G >0) = Nonspontaneous(delta G > 0)
    - H2O(l) -> H2O(s)
42
Q

delta H(+), delta S(+)

A
  • Low temp(delta G >0) = Nonspontaneous(delta G >0)
    - High Temp(delta G < 0)
    - H2O(l) -> H2O(g)
43
Q

when deltaH and deltaS have opposite signs the spontaneity of the reaction

A

does not depend on temperature

44
Q

when deltaH and deltaS have the same sign the spontaneity

A

depends on temperature

45
Q

the point where delta G changes sign is where the reaction

A

changes from spontaneous to nonspontaneous

46
Q

standard state definitions:

Gas

A

the standard state is the pure gas at a pressure of exactly 1atm

47
Q

standard state definitions:

Solid and Liquid

A

the standard state in pure substance force is pressure of 1atm and at the temperature of interest

48
Q

standard state definitions:

Solutions

A

standard state is 1M

49
Q

standard entropy change for a reaction(deltaSrxn)

A

the change of entropy for a process in which all reactants and products are in their standard states

	- deltaSrxn = Sproducts – Sreactants
		- standard molar entopies(S) are used in calculating deltaSrxn
50
Q

for entropy there is

A

an absolute zero

51
Q

third law of thermodynamics

A

the entropy of a perfect crystal at absolute zero(0K) is zero

52
Q

standard entropy values are listed in

A

J/mol*K

53
Q

entropy is an extensive property

A

depends on the amount of the substance

54
Q

allotrope

A

a substance existing in two or more forms within the same state

55
Q

5 factors on the number of places to put energy and the standard entropy

A
  • state of substance
    - molar mass
    - particular allotrope
    - molecular complexity
    - extent of dissolution
56
Q

the entropy of a gas is

A

generally higher than a liquid than a solid

57
Q

the greater the molar mass the

A

greater the entropy in the same state

58
Q

the less constrained the allotrope results in

A

more places to put energy and therefore greater entropy(graphite has more entropy than diamond)

59
Q

for a given state of matter entropy generally

A

increases with increasing molecular complexity

60
Q

the dissolution of a crystalline solid into solution usually results in

A

an increase in entropy become of greater energy dispersal

61
Q

to calculate deltaSrxn

A

subtract the standard entropies if the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multipled by their stoichiometric coefficients

62
Q

deltaSrxn =

A

the sum of nS(products) – sum of nS(reactants)

63
Q

unlike enthalpies of formation which are 0 for elements in their standard states

A

entropies are always nonzero at 25C.

64
Q

standard change in free energy(deltaGrxn)

A

the criterion for spontaneity at standard conditions

65
Q

the more negative deltaGrxn the more spontaneous the process is

A

the further it will go toward products to reach equilibrium

66
Q

using deltaGrxn = delta Hrxn – T(deltaSrxn)

A
  • deltaHrxn = H,productsmoles – H,reactantsmoles
    - deltaSrxn = S,productsmoles – S,reactantsmoles
    - T in Kelvin
67
Q

using tabulated values of free energies of formation

A
  • free energy of formation(deltaGf) – the free energy of formation(deltaGf) is the change in free energy when 1 mol of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is 0.
    - deltaGrxn = molesdeltaGf(products) – molesdeltaGf(reactants)
68
Q

free energy of formation(deltaGf)

A

the free energy of formation(deltaGf) is the change in free energy when 1 mol of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is 0.

69
Q

deltaGrxn is a stepwise reaction so you can combine reverse and switch parts as long as you

A

multiply the deltaGrxn by the appropriate factor

70
Q

free energy is the amount of energy available

A

to do work

71
Q

the deltaGrxn is the theoretical maximum amount of free energy available to do work

A

in reality the free energy is even less because additional energy is lost to the surroundings as heat

72
Q

reversible reaction

A

a reaction that achieve the theoretical limit with respect to free energy
- occurs infinitesimally slowly and the free energy can only be drawn out in infinitesimally small increments that exactly match the amount of energy that the reaction is producing during that increment

73
Q

irreversible reactions

A

all real reactions do not achieve the theoretical limit of available free energy

74
Q

if the change in free energy of a chemical reaction is positive then

A

deltaGrxn represents the minimum amount of energy required to make the reaction occur

75
Q

deltaGrxn represents a theoretical limit

A

making a real nonspontaneous reaction occur always required more energy than the theoretical limit

76
Q

ordinary life conditions are not

A

standard conditions

77
Q

deltaGrxn(degree symbol) only applies at

A

standard conditions

78
Q

for nonstandard conditions:

deltaGrxn = deltaGrxn(degree sign) + RT(lnQ)

A
  • Q = reaction quotient(partial pressure in atm)
    - T = kelvin
    - R = gas constant(8.314 J/mol*K)
79
Q

under standard conditions Q always equals 1

A

ln1 = 0 so its ignored

80
Q

under equilibrium conditions the value of RT(lnQ) is always equal in magnitude but opposite in sign to the value of deltaGrxn(degree sign)

A
  • making deltaGrxn = 0

- nonspontaneous in either direction of the reaction

81
Q

deltaGrxn<0 is spontaneous in

A

the forward direction

82
Q

deltaGrxn>0 is spontaneous in

A

the reverse direction

83
Q

the equilibrium constant becomes larger as

A

the standard free energy change becomes more negative

84
Q

large negative free energy change = large equilibrium constant =

A

product strongly favored at equilibrium

85
Q

large positive free energy change = small equilibrium constant =

A

reactants strongly favored at equilibrium

86
Q

deltaGrxn(degree sign) = -RT(lnK)

A

use Kp for reactions involving gases and use Kc for reactions involving substances dissolved in solution

87
Q

K< 1, lnK is negative and deltaGrxn(degree sign) is positive

A

under standard conditions(Q=1) the reaction is spontaneous in the reverse direction

88
Q

K> 1, lnK is positive and deltaGrxn(degree sign) is negative

A

under standard conditions(Q=1) the reaction is spontaneous in the forward direction

89
Q

K= 1, lnK is zero and deltaGrxn(degree sign) is zero

A

under standard conditions(Q=1) the reaction is in equilibrium

90
Q

lnK = (-deltaHrxn/R)(1/T)+(deltaSrxn/R)

A

ln(K2/K1)=(-deltaHrxn/R)(1/T2-1/T1)

Chem 2 Heider (9 decks)

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