Chem 112B Exam 4 Flashcards
A single-step reaction is called an __________ reaction
elementary
How to find rate law of a table?
Compare the rates of 1-2 experiments to see if rates doubled or halved. The exponent 2 would be on a rate that doubled.
We can directly derive rate laws of elementary reactions from _________
stoichiometric coefficients
Intermediates
are produced in an early step and consumed in a later step
DO NOT appear in the overall reaction or rate law
Rate-determining step of a multi-step reaction
Slowest step (Largest activate energy Ea)
Integrated rate laws describe
the reactant concentration as a function of time
0th reaction order integrated rate law
Law: [A]t= -kt +[A]0
Linear Plot: [A]t vs. t
Slope: -k
Units of k: M/s
1st reaction order integrated rate law
Law: ln([A]t)= -kt +ln([A]0)
Linear Plot: ln([A]t) vs. t
Slope: -k
Units of k:s^-1
2nd reaction order integrated rate law
Law:1/[A]t =kt + 1/[A]0
Linear Plot: 1/[A]t vs. t
Slope: k
Units of k:M-1s-1
Half-life remains constant for
1st order reactions
Half life increases over time for
second order reactions
4 factors that affect reaction rates
1) Concentrations/partial pressures of reactants
2)Physical state (phase)
3)Temperature
4)Catalyst presence
Catalysts speed up reactions by
lowering activation energy
Catalysts do not affect
thermodynamic values like H and E
Catalysts can
appear in the rate law and change the reaction mechanism
Catalysts cannot
appear in the overall reaction
The active site for an enzyme is based on
shape, size, and chemical properties
To find a catalyst,
species if consumed on reactants side then reproduced
Phosphorylation
The addition of a phosphoryl group onto a chemical structure
Which functional groups are most commonly phosphorylated?
Alcohols or amines in amino side chains
Phosphorylation will completely change
the shape of a protein, which causes a new function (turn on/off, enzyme location)
ATP provides ____________ in phosphorylation reactions
the phosphoryl group
Enzyme catalysis requires
enzyme-substrate binding
kcat
Rate constant for the conversion of the enzyme/substrate complex to product
Large kcat
Faster conversion of complex to products
Small kcat
Slower conversion of complex to products
Km tells us
how tightly enzymes bind
Large kM
Takes more substrate to reach 1/2 Vmax
To maximize catalytic efficiency, we want
large kcat and small kM
Chemical equilibrium factors
Concentrations of reactants/products are equal
Rates of forward and reverse reactions are =
Chemical equilibrium can be reached…
from either direction
Equilibrium constant is a
unitless value (Keq)
Kc is expressed using
molarity (concentration)
Kp is expressed using
partial pressures
What does not appear in the equilibrum (Keq) expression?
Liquids and solids
Keq=
Kc = Kp = products/reactants
In the Kp = Kc (RT)^n equation, what is R?
0.08206 Latmmol-1*K-1
When delta n =0,
Kp=Kc
molarity=
moles/liters
Keq»_space; 1
Products dominate, forward favored
Keq = 1
equilibrium
Keq «_space;1
Reactants dominate, reverse favored
When a reaction is multiplied by a number
Raise keq to the power of that number
When a reaction is reversed
Keq is the inverse of Keq for the forward reaction
If a net reaction is the addition of multiple steps
Keq is the product of the Keq values of the individual steps
Q < Keq
Too many reactants, so proceed right/forward
Q>Keq
Too many products, so proceed left/reverse
Q is calculated using __, but Keq must be calculated using ____
(initial) any concentrations, equilibrium concentrations
Le’Chatelier’s Principle
If a system at equilibrium is disturbed, the system will shift its position to minimize the disturbance
Disturbing equilbrium means Q
does not equal Keq
What type of changes cause a shift? (Le’Chatelier’s)
Change in component concentration
Change in volume
Change in partial pressures
Change in temperature
Endothermic reactions have heat as a (take in heat)
reactant
Exothermic reactions have heat as a (heat released)
product
Endothermic reactions
Melting, Evaporation, Sublimation
Exothermic reactions
Freezing, Condensation, Deposition
Ligand
Any molecule that binds to protein
Binding constant (Kb)
Kb = Kon/Koff
Dissociation constant (Kd)
Kd = Koff/Kon = [protein][ligand] / [protein-ligand complex]
