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PSU Freshman > Chem 112B Exam 4 > Flashcards

Chem 112B Exam 4 Flashcards

1
Q

A single-step reaction is called an __________ reaction

A

elementary

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2
Q

How to find rate law of a table?

A

Compare the rates of 1-2 experiments to see if rates doubled or halved. The exponent 2 would be on a rate that doubled.

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3
Q

We can directly derive rate laws of elementary reactions from _________

A

stoichiometric coefficients

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4
Q

Intermediates

A

are produced in an early step and consumed in a later step
DO NOT appear in the overall reaction or rate law

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5
Q

Rate-determining step of a multi-step reaction

A

Slowest step (Largest activate energy Ea)

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6
Q

Integrated rate laws describe

A

the reactant concentration as a function of time

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7
Q

0th reaction order integrated rate law

A

Law: [A]t= -kt +[A]0
Linear Plot: [A]t vs. t
Slope: -k
Units of k: M/s

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8
Q

1st reaction order integrated rate law

A

Law: ln([A]t)= -kt +ln([A]0)
Linear Plot: ln([A]t) vs. t
Slope: -k
Units of k:s^-1

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9
Q

2nd reaction order integrated rate law

A

Law:1/[A]t =kt + 1/[A]0
Linear Plot: 1/[A]t vs. t
Slope: k
Units of k:M-1s-1

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10
Q

Half-life remains constant for

A

1st order reactions

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11
Q

Half life increases over time for

A

second order reactions

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12
Q

4 factors that affect reaction rates

A

1) Concentrations/partial pressures of reactants
2)Physical state (phase)
3)Temperature
4)Catalyst presence

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13
Q

Catalysts speed up reactions by

A

lowering activation energy

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14
Q

Catalysts do not affect

A

thermodynamic values like H and E

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15
Q

Catalysts can

A

appear in the rate law and change the reaction mechanism

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16
Q

Catalysts cannot

A

appear in the overall reaction

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17
Q

The active site for an enzyme is based on

A

shape, size, and chemical properties

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18
Q

To find a catalyst,

A

species if consumed on reactants side then reproduced

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19
Q

Phosphorylation

A

The addition of a phosphoryl group onto a chemical structure

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20
Q

Which functional groups are most commonly phosphorylated?

A

Alcohols or amines in amino side chains

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21
Q

Phosphorylation will completely change

A

the shape of a protein, which causes a new function (turn on/off, enzyme location)

