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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
Q

Large kcat

A

Faster conversion of complex to products

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26
Q

Small kcat

A

Slower conversion of complex to products

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27
Q

Km tells us

A

how tightly enzymes bind

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28
Q

Large kM

A

Takes more substrate to reach 1/2 Vmax

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29
Q

To maximize catalytic efficiency, we want

A

large kcat and small kM

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30
Q

Chemical equilibrium factors

A

Concentrations of reactants/products are equal
Rates of forward and reverse reactions are =

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31
Q

Chemical equilibrium can be reached…

A

from either direction

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32
Q

Equilibrium constant is a

A

unitless value (Keq)

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33
Q

Kc is expressed using

A

molarity (concentration)

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34
Q

Kp is expressed using

A

partial pressures

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35
Q

What does not appear in the equilibrum (Keq) expression?

A

Liquids and solids

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36
Q

Keq=

A

Kc = Kp = products/reactants

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37
Q

In the Kp = Kc (RT)^n equation, what is R?

A

0.08206 Latmmol-1*K-1

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38
Q

When delta n =0,

A

Kp=Kc

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39
Q

molarity=

A

moles/liters

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40
Q

Keq&raquo_space; 1

A

Products dominate, forward favored

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41
Q

Keq = 1

A

equilibrium

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42
Q

Keq &laquo_space;1

A

Reactants dominate, reverse favored

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43
Q

When a reaction is multiplied by a number

A

Raise keq to the power of that number

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44
Q

When a reaction is reversed

A

Keq is the inverse of Keq for the forward reaction

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45
Q

If a net reaction is the addition of multiple steps

A

Keq is the product of the Keq values of the individual steps

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46
Q

Q < Keq

A

Too many reactants, so proceed right/forward

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47
Q

Q>Keq

A

Too many products, so proceed left/reverse

48
Q

Q is calculated using __, but Keq must be calculated using ____

A

(initial) any concentrations, equilibrium concentrations

49
Q

Le’Chatelier’s Principle

A

If a system at equilibrium is disturbed, the system will shift its position to minimize the disturbance

50
Q

Disturbing equilbrium means Q

A

does not equal Keq

51
Q

What type of changes cause a shift? (Le’Chatelier’s)

A

Change in component concentration
Change in volume
Change in partial pressures
Change in temperature

52
Q

Endothermic reactions have heat as a (take in heat)

A

reactant

53
Q

Exothermic reactions have heat as a (heat released)

A

product

54
Q

Endothermic reactions

A

Melting, Evaporation, Sublimation

55
Q

Exothermic reactions

A

Freezing, Condensation, Deposition

56
Q

Ligand

A

Any molecule that binds to protein

57
Q

Binding constant (Kb)

A

Kb = Kon/Koff

58
Q

Dissociation constant (Kd)

A

Kd = Koff/Kon = [protein][ligand] / [protein-ligand complex]

59
Q

Large kbinding means

A

Strong binding

60
Q

Large Kd means

A

Weak binding

61
Q

Kon

A

Rate constant for binding
2nd order process
Typically 10^7

62
Q

Koff

A

Rate constant for dissociation
1st order process

63
Q

Large Koff means

A

Weak binding

64
Q

% bound =

A

[protein-ligand complex] / [protein] + [protein-ligand complex (PL)]

65
Q

Positive deltaH is

A

endothermic

66
Q

Negative deltaH is

A

exothermic

67
Q

If volume of container is increased, the reaction

A

shifts towards more moles of gas

68
Q

Competitive inhibition

A

Inhibitor binds to E, but not the E-S complex, and it prevents S binding

69
Q

Noncompetitive inhibitor

A

Inhibitor binds to E or the E-S complex

70
Q

In comp inhibition, high substrate will cause

A

equilibrium to shift towards E-S complex

71
Q

In non-comp inhibition, high substrate will cause

A

equilibrium to shift towards [E-S-I] complex (still inhibited)

72
Q

Competitive inhibitors effect on Vmax and Km

A

Vmax is not affected
Km increases because higher [S] is needed to outcompete the inhibitor achieve 1/2 Vmax

73
Q

Noncompetitive inhibitors effect on Vmax and Km

A

Vmax decreased since fewer enzyme molecules available
Km stays the same bc raising [S] cannot relieve inhibition

