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Chemistry 1 > Lectures 26-35 > Flashcards

Lectures 26-35 Flashcards

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1
Q

chemical equilibrium

A

when molecules form a state in which the composition of the reaction mixture remains constant and concentrations/partial pressures of reactants and products no longer change

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2
Q

at equilibrium, forward reaction rate =

A

reverse reaction rate

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3
Q

K(p)

A

p(products)^coefficients/p(reactants)^coefficients

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4
Q

K(c)

A

[products]^coefficients/[reactants]^coefficients

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5
Q

relationship between forward and reverse reactions for K

A

K(reverse) = 1/K(forward)

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6
Q

doubling reactions

A

K^2

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7
Q

activities

A

describe ‘effective’ concentration

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8
Q

for pure liquids and solids, concentrations of these components

A

do not change and are given an activity of 1 (not included in calculations)

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9
Q

solubility product

A

normal K(c) with solid not included; only included solvated ions/molecules

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10
Q

smaller K(sp), harder it is

A

to dissolve a substance

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11
Q

product-favoured

A

K > 1

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12
Q

reactant-favoured

A

K < 1

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13
Q

reaction quotient

A

tells us how far we are from equilibrium

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14
Q

at equilibrium concentrations, Q =

A

K

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15
Q

if Q < K, then

A

forward reaction is faster until equilibrium attained

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16
Q

if Q > K, then

A

reverse reaction is faster until equilibrium attained

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17
Q

Q = 1

A

standard-state reaction mixture

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18
Q

equation relating Gibbs free energy and reaction quotient

A

deltarG = deltarG^o + RTln(Q)

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19
Q

in standard-state reaction mixture, Q = 1 and ln(Q) =

A

zero, so deltarG = deltarG^o

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20
Q

at equilibrium, deltarG = 0 =

A

deltarG^o + RTln(K)

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21
Q

at equilibrium, deltarG^o =

A

-RTln(K)

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22
Q

Le Chatelier’s Principle

A

when a change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change

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23
Q

if reactant is added,

A

equilibrium is re-established by consuming some of the added reactant

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24
Q

if product is added,

A

equilibrium is re-established by consuming some of the added product

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25
Q

for gaseous reactions, increasing pressure will push equilibrium

A

in the direction of fewer molecules

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26
Q

if a reaction is exothermic, a temperature increase will favour

A

reactants

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27
Q

if a reaction is endothermic, a temperature increase will favour

A

products

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28
Q

linking entropy with equilibrium constants

A

ln(K) = -deltaH/RT + deltaS/R

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29
Q

van’T Hoff equation

A

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

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30
Q

ICE tables

A

initial, change, equilibrium

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31
Q

equation connecting K(p) and K(c)

A

K(p) = K(c)(RT)^delta(n)(gas)

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32
Q

other value (not 8.314)

A

0.0821 L atm/mol K

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33
Q

if K > 1, deltaG is expected to be

A

negative

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34
Q

if K < 1, deltaG is expected to be

A

positive

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35
Q

Lewis acids and bases

A

electron acceptors (acids) and donors (bases)

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36
Q

Bronsted-Lowry acids and bases

A

proton donors (acids) and bases (acceptors)

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37
Q

all ions/molecules are surrounded by solvent (water) to form a

A

solvation shell

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38
Q

HA is acid,

A

A^- is the conjugate base

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39
Q

B is the base

A

HB+ is the conjugate acid

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40
Q

weak acid-base pair is always

A

favoured in the equilibrium

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41
Q

water undergoes

A

self-ionisation and is a weak acid and weak base (reaction lies to left)

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42
Q

ionisation constant, K(w)

A

extent of ionisation of water; [H+][OH-] = 10^-14

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43
Q

in pure water, [H+] =

A

[OH-] = 10^-7

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44
Q

solution is neutral when

A

[H+] = [OH-]

45
Q

ionisation of water is an

A

endothermic process

46
Q

addition of other substances may change [H+] and [OH-], but the

A

product of their concentrations is always constant

47
Q

[H+] > 10^-7 M

A

acidic

48
Q

[H+] < 10^-7

A

basic

49
Q

pH

A

—log[H+]

50
Q

pOH

A

—log[OH-]

51
Q

pH + pOH =

A

14

52
Q

K(a), acid dissociation constant

A

[H3O+][A-]/[HA]

53
Q

pK(a)

A

—log[K(a)]

54
Q

strong acids have

A

high K(a) and small pK(a)

55
Q

K(b), base equilibrium constant

A

[OH-][BH+]/[B]

56
Q

pK(b)

A

—log[K(b)]

