Brainscape's Knowledge GenomeTM


Top Flashcards (Certified)
All Flashcards

Chem 2 Heider > Ch 17 - Thermodynamics > Flashcards

Ch 17 - Thermodynamics Flashcards

1
Q

entropy

A

the dispersion(spreading out) of energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

in our universe entropy always

A

increases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

heat tax

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

in general, the most efficient use of energy occurs with

A

the smallest number of transactions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

there is always a loss of energy as

A

heat

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

fundamental goal of thermodynamics is to

A

predict spontaneity

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

spontaneous process

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

spontaneity of a process does not depend on

A

the speed of a reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

the more microstates present the greater entropy

A

more energetically equivalent ways to arrange the components of the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

the state with the highest entropy also has the greatest

A

dispersal of energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

second law of thermodynamics

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

the criterion for spontaneity is the

A

entropy of the universe

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

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

A

deltaS = Sfinal – Sinitial

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

solids -> gas

A

entropy increases as state becomes more fluid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

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

A

increases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

deltaSuniv =

A

deltaSsys+ deltaSsurr

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

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

A

increasing the entropy of the surroundings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

An exothermic process increases

A

the entropy of the surroundings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

an endothermic process decreases

A

the entropy of the surroundings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

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

A

the entropy of the surroundings(+ deltaSsurr)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
a process that absorbs heat from the surroundings(Qsys positive) decreases
the entropy of the surroundings(- deltaSsurr)
26
the magnitude of the change in entropy of the surroundings is
proportional to the magnitude of Qsys
27
in general the higher the temperature the lower the
magnitude of deltaSsurr(inverse relationship)
28
-Qsys = Qsurr as long as there is
constant temperature and pressure
29
as temperature increases in exothermic process the spontaneity tends to
decrease and being exothermic becomes less of a determining factor for spontaneity as temperature increases
30
at constant pressure Qsys = deltaHsys
deltaSsurr = (-deltaHsys)/T | - constant P,T
31
Gibbs free energy(G)
- G = H – TS - H = enthalpy - T = temp in K - S = entropy - deltaG = deltaH – T(deltaS)
32
the change in Gibbs free energy for a process occurring at constant temperature and pressure is proportional to
the negative deltaSuniv
33
deltaG is also a criterion for spontaneity
although opposite in sign
34
Gibbs free energy is also called chemical potential because
it is analogous to mechanical potential | - chemical systems tend toward lower Gibbs free energy(lower chemical potential)
35
deltaG is proportional to the
negative of deltaSuniv
36
a decrease in Gibbs free energy(deltaG<0) corresponds to
a spontaneous process
37
an increase in Gibbs free energy(deltaG>0) corresponds to
a nonspontaneous process
38
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)
combined you can calculate deltaG = deltaH – T(deltaSsys)
39
delta H(-), delta S(+)
- Low temp(delta G < 0) - High Temp(delta G < 0) - 2N2O(g) -> 2N2(g) + O2(g)
40
delta H(+), delta S(-)
- Low temp(delta G >0) = Nonspontaneous(delta G > 0) - High Temp(delta G >0) = Nonspontaneous(delta G > 0) - 3O2(g) -> 2O3(g)
41
delta H(-), delta S(-)
- Low temp(delta G < 0) - High Temp(delta G >0) = Nonspontaneous(delta G > 0) - H2O(l) -> H2O(s)
42
delta H(+), delta S(+)
- Low temp(delta G >0) = Nonspontaneous(delta G >0) - High Temp(delta G < 0) - H2O(l) -> H2O(g)
43
when deltaH and deltaS have opposite signs the spontaneity of the reaction
does not depend on temperature
44
when deltaH and deltaS have the same sign the spontaneity
depends on temperature
45
the point where delta G changes sign is where the reaction
changes from spontaneous to nonspontaneous
46
standard state definitions: Gas