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22
Q

ATP provides ____________ in phosphorylation reactions

A

the phosphoryl group

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23
Q

Enzyme catalysis requires

A

enzyme-substrate binding

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24
Q

kcat

A

Rate constant for the conversion of the enzyme/substrate complex to product

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25
Large kcat
Faster conversion of complex to products
26
Small kcat
Slower conversion of complex to products
27
Km tells us
how tightly enzymes bind
28
Large kM
Takes more substrate to reach 1/2 Vmax
29
To maximize catalytic efficiency, we want
large kcat and small kM
30
Chemical equilibrium factors
Concentrations of reactants/products are equal Rates of forward and reverse reactions are =
31
Chemical equilibrium can be reached...
from either direction
32
Equilibrium constant is a
unitless value (Keq)
33
Kc is expressed using
molarity (concentration)
34
Kp is expressed using
partial pressures
35
What does not appear in the equilibrum (Keq) expression?
Liquids and solids
36
Keq=
Kc = Kp = products/reactants
37
In the Kp = Kc (RT)^n equation, what is R?
0.08206 L*atm*mol-1*K-1
38
When delta n =0,
Kp=Kc
39
molarity=
moles/liters
40
Keq >> 1
Products dominate, forward favored
41
Keq = 1
equilibrium
42
Keq << 1
Reactants dominate, reverse favored
43
When a reaction is multiplied by a number
Raise keq to the power of that number
44
When a reaction is reversed
Keq is the inverse of Keq for the forward reaction
45
If a net reaction is the addition of multiple steps
Keq is the product of the Keq values of the individual steps
46
Q < Keq
Too many reactants, so proceed right/forward
47
Q>Keq
Too many products, so proceed left/reverse
48
Q is calculated using __, but Keq must be calculated using ____
(initial) any concentrations, equilibrium concentrations
49
Le'Chatelier's Principle
If a system at equilibrium is disturbed, the system will shift its position to minimize the disturbance
50
Disturbing equilbrium means Q
does not equal Keq
51
What type of changes cause a shift? (Le'Chatelier's)
Change in component concentration Change in volume Change in partial pressures Change in temperature
52
Endothermic reactions have heat as a (take in heat)
reactant
53
Exothermic reactions have heat as a (heat released)
product
54
Endothermic reactions
Melting, Evaporation, Sublimation
55
Exothermic reactions
Freezing, Condensation, Deposition
56
Ligand
Any molecule that binds to protein
57
Binding constant (Kb)
Kb = Kon/Koff
58
Dissociation constant (Kd)
Kd = Koff/Kon = [protein][ligand] / [protein-ligand complex]
59
Large kbinding means
Strong binding
60
Large Kd means
Weak binding
61
Kon
Rate constant for binding 2nd order process Typically 10^7
62
Koff
Rate constant for dissociation 1st order process
63
Large Koff means
Weak binding
64
% bound =
[protein-ligand complex] / [protein] + [protein-ligand complex (PL)]
65
Positive deltaH is
endothermic
66
Negative deltaH is
exothermic
67
If volume of container is increased, the reaction
shifts towards more moles of gas
68
Competitive inhibition
Inhibitor binds to E, but not the E-S complex, and it prevents S binding
69
Noncompetitive inhibitor
Inhibitor binds to E or the E-S complex
70
In comp inhibition, high substrate will cause
equilibrium to shift towards E-S complex
71
In non-comp inhibition, high substrate will cause
equilibrium to shift towards [E-S-I] complex (still inhibited)
72
Competitive inhibitors effect on Vmax and Km
Vmax is not affected Km increases because higher [S] is needed to outcompete the inhibitor achieve 1/2 Vmax
73
Noncompetitive inhibitors effect on Vmax and Km
Vmax decreased since fewer enzyme molecules available Km stays the same bc raising [S] cannot relieve inhibition
74
app = opp =
inhibited
75
Spontaneous process
Capable of proceeding in the forward direction on its own without external intervention
76
If the process is spontaneous in one direction, the reverse direction will be
nonspontaneous
77
A spontaneous process can perform
work
78
Processes that are spontaneous at one temp can be
nonspontaneous at other temps
79
We cannot predict spontaneity strictly with change in enthalpy (deltaH)
Need to take entropy change into account as well (deltaS)
80
Entropy
The degree of spreading and thermal energies within a system
81
Second Law of Thermodynamics
For a process to be spontaneous, deltaS for the universe must always be +
82
Standard molar entropy (S*)
Entropy of 1 mole of substance in standard state (always positive for pure substances and units are J/mol*K)
83
deltaSuniverse < 0
Nonspontaneous in forward direction Spontaneous in reverse direction
84
deltaSuniverse > 0
Spontaneous in forward direction
85
deltaSuniverse = 0
Reversible (at equilibrium) with no preferred direction
86
Entropy is higher when (comparing molar entropy for 2 substances)
temperature is higher Volume is larger or pressure is lower (for gases)
87
Entropy increases when (in a reaction)
Number of moles of gas increase Endothermic change (S->L->G) Solids dissolve Number of independently moving particles increase during a reaction
88
When looking at chemical structures, the highest molar entropy is seen when
No ring No double bonds Easily rotated (single bonds, Hydrocarbon chain)
89
Gibbs free energy
allows us to predict spontaneity of different processes
90
delta G < 0
Spontaneous in forward direction
91
deltaG > 0
Nonspontaneous in forward direction
92
deltaG=0
System is at equilibrium Free energy is minimized
93
spontaneous process can do work while non-spontaneous process
must have work done on them in order to proceed
94
W max = delta G
Maximum amount of work a spontaneous process can perofrm
95
Wmin = delta G
minimum amount of work required to make a nonspontaneous process occur
96
If deltaH is negative, delta S is positive
delta G is always negative Always spontaneous
97
If deltaH is positive, delta S is negative
deltaG is always positive Never spontaneous
98
If deltaH is negative, delta S is negative
deltaG is negative at low T Spontaneous at low temps
99
If deltaH is positive, delta S is positive
deltaG positive at lowT Spontaneous at high temp
100
In the gibbs free energy equation linear, deltaH is and deltaS is
y-intercept, slope
101
DeltaG standard units
kj/mol
102
DeltaH standard units
kj/mol
103
DeltaS standard units
J/(mol*K)
104
Crossover temperature
Can be calculated at phase changes using the Gibbs Free energy equation (crossing over from no spontaneity to spontaneity or other way around)
105
Crossover temperature
T= deltaH standard / deltaS standard
106
From kJ to joules
Multiply by 1000
107
deltaG formation is
the free energy change in forming one mole of a compound from elements in their standard states
108
delta G < 0 Q < Keq
Spontaneous in forward direction
109
delta G > 0 Q > Keq
Spontaneous in reverse direction
110
delta Gstandard < 0 Keq > 1
Products predominate
111
delta Gstandard > 0 Keq < 1
Reactants predominate
112
Perturbations that cause a reaction to shift right
Decrease Q deltaG negative Spontaneous in forward direction
113
Perturbations that cause a reaction to shift left
Increase Q deltaG positive Reaction spontaneous in reverse direction
114
Enzymes drive unfavorable reactions by
coupling them with favorable reactions
115
deltaG standard of a reaction is the sum of the
deltaGrxn for each component step
116

PSU Freshman (12 decks)

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