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
2
Q
rate =
A
change in concentration/change in time
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)
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
5
Q
describe the rate law in words
A
an equation that shows how the reaction rate depends on the concentration of each reactant
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
7
Q
what si the overall order of a reaction?
A
the sum of the individual orders
8
Q
units of rate
A
always M/s
9
Q
units of k
A
depends on the rate law expression
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)
11
Q
how are the order of reactions (m, n) and the rate constant (k) determined?
A
experimentally
12
Q
give 2 ways of determining order of reaction and rate constant
A
- 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 - graphical method/integrated rate law
- monitor the course of a reaction over time
- plot data
- the shape of the curve reveals reaction order
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)
14
Q
half life
A
the time required for reactant concentration to reach half of its original value
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)
16
Q
how are 0th order reactions possible?
A
in reactions whether the kinetics are governed by the availability of a catalyst
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)
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
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
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
21
Q
draw graphs for zeroth order, first order, and second order. also draw straight-line plots used to determine rate constant
A
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
23
Q
reaction intermediate
A
a species formed in one step of a reaction mechanism and consumed in a later step
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
25
unimolecular reaction
an elementary reaction that involves a single reactant molecule
eg rate = k[O3]
26
bimolecular reaction
an elementary reaction that results from an energetic collision of two reactant molecules
eg rate = k[O3][O]
27
termolecular reaction
an elementary reaction that involves three atoms or molecules (rare)
eg rate = k[O]^2[M]
28
how can you find a plausible mechanism for a reaction?
- elementary steps must add up to the overall reaction
- mechanism must correlate with the experimental rate law
29
the rate law of the overall reaction must be the rate law of
the rate limiting step
30
how do you determine the rate law for a reaction where the rate limiting step is the second step?
31
factors affecting reaction rates
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
32
collision theory
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
33
k =
Zpf
Z = volumetric collision frequency
p = fraction with correct orientation
f = fraction with sufficient energy
34
k depends strongly on temperature - so which of Z, p, and f depend on temperature?
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
35
transition state
the unstable group of atoms that are the highest-energy species along the pathway from reactants to products
36
activation energy
the minimum energy required for a successful reaction
37
f =
e^(-Ea/RT)
38
as T increases, how does f change?
increases exponentially
39
draw a graph of collision energy vs fraction of collisions for two temperatures
40
Arrhenius equation
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
41
Arrhenius plot
y intercept = ln(A)
slope = -Ea/R
42
catalysts
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
43
Michaelis-Menten kinetics
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
44
define chemical equilibrium
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
45
Kc =
aA + bB <--> cC + dD
Kc = [C]^c[D]^d/[A]^a[B]^b
concentrations at equilibrium
46
Kc when you reverse the reaction
1/Kc
47
Kc when you multiply the reaction by 2
Kc^2
48
Kc when you add two reactions
Kc1 x Kc2
49
units of the equilibrium constant
equilibrium constants are defined as unites quantities
50
homogeneous equilibria
all species are present in the same phase.
51
heterogeneous equilibria
system's state of equilibrium contains components from multiple phases
52
why are pure solids and liquids excluded from equilibrium expressions?
they are always in their standard state, where a=1: their concentrations don't change
53
equilibrium constant Kp
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
54
Kc > 10^3
reaction proceeds nearly to completion, products favoured
55
Kc < 10^-3
reaction proceeds hardly at all, reactants favoured
56
10^-3
appreciable concentrations of both reactants and products
57
reaction quotient, Qc
defined similarly to an equilibrium constant Kc except that the []'s in Qc can have ant values, not necessarily equilibrium values
58
if Qc = Acc
no net reaction occurs, already at equilibrium
59
if Qc < Kc
net reaction goes from left to right
60
if Qc > Kc
net reaction goes from right to left
61
le Chatelier's principle
if a stress is applied to a reaction mixture at equilibrium, a net reaction then occurs in the direction that relieves the stress
62
in general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Chateliers principle predicts that
- 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
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
- 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
adding inert gas
no effect, because the partial pressures (molar concentrations) don't change
65
temperature can alter the equilibrium concentrations, but for a different reason:
it changes the value of Kc
66
in general, when an equilibrium is disturbed by a change in temperature,
Kc for an exothermic reaction (-ΔH) decreases as the temperature increases
Kc for an endothermic reaction (+ΔH) increases as the temperature increases
67
linking chemical equilibrium and kinetics
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
if Kc is really large
if Kc is nearly unity
if Kc is really small
Kf >> Kr and the reaction goes almost to completion
Kf = Kr and both reactants and products exist at equilibrium
Kr >> Kf and the reaction mixture consists mostly of reactants
69
effect of a catalyst on equilibrium
- 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
Haber proces
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
Arrhenius acid
