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CHM135 > midterm 2 > Flashcards

midterm 2 Flashcards

1
Q

rates of reaction can be determined using the two formulas

A

A->B
rate of production of B = Δ[B]/Δt
rate of consumption of A = -Δ[A]/Δt

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2
Q

rate =

A

change in concentration/change in time

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3
Q

The General Reaction Rate

A

for a balanced reaction:

aA -> bB + cC

rate = -1/a(Δ[A]/Δt) = 1/b(Δ[B]/Δt) = 1/c*(Δ[C]/Δt)

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4
Q

Initial instantaneous rate vs instantaneous rate

A

initial instantaneous rate is at t=0.
instantaneous rate can be at any point on the graph

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5
Q

describe the rate law in words

A

an equation that shows how the reaction rate depends on the concentration of each reactant

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6
Q

Rate Law

A

aA + bB -> cC + dD

rate = k[A]^m[B]^n

[A], [B] - concentration in M
k - rate constant
m - reaction order in A
n - reaction order in B

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7
Q

what si the overall order of a reaction?

A

the sum of the individual orders

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8
Q

units of rate

A

always M/s

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9
Q

units of k

A

depends on the rate law expression

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10
Q

units of k if:
rate = k
rate = k[A]
rate = k[A][B]
rate = k[A][B]^2

A

M/s
1/s
1/(Ms)
1/(M^2s)

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11
Q

how are the order of reactions (m, n) and the rate constant (k) determined?

A

experimentally

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12
Q

give 2 ways of determining order of reaction and rate constant

A
  1. method of initial rates:
    - vary initial concentration of one reactant at a time
    - measure the initial reaction rate for each reaction
    - solve a system of rate law equations to determine the order
  2. graphical method/integrated rate law
    - monitor the course of a reaction over time
    - plot data
    - the shape of the curve reveals reaction order
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13
Q

0th order reaction

A

A –(k)–> product

[A]t = -kt + [A]0
[A]0 is the initial concentration
[A]t is the concentration at time t

  • plot the experimental [A] vs time
  • if the graph is linear, the reaction is zero-order
  • rate constant k = (-slope)
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14
Q

half life

A

the time required for reactant concentration to reach half of its original value

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15
Q

half life for a 0th order reaction

A

t(1/2) = [A]0/2k
- dependent on the (initial) concentration
- gets shorter over the course of the reaction (each successive half-life is half as long)

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16
Q

how are 0th order reactions possible?

A

in reactions whether the kinetics are governed by the availability of a catalyst

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17
Q

1st order reaction

A

A –(k)–> products

[A]t = [A]0e^-kt
ln[A]t - ln[A]0 = -kt

  • plot the experimental ln[A] vs time
  • if the graph is linear, the reaction is first-order
  • the rate constant k = (-slope)
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18
Q

half life of 1st order reaction

A

t(1/2) = ln(2)/k
- t(1/2) is independent of the concentration
- every half-life is an equal period of time

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19
Q

2nd order reaction

A

A –(k)–> products

1/[A]t = kt + 1/[A]0

  • plot the experimental 1/[A] vs. time
  • if the graph is linear, the reaction is second-order
  • the rate constant k = slope
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20
Q

half life of 2nd order reaction

A

t(1/2) = 1/k[A]0
- t(1/2) is dependent on the (initial) concentration
- t(1/2) gets longer over the course of the reaction, so each successive t(1/2) doubles in length

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21
Q

draw graphs for zeroth order, first order, and second order. also draw straight-line plots used to determine rate constant

A
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22
Q

how can we find the overall reaction from the elementary steps?

A

the elementary steps in the true reaction mechanism must sum to give the overall (stoichiometric) reaction

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23
Q

reaction intermediate

A

a species formed in one step of a reaction mechanism and consumed in a later step

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24
Q

molecularity

A

a classification of an elementary reaction based on the number of molecules (or atoms) on the reactant side of the chemical equation

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25
Q

unimolecular reaction

A

an elementary reaction that involves a single reactant molecule
eg rate = k[O3]

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26
Q

bimolecular reaction

A

an elementary reaction that results from an energetic collision of two reactant molecules
eg rate = k[O3][O]

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27
Q

termolecular reaction

A

an elementary reaction that involves three atoms or molecules (rare)
eg rate = k[O]^2[M]

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28
Q

how can you find a plausible mechanism for a reaction?