the standard state is the pure gas at a pressure of exactly 1atm
47
standard state definitions: Solid and Liquid
the standard state in pure substance force is pressure of 1atm and at the temperature of interest
48
standard state definitions: Solutions
standard state is 1M
49
standard entropy change for a reaction(deltaSrxn)
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
for entropy there is
an absolute zero
51
third law of thermodynamics
the entropy of a perfect crystal at absolute zero(0K) is zero
52
standard entropy values are listed in
J/mol*K
53
entropy is an extensive property
depends on the amount of the substance
54
allotrope
a substance existing in two or more forms within the same state
55
5 factors on the number of places to put energy and the standard entropy
- state of substance - molar mass - particular allotrope - molecular complexity - extent of dissolution
56
the entropy of a gas is
generally higher than a liquid than a solid
57
the greater the molar mass the
greater the entropy in the same state
58
the less constrained the allotrope results in
more places to put energy and therefore greater entropy(graphite has more entropy than diamond)
59
for a given state of matter entropy generally
increases with increasing molecular complexity
60
the dissolution of a crystalline solid into solution usually results in
an increase in entropy become of greater energy dispersal
61
to calculate deltaSrxn
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
deltaSrxn =
the sum of nS(products) – sum of nS(reactants)
63
unlike enthalpies of formation which are 0 for elements in their standard states
entropies are always nonzero at 25C.
64
standard change in free energy(deltaGrxn)
the criterion for spontaneity at standard conditions
65
the more negative deltaGrxn the more spontaneous the process is
the further it will go toward products to reach equilibrium
66
using deltaGrxn = delta Hrxn – T(deltaSrxn)
- deltaHrxn = H,products*moles – H,reactants*moles - deltaSrxn = S,products*moles – S,reactants*moles - T in Kelvin
67
using tabulated values of free energies of formation
- 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 = moles*deltaGf(products) – moles*deltaGf(reactants)
68
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.
69
deltaGrxn is a stepwise reaction so you can combine reverse and switch parts as long as you
multiply the deltaGrxn by the appropriate factor
70
free energy is the amount of energy available
to do work
71
the deltaGrxn is the theoretical maximum amount of free energy available to do work
in reality the free energy is even less because additional energy is lost to the surroundings as heat
72
reversible reaction
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
irreversible reactions
all real reactions do not achieve the theoretical limit of available free energy
74
if the change in free energy of a chemical reaction is positive then
deltaGrxn represents the minimum amount of energy required to make the reaction occur
75
deltaGrxn represents a theoretical limit
making a real nonspontaneous reaction occur always required more energy than the theoretical limit
76
ordinary life conditions are not
standard conditions
77
deltaGrxn(degree symbol) only applies at
standard conditions
78
for nonstandard conditions: deltaGrxn = deltaGrxn(degree sign) + RT(lnQ)
- Q = reaction quotient(partial pressure in atm) - T = kelvin - R = gas constant(8.314 J/mol*K)
79
under standard conditions Q always equals 1
ln1 = 0 so its ignored
80
under equilibrium conditions the value of RT(lnQ) is always equal in magnitude but opposite in sign to the value of deltaGrxn(degree sign)
- making deltaGrxn = 0 | - nonspontaneous in either direction of the reaction
81
deltaGrxn<0 is spontaneous in
the forward direction
82
deltaGrxn>0 is spontaneous in
the reverse direction
83
the equilibrium constant becomes larger as
the standard free energy change becomes more negative
84
large negative free energy change = large equilibrium constant =
product strongly favored at equilibrium
85
large positive free energy change = small equilibrium constant =
reactants strongly favored at equilibrium
86
deltaGrxn(degree sign) = -RT(lnK)
use Kp for reactions involving gases and use Kc for reactions involving substances dissolved in solution
87
K< 1, lnK is negative and deltaGrxn(degree sign) is positive
under standard conditions(Q=1) the reaction is spontaneous in the reverse direction
88
K> 1, lnK is positive and deltaGrxn(degree sign) is negative
under standard conditions(Q=1) the reaction is spontaneous in the forward direction
89
K= 1, lnK is zero and deltaGrxn(degree sign) is zero
under standard conditions(Q=1) the reaction is in equilibrium
90
lnK = (-deltaHrxn/R)(1/T)+(deltaSrxn/R)
ln(K2/K1)=(-deltaHrxn/R)(1/T2-1/T1)

Chem 2 Heider (9 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2025 Bold Learning Solutions. Terms and Conditions