a substance that dissociates in water to produce hydrogen ions, H+
72
Arrhenius base
a substance that dissociates in water to produce hydroxide ions, OH-
73
shortfalls of Arrhenius theory
works with many compounds in water, but misses some (particularly bases), and clearly doesn't describe non-aqueous solutions
74
bronsted Lowry acid
a substance that can give a hydrogen ion (proton donor)
75
bronsted Lowry base
a substance that can take a hydrogen ion (proton acceptor)
76
difference/similarity between BL bases and Arrhenius bases
D: B-L bases do not need to contain OH-
S: they can generate OH- when in water
77
general equation for a Bronsted Lowry acid and base interacting
HA + B <---> BH+ + A-
78
conjugate acid-base pairs
chemical species whose formulas differ by only one H+
79
acid dissociation
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
base dissociation
B + H2O <--> OH- + BH+
- stronger the base, the larger the Kb
81
amphiprotic
can both donate and accept protons
82
amphoteric
can act as both an acid and a base
83
dissociation of water
H2O + H2O <--> H3O+ + OH-
Kw = [H3O+][OH-] = 1 x 10^-14
84
pH =
-log[H3O+]
85
[H3O+] =
10^-pH
86
pOH =
-log[OH-]
87
[OH-] =
10^-pOH
88
basic solution
neutral solution
acidic solution
- [H3O+]<[OH-], pH > 7
- [H3O+] = [OH-], pH = 7
- [H3O+]>[OH-], pH < 7
89
strong acid
fully dissociates in water
Ka >> 1
90
weak acid
partially dissociates in water
Ka = 1
91
inert acid
does not dissociate in water
92
conjugates:
strong acid <-> 'inert' base
weak acid <-> weak base
'inert' acid <-> strong base
93
compare strong acids/bases and weak acids/bases as electrolytes
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
The levelling effect
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
pKa + pKb =
14
96
distinguish between pH, pKa and Ka
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
percent dissociation
a common and useful description of a weak acid in solution
% dissociation = [HA]dissociated/[HA]initial x 100
98
factors that affect acid strength
1. degree of polarity of H-A bond
2. strength of H-A bond
3. oxoacids
99
degree of polarity of H-A bond
depends upon the electronegativity of A
the more polar the H-A bond, the stronger the aid
100
strength of H-A bond
depends on the size of the A atom. the larger the A atom, the longer/weaker the bond, the stronger the acid
101
along the period, the most important factor is
electronegativity
102
down the group, the most important factor is
the H-A bond strength
103
oxoacids
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
salt solutions from conjugates of strong acids and strong bases are
neutral
105
strong acids
HClO4, H2SO4, HNO3, HCl, HBr, HI
106
strong bases
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
107
salts that are derived from a weak base and strong acid yield
acidic solutions
108
why does weak base + strong acid give acidic solutions?
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
salts, such as NaCN, that are derived from a strong base (eg NaOH) and a weak acid (eg HCN) yield basic solutions
see slides for example
110
Ka>Kb
Ka
solution contains excess H3O+ ions so pH < 7
solution will contain excess OH- ions so pH>7
111
Lewis acid
a species that accepts an electron pair
112
Lewis base
a species that donates an electron pair
113
hydrated metal cations
small, highly-charged metal ions (eg Al3+) form complexes in water
the resulting complexes are proton donors
114
acid rain
- 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
state the 2 equations for acid rain
116
ions that do not react appreciably with water to produce either H3O+ or OH-
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
common ion effect
the shift in the position of an equilibrium upon addition of a substance that provides an ion already involved in that equilibrium
118
buffers
solutions that resist changes in pH when limited amounts of acid or base are added
119
where does the buffer's resistance to change in pH arise from?
the presence of appreciable concentrations of weak acids and its conjugate (weak) base
120
addition of OH- to a buffer
the acid in the buffer will neutralise added strong base
121
addition of H3O+ to a buffer
the base in the buffer will neutralise added strong acid
122
a good buffer contains
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
Henderson-Hasselbach equation
the equation highlights that the pH of a buffer solution has a value close to the pKa of the weak acid
124
how to make a buffer
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
buffer capacity
the molar amount of acid or base which the buffer can handle without significant changes in pH
126
why can't buyers tolerate the addition of infinite amounts of strong acid or base?
after enough external acid or base has been added to deplete the base or acid in the system, the buffer is destroyed
127
how would you approach quantitative problems of neutralisation and buffers?
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
titrant
known concentration
129
analyte
unknown concentration solution
130
equivalence point
point at which stoichiometrically equivalent quantities of acid and base have been mixed together
131
strong acid/strong base titration
132
weak acid/strong base
133
acid-base indicator
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
strong base/strong acid
135
weak base/strong acid
136
solubility
amount of solute that dissolves in a given amount of solvent (mol/L)
137
formula for ionic compounds dissociating into ions in solution as they dissolve
138
saturated solution
we've added enough solid so that some remains at the equilibrium, where ions are (re)crystallising and dissolving at the same rates.
139
equilibrium constant (ion-product) expression
solubility product, Ksp
140
Ksp
a measure of how much of an ionic compound has dissolved at equilibrium
= [M^n+]^m[X^y-]^x
141
Precipitation of ionic compounds
AB (s) <--> A+ (aq) + B-(aq)
-> dissolution
<- precipitation
Ksp = [A+]eq[B-]eq
Qsp = [A+]t[B-]t
142
compare Qsp to Ksp
Qsp>Ksp: solution is supersaturated and precipitation will occur
Qsp=Ksp: solution is saturated and equilibrium exists already
Qsp
143
how does pH affect solubility
if the compound contains the conjugate anion of a weak acid, the addition of a strong acid will increase solubility
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