74
Q

app = opp =

A

inhibited

75
Q

Spontaneous process

A

Capable of proceeding in the forward direction on its own without external intervention

76
Q

If the process is spontaneous in one direction, the reverse direction will be

A

nonspontaneous

77
Q

A spontaneous process can perform

A

work

78
Q

Processes that are spontaneous at one temp can be

A

nonspontaneous at other temps

79
Q

We cannot predict spontaneity strictly with change in enthalpy (deltaH)

A

Need to take entropy change into account as well (deltaS)

80
Q

Entropy

A

The degree of spreading and thermal energies within a system

81
Q

Second Law of Thermodynamics

A

For a process to be spontaneous, deltaS for the universe must always be +

82
Q

Standard molar entropy (S*)

A

Entropy of 1 mole of substance in standard state (always positive for pure substances and units are J/mol*K)

83
Q

deltaSuniverse < 0

A

Nonspontaneous in forward direction
Spontaneous in reverse direction

84
Q

deltaSuniverse > 0

A

Spontaneous in forward direction

85
Q

deltaSuniverse = 0

A

Reversible (at equilibrium) with no preferred direction

86
Q

Entropy is higher when (comparing molar entropy for 2 substances)

A

temperature is higher
Volume is larger or pressure is lower (for gases)

87
Q

Entropy increases when (in a reaction)

A

Number of moles of gas increase
Endothermic change (S->L->G)
Solids dissolve
Number of independently moving particles increase during a reaction

88
Q

When looking at chemical structures, the highest molar entropy is seen when

A

No ring
No double bonds
Easily rotated (single bonds, Hydrocarbon chain)

89
Q

Gibbs free energy

A

allows us to predict spontaneity of different processes

90
Q

delta G < 0

A

Spontaneous in forward direction

91
Q

deltaG > 0

A

Nonspontaneous in forward direction

92
Q

deltaG=0

A

System is at equilibrium
Free energy is minimized

93
Q

spontaneous process can do work while non-spontaneous process

A

must have work done on them in order to proceed

94
Q

W max = delta G

A

Maximum amount of work a spontaneous process can perofrm

95
Q

Wmin = delta G

A

minimum amount of work required to make a nonspontaneous process occur

96
Q

If deltaH is negative, delta S is positive

A

delta G is always negative
Always spontaneous

97
Q

If deltaH is positive, delta S is negative

A

deltaG is always positive
Never spontaneous

98
Q

If deltaH is negative, delta S is negative

A

deltaG is negative at low T
Spontaneous at low temps

99
Q

If deltaH is positive, delta S is positive

A

deltaG positive at lowT
Spontaneous at high temp

100
Q

In the gibbs free energy equation linear, deltaH is
and deltaS is

A

y-intercept, slope

101
Q

DeltaG standard units

A

kj/mol

102
Q

DeltaH standard units

A

kj/mol

103
Q

DeltaS standard units

A

J/(mol*K)

104
Q

Crossover temperature

A

Can be calculated at phase changes using the Gibbs Free energy equation (crossing over from no spontaneity to spontaneity or other way around)

105
Q

Crossover temperature

A

T= deltaH standard / deltaS standard

106
Q

From kJ to joules

A

Multiply by 1000

107
Q

deltaG formation is

A

the free energy change in forming one mole of a compound from elements in their standard states

108
Q

delta G < 0
Q < Keq

A

Spontaneous in forward direction

109
Q

delta G > 0
Q > Keq

A

Spontaneous in reverse direction

110
Q

delta Gstandard < 0
Keq > 1

A

Products predominate

111
Q

delta Gstandard > 0
Keq < 1

A

Reactants predominate

112
Q

Perturbations that cause a reaction to shift right

A

Decrease Q
deltaG negative
Spontaneous in forward direction

113
Q

Perturbations that cause a reaction to shift left

A

Increase Q
deltaG positive
Reaction spontaneous in reverse direction

114
Q

Enzymes drive unfavorable reactions by

A

coupling them with favorable reactions

115
Q

deltaG standard of a reaction is the sum of the

A

deltaGrxn for each component step

116
Q
A

PSU Freshman (12 decks)

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