57
Q

strong bases have

A

high K(b) and small pK(b)

58
Q

strong acids

A

ionise almost completely in solution; reaction lies to right

59
Q

conjugate base of a strong acid is very

A

weak

60
Q

for strong acids, the [H3O+] concentration equals

A

the acid concentration

61
Q

for strong bases, we can assume the concentration of the base equals

A

the [OH-] concentration

62
Q

for weak acid, equilibrium lies well to

A

left

63
Q

calculating pH of solutions of weak acids

A

[H3O+] = (K(a)[HA])^1/2 and go from there

64
Q

[weak acid initially in solution ] = very close to

A

[weak acid concentration at equilibrium]

65
Q

5% Rule

A

if dissociation is <5%, approximation is valid

66
Q

% dissociation

A

[conjugate base]/[acid] x 100

67
Q

calculating pH of solutions of weak bases

A

write equation, write basicity constant, write ICE table (if only base added, [BH+] = [OH-]), decide if base can be assumed weak and solve for x ([OH-]), check assumption is correct

68
Q

buffered solution is one that

A

resists a change in pH when either OH- or H+ ions are added

69
Q

good buffers will have close to equal

A

proportions of acid and its conjugate base

70
Q

if too much acid or base is added to the buffer, it will

A

fail

71
Q

buffer strength is at its strongest when

A

pH = pK(a)

72
Q

Henderson-Hasselbach equation

A

pH = pK(a) + log([A-]/[HA])

73
Q

solving buffer problems without using Henderson-Hasselbach equation

A

write out the chemical equilibrium, use an ICE table, write out the equilibrium constant and define in terms of x, solve for x

74
Q

buffers may be prepared by adding a weak acid

A

and its conjugate base to a solution

75
Q

buffers may be prepared by adding enough strong base to

A

neutralise the weak acid

76
Q

polyprotic acids

A

molecules able to be ionised more than once

77
Q

negative charge on an acid makes it

A

harder to ionise

78
Q

lewis acids are

A

cations, incomplete octet, polar bonds

79
Q

lewis bases are

A

anions, H2O, lone pairs

80
Q

conjugate acids and bases in a buffer solution should have similar concentrations

A

within a factor of 10

81
Q

when creating buffers, if not adding conjugate acid/base and simply adding a random STRONG acid/base, only add

A

HALF original concentration

82
Q

factors deciding whether an acid is strong or not

A

electronegativity, hybridisation, resonance stabilisation of anions

83
Q

more electronegative the non-hydrogen atom,

A

the more acidic

84
Q

the larger the ion,

A

the more the products are favoured and the stronger the acid (charge is stabilised)

85
Q

loss of charge on dissociation

A

favours that reaction

86
Q

increasing s orbital contribution

A

sp^3 < sp^2 < sp

87
Q

acidity is judged by stability of

A

anion formed after releasing proton

88
Q

increasing acidity (orbitals)

A

sp^3, sp^2, sp

89
Q

stabilised anion -> resonance

A

stronger acid

90
Q

cations of bases NOT HAVING resonance

A

stronger base

91
Q

Mesomeric

A

+M or -M

92
Q

electron-donating groups

A

+M; have lone pair of electrons; stabilise positive charges and destabilise conjugate base

93
Q

electron-withdrawing groups

A

-M; have full/partial positive charge and stabilise negative charges; stabilise conjugate base

94
Q

if Ksp is very small,

A

solid is only sparingly soluble

95
Q

solubility

A

amount of a salt in a specified volume of a saturated solution. Measured in M. Find concentration of salt at equilibrium!

96
Q

presence of a common ion

A

decreases solubility of the salt

97
Q

if Q = Ksp

A

saturated; at equilibrium

98
Q

if Q < Ksp

A

unsaturated; more salt can dissolve

99
Q

Q > Ksp

A

supersaturated; precipitation will occur to bring system to equilibrium

100
Q

equilibrium constant for soluble reactions

A

> 1

101
Q

equilibrium constant for insoluble reactions

A

< 1

102
Q

precipitation reaction equation

A

creating the solid

103
Q

higher value for solubility,

A

less soluble (see Ksp)

104
Q

finding pH for weak acid

A

pH = 1/2pK(a) -1/2log[HA]

105
Q

K(a) x K(b) =

A

10^-14

106
Q

if Ka(2) for second reaction is very small, assume

A

this reaction is negligible when calculating pH and [conjugate base] = [overall H3O+]

107
Q

diluting a buffer solution does not change [base] : [acid] ratio, so

A

pH of solution will not change

108
Q

pK(a) + pK(b)

A

= 14

Chemistry 1 (4 decks)

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