A
  • elementary steps must add up to the overall reaction
  • mechanism must correlate with the experimental rate law
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29
Q

the rate law of the overall reaction must be the rate law of

A

the rate limiting step

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30
Q

how do you determine the rate law for a reaction where the rate limiting step is the second step?

A
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31
Q

factors affecting reaction rates

A
  1. concentration; increased concentration of reactants = increased rate
    - moving molecules are closer together
    - increased likelihood that reactants will collide
  2. temperature: observed increase in rate wen there is an increase in temperature
    - due to collision and transition state theory
    - molecules collide with energy greater than Ea more often
  3. catalyst: a catalyst provides an alternative reaction mechanism that proceeds faster than the original (uncatalysed) mechanism
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32
Q

collision theory

A

for a bimolecular reaction to take place, reactants A and B must collide with proper orientation, and an energy greater than the activation energy, Ea

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33
Q

k =

A

Zpf
Z = volumetric collision frequency
p = fraction with correct orientation
f = fraction with sufficient energy

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34
Q

k depends strongly on temperature - so which of Z, p, and f depend on temperature?

A

collision frequency (Z)
- increase in temperature -> reactants collide more often (but not that more often)
fraction with correct orientation (p)
- increase T -> no change
fraction with sufficient energy (f)
- increase T -> reactants collide with greater kinetic energy

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35
Q

transition state

A

the unstable group of atoms that are the highest-energy species along the pathway from reactants to products

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36
Q

activation energy

A

the minimum energy required for a successful reaction

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37
Q

f =

A

e^(-Ea/RT)

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38
Q

as T increases, how does f change?

A

increases exponentially

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39
Q

draw a graph of collision energy vs fraction of collisions for two temperatures

A
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40
Q

Arrhenius equation

A

k = Ae^(-Ea/RT)
A = zp : the frequency factor (aka pre-exponential factor)

ln(k) = -Ea/R (1/T) + lnA
ln(k) = y
-Ea/R = m
1/T = x
lnA = b

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41
Q

Arrhenius plot

A

y intercept = ln(A)
slope = -Ea/R

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42
Q

catalysts

A

a substance that provides an alternative reaction mechanism that has a lower activation energy than the uncatalyzed mechanism -> faster reaction
- usually involved in the rate-limiting step of the new pathway
- often appears in the rate law of the catalysed reaction
- react early in a multi-step mechanism; regenerated in a later step, so not consumed
- do not show up in balanced, overall reaction
- catalysed reaction has the same endo/exothermicity as the uncatalysed reaction

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43
Q

Michaelis-Menten kinetics

A

enzymes are catalysts of biological organisms, typically protein molecules with large molecular weights

at low substrate concentration
- first order behaviour: rate increases linearly with [S]

at high substrate concentration
- zeroth order behaviour: once all the enzyme is complexed, rate of reaction saturates

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44
Q

define chemical equilibrium

A

the state reached when the concentrations of reactants and products remain constant over time
- forward and reverse reactions occur at the same rate
- reactions are still occurring, but there is no net reaction
- chemical equilibria are dynamic and reversible

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45
Q

Kc =

A

aA + bB <–> cC + dD

Kc = [C]^c[D]^d/[A]^a[B]^b
concentrations at equilibrium

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46
Q

Kc when you reverse the reaction

A

1/Kc

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47
Q

Kc when you multiply the reaction by 2

A

Kc^2

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48
Q

Kc when you add two reactions

A

Kc1 x Kc2

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49
Q

units of the equilibrium constant

A

equilibrium constants are defined as unites quantities

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50
Q

homogeneous equilibria

A

all species are present in the same phase.

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51
Q

heterogeneous equilibria

A

system’s state of equilibrium contains components from multiple phases

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52
Q

why are pure solids and liquids excluded from equilibrium expressions?