Large kbinding means
Strong binding
Large Kd means
Weak binding
Kon
Rate constant for binding
2nd order process
Typically 10^7
Koff
Rate constant for dissociation
1st order process
Large Koff means
Weak binding
% bound =
[protein-ligand complex] / [protein] + [protein-ligand complex (PL)]
Positive deltaH is
endothermic
Negative deltaH is
exothermic
If volume of container is increased, the reaction
shifts towards more moles of gas
Competitive inhibition
Inhibitor binds to E, but not the E-S complex, and it prevents S binding
Noncompetitive inhibitor
Inhibitor binds to E or the E-S complex
In comp inhibition, high substrate will cause
equilibrium to shift towards E-S complex
In non-comp inhibition, high substrate will cause
equilibrium to shift towards [E-S-I] complex (still inhibited)
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
Noncompetitive inhibitors effect on Vmax and Km
Vmax decreased since fewer enzyme molecules available
Km stays the same bc raising [S] cannot relieve inhibition
app = opp =
inhibited
Spontaneous process
Capable of proceeding in the forward direction on its own without external intervention
If the process is spontaneous in one direction, the reverse direction will be
nonspontaneous
A spontaneous process can perform
work
Processes that are spontaneous at one temp can be
nonspontaneous at other temps
We cannot predict spontaneity strictly with change in enthalpy (deltaH)
Need to take entropy change into account as well (deltaS)
Entropy
The degree of spreading and thermal energies within a system
Second Law of Thermodynamics
For a process to be spontaneous, deltaS for the universe must always be +
Standard molar entropy (S*)
Entropy of 1 mole of substance in standard state (always positive for pure substances and units are J/mol*K)
deltaSuniverse < 0
Nonspontaneous in forward direction
Spontaneous in reverse direction
deltaSuniverse > 0
Spontaneous in forward direction
deltaSuniverse = 0
Reversible (at equilibrium) with no preferred direction
Entropy is higher when (comparing molar entropy for 2 substances)
temperature is higher
Volume is larger or pressure is lower (for gases)
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
When looking at chemical structures, the highest molar entropy is seen when
No ring
No double bonds
Easily rotated (single bonds, Hydrocarbon chain)
Gibbs free energy
allows us to predict spontaneity of different processes
delta G < 0
Spontaneous in forward direction
deltaG > 0
Nonspontaneous in forward direction
deltaG=0
System is at equilibrium
Free energy is minimized
spontaneous process can do work while non-spontaneous process
must have work done on them in order to proceed
W max = delta G
Maximum amount of work a spontaneous process can perofrm
Wmin = delta G
minimum amount of work required to make a nonspontaneous process occur
If deltaH is negative, delta S is positive
delta G is always negative
Always spontaneous
If deltaH is positive, delta S is negative
deltaG is always positive
Never spontaneous
If deltaH is negative, delta S is negative
deltaG is negative at low T
Spontaneous at low temps
If deltaH is positive, delta S is positive
deltaG positive at lowT
Spontaneous at high temp
In the gibbs free energy equation linear, deltaH is
and deltaS is
y-intercept, slope
DeltaG standard units
kj/mol
DeltaH standard units
kj/mol
DeltaS standard units
J/(mol*K)
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)
Crossover temperature
T= deltaH standard / deltaS standard
From kJ to joules
Multiply by 1000
deltaG formation is
the free energy change in forming one mole of a compound from elements in their standard states
delta G < 0
Q < Keq
Spontaneous in forward direction
delta G > 0
Q > Keq
Spontaneous in reverse direction
delta Gstandard < 0
Keq > 1
Products predominate
delta Gstandard > 0
Keq < 1
Reactants predominate
Perturbations that cause a reaction to shift right
Decrease Q
deltaG negative
Spontaneous in forward direction
Perturbations that cause a reaction to shift left
Increase Q
deltaG positive
Reaction spontaneous in reverse direction
Enzymes drive unfavorable reactions by
coupling them with favorable reactions
deltaG standard of a reaction is the sum of the
deltaGrxn for each component step