A

they are always in their standard state, where a=1: their concentrations don’t change

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53
Q

equilibrium constant Kp

A

for gases, we can use pressures instead of concentrations

Kp = Kc(RT)^Δn
Kp = Pc^c x Pd^d/Pa^a x Pb^b

Δn is the number of moles of gaseous products minus the number of moles of gaseous reactants

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54
Q

Kc > 10^3

A

reaction proceeds nearly to completion, products favoured

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55
Q

Kc < 10^-3

A

reaction proceeds hardly at all, reactants favoured

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56
Q

10^-3<Kc<10^3

A

appreciable concentrations of both reactants and products

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57
Q

reaction quotient, Qc

A

defined similarly to an equilibrium constant Kc except that the []’s in Qc can have ant values, not necessarily equilibrium values

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58
Q

if Qc = Acc

A

no net reaction occurs, already at equilibrium

59
Q

if Qc < Kc

A

net reaction goes from left to right

60
Q

if Qc > Kc

A

net reaction goes from right to left

61
Q

le Chatelier’s principle

A

if a stress is applied to a reaction mixture at equilibrium, a net reaction then occurs in the direction that relieves the stress

62
Q

in general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Chateliers principle predicts that

A
  • the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance
  • the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance
63
Q

in general, when an equilibrium is disturbed by a change in volume that results in a corresponding change in pressure, Le Chateliers principle predicts that

A
  • reduction in volume increases the partial pressures of all gases, which provokes a net reaction in the direction that decreases the total moles of gas
  • enlargement in volume decreases the partial pressures of all gases, which provokes a net reaction in the direction that increases the total moles of gas
64
Q

adding inert gas

A

no effect, because the partial pressures (molar concentrations) don’t change

65
Q

temperature can alter the equilibrium concentrations, but for a different reason:

A

it changes the value of Kc

66
Q

in general, when an equilibrium is disturbed by a change in temperature,

A

Kc for an exothermic reaction (-ΔH) decreases as the temperature increases
Kc for an endothermic reaction (+ΔH) increases as the temperature increases

67
Q

linking chemical equilibrium and kinetics

A

A + B <-(Kf)-(Kr)-> C + D

rate forward = Kf[A][B]
rate backward = Kr[C][D]
at equilibrium: Kf[A][B]=Kr[C][D]

Kf/Kr = [C][D]/[A][B] = Kc
Kc = kf/kr

68
Q

if Kc is really large
if Kc is nearly unity
if Kc is really small

A

Kf&raquo_space; Kr and the reaction goes almost to completion
Kf = Kr and both reactants and products exist at equilibrium
Kr&raquo_space; Kf and the reaction mixture consists mostly of reactants

69
Q

effect of a catalyst on equilibrium

A
  • because the forward and reverse reactions pass through the same transition state, the catalyst lowers the activation energy barrier by the same amount
  • the catalyst accelerates the forward and reverse reactions by the same factor, so the composition of the equilibrium mixture is unchanged
  • Kc is not affected by the catalyst
  • we reach the same equilibrium, only faster
70
Q

Haber proces

A

N2 + 3H2 <-> 2NH3
exothermic

reaction conditions:
- catalyst: iron mixed with metal oxides increases the rate
- temperature: 400-500’C
- pressure: yield improved by running the reaction at high pressures (130-300atm)

71
Q

Arrhenius acid

A

a substance that dissociates in water to produce hydrogen ions, H+

72
Q

Arrhenius base

A

a substance that dissociates in water to produce hydroxide ions, OH-

73
Q

shortfalls of Arrhenius theory

A

works with many compounds in water, but misses some (particularly bases), and clearly doesn’t describe non-aqueous solutions

74
Q

bronsted Lowry acid

A

a substance that can give a hydrogen ion (proton donor)

75
Q

bronsted Lowry base

A

a substance that can take a hydrogen ion (proton acceptor)

76
Q

difference/similarity between BL bases and Arrhenius bases

A

D: B-L bases do not need to contain OH-
S: they can generate OH- when in water

77
Q

general equation for a Bronsted Lowry acid and base interacting

A

HA + B <—> BH+ + A-

78
Q

conjugate acid-base pairs

A

chemical species whose formulas differ by only one H+

79
Q

acid dissociation

A

HA + H2O <–> H3O+ + A-
- an equilibrium
- H2O solvent is an (almost) pure liquid so is not in equilibrium expression
- Ka only describes the reaction of an acid with the solvent H2O as the base
- the stronger the acid, the larger the Ka

80
Q

base dissociation

A

B + H2O <–> OH- + BH+
- stronger the base, the larger the Kb

81
Q

amphiprotic

A

can both donate and accept protons

82
Q

amphoteric

A

can act as both an acid and a base

83
Q

dissociation of water

A

H2O + H2O <–> H3O+ + OH-
Kw = [H3O+][OH-] = 1 x 10^-14

84
Q

pH =

A

-log[H3O+]

85
Q

[H3O+] =

A

10^-pH

86
Q

pOH =

A

-log[OH-]

87
Q

[OH-] =

A

10^-pOH

88
Q

basic solution
neutral solution
acidic solution

A
  • [H3O+]<[OH-], pH > 7
  • [H3O+] = [OH-], pH = 7
  • [H3O+]>[OH-], pH < 7
89
Q

strong acid

A

fully dissociates in water
Ka&raquo_space; 1

90
Q

weak acid

A

partially dissociates in water
Ka = 1

91
Q

inert acid

A

does not dissociate in water

92
Q

conjugates:

A

strong acid <-> ‘inert’ base
weak acid <-> weak base
‘inert’ acid <-> strong base

93
Q

compare strong acids/bases and weak acids/bases as electrolytes

A

strong acids and strong bases are strong electrolytes. we assume they ionise completely in water.

weak acids and weak bases are weak electrolytes. they ionise to a limited but detectable extent in water

94
Q

The levelling effect

A

for a strong acid in water, the ‘active proton donor’ isn’t HA, its H3O+
for a strong base in water, the active proton acceptor is OH-
- different strong acids usually have different Ka’s
- however, in water, they exhibit the similar acidic properties, i.e. the acid strength of H3O+

95
Q

pKa + pKb =

A

14

96
Q

distinguish between pH, pKa and Ka

A

pH measures the acidity of the solution and depends on the absolute [H3O+]

pKa and Ka reflect the strength of the acid molecule and depends on the relative concentration at equilibrium

97
Q

percent dissociation

A

a common and useful description of a weak acid in solution

% dissociation = [HA]dissociated/[HA]initial x 100

98
Q

factors that affect acid strength

A
  1. degree of polarity of H-A bond
  2. strength of H-A bond
  3. oxoacids
99
Q

degree of polarity of H-A bond

A

depends upon the electronegativity of A
the more polar the H-A bond, the stronger the aid

100
Q

strength of H-A bond

A

depends on the size of the A atom. the larger the A atom, the longer/weaker the bond, the stronger the acid

101
Q

along the period, the most important factor is

A

electronegativity

102
Q

down the group, the most important factor is

A

the H-A bond strength

103
Q

oxoacids

A

YOm(HO)n
- eg H3PO4
- if same structure, different Y: acid strength increases as the electronegativity of Y increases
- a more electronegative Y pulls electron density away from O-H bonds, making it easier for H+ to dissociate
- the strength of oxoacids also increases with m, the number of lone oxygen atoms
- the electronegative oxygen atoms pull electron density from the chlorine, making it more positive, which in turn weakens the O-H bond

104
Q

salt solutions from conjugates of strong acids and strong bases are

A

neutral

105
Q

strong acids

A

HClO4, H2SO4, HNO3, HCl, HBr, HI

106
Q

strong bases

A

LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

107
Q

salts that are derived from a weak base and strong acid yield

A

acidic solutions

108
Q

why does weak base + strong acid give acidic solutions?

A

forms weak conjugate acid of a weak base which then dissociates
eg NH4Cl (see slides)
- solutions of NH4Cl are acidic because the ammonium ion (NH4+, a weak acid) will dissociate while the chloride ion does not

109
Q

salts, such as NaCN, that are derived from a strong base (eg NaOH) and a weak acid (eg HCN) yield basic solutions

A

see slides for example

110
Q

Ka>Kb
Ka<Kb

A

solution contains excess H3O+ ions so pH < 7
solution will contain excess OH- ions so pH>7

111
Q

Lewis acid

A

a species that accepts an electron pair

112
Q

Lewis base

A

a species that donates an electron pair

113
Q

hydrated metal cations

A

small, highly-charged metal ions (eg Al3+) form complexes in water
the resulting complexes are proton donors

114
Q

acid rain

A
  • pollutants such as sulfur oxides and nitrogen oxides dissolve in rain to form acids
  • buildings and monuments made of marble and limestone eroded by acid rain
115
Q

state the 2 equations for acid rain

A
116
Q

ions that do not react appreciably with water to produce either H3O+ or OH-

A

conjugate cations from strong bases:
- alkali metal cations (group 1)
- alkaline earth metal cations (group 2)

conjugate anions from strong monoprotic acids
- Cl-, Br-, I-, NO3-, ClO4-

because they are inert

117
Q

common ion effect

A

the shift in the position of an equilibrium upon addition of a substance that provides an ion already involved in that equilibrium

118
Q

buffers

A

solutions that resist changes in pH when limited amounts of acid or base are added

119
Q

where does the buffer’s resistance to change in pH arise from?

A

the presence of appreciable concentrations of weak acids and its conjugate (weak) base

120
Q

addition of OH- to a buffer

A

the acid in the buffer will neutralise added strong base

121
Q

addition of H3O+ to a buffer

A

the base in the buffer will neutralise added strong acid

122
Q

a good buffer contains

A

the conjugate acid and base in similar amounts
- source of protons: eg HA to neutralise incoming bases
- sink of protons : eg A- to neutralise incoming acids

123
Q

Henderson-Hasselbach equation

A

the equation highlights that the pH of a buffer solution has a value close to the pKa of the weak acid

124
Q

how to make a buffer

A
  1. select a weak acid with a pJa similar to desired pH
  2. A) mix equal amounts of acid and its conjugate base or B) start with the weak acid and neutralise half of it with a strong base
  3. adjust to desired pH by adding small amounts of strong acid/base
125
Q

buffer capacity

A

the molar amount of acid or base which the buffer can handle without significant changes in pH

126
Q

why can’t buyers tolerate the addition of infinite amounts of strong acid or base?

A

after enough external acid or base has been added to deplete the base or acid in the system, the buffer is destroyed

127
Q

how would you approach quantitative problems of neutralisation and buffers?

A
  1. figure out which major species remain after any strong acids/bases act
    - assume these reactions go to completion
    - it’s often convenient to work in amount (moles)
  2. determine the concentrations of any minor species that the major species generate via equilibrium reactions
    - work in concentrations (volume is constant)

how?
have both an acid and its conjugate base -> buffer -> HH
have only a weak acid/base -> acid/base dissociation -> ICE
have either H3O+ or OH- directly remaining from step 1 -> done

128
Q

titrant

A

known concentration

129
Q

analyte

A

unknown concentration solution

130
Q

equivalence point

A

point at which stoichiometrically equivalent quantities of acid and base have been mixed together

131
Q

strong acid/strong base titration

A
132
Q

weak acid/strong base

A
133
Q

acid-base indicator

A

a substance that changes colour in a specific pH range. indicators exhibit pH-dependent colour changes because they are weak acids and have different colours in their acid (Hln) and conjugate base (In-) forms

HIn + H2O <–> H3O+ + In-

134
Q

strong base/strong acid

A
135
Q

weak base/strong acid

A
136
Q

solubility

A

amount of solute that dissolves in a given amount of solvent (mol/L)

137
Q

formula for ionic compounds dissociating into ions in solution as they dissolve

A
138
Q

saturated solution

A

we’ve added enough solid so that some remains at the equilibrium, where ions are (re)crystallising and dissolving at the same rates.

139
Q

equilibrium constant (ion-product) expression

A

solubility product, Ksp

140
Q

Ksp

A

a measure of how much of an ionic compound has dissolved at equilibrium
= [M^n+]^m[X^y-]^x

141
Q

Precipitation of ionic compounds

A

AB (s) <–> A+ (aq) + B-(aq)
-> dissolution
<- precipitation

Ksp = [A+]eq[B-]eq
Qsp = [A+]t[B-]t

142
Q

compare Qsp to Ksp

A

Qsp>Ksp: solution is supersaturated and precipitation will occur
Qsp=Ksp: solution is saturated and equilibrium exists already
Qsp<Ksp: solution is unsaturated and dissolution will occur

143
Q

how does pH affect solubility

A

if the compound contains the conjugate anion of a weak acid, the addition of a strong acid will increase solubility

CHM135 (23 decks)

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