Gen chem 6 Flashcards
what is the bicarbonate buffer equation
what principle is taking place in this equation
CO2 (g) + H2O <> H2CO3 (aq) <>H+ (aq) + HCO3- (aq)
Le Chatelier principle(review this concept in study videos)
in regards to the bicarbonate buffer equation, what is the resp system trying to compensate for? what two things happen to cause an equilibrium shift and what direction is the shift? How do hydrogen and bicarbonate ions and CO2 gas play a role here, and how is the H+ conc affected, and how does this affect pH
the resp system is trying to compensate for the metab acidosis; breathing rate goes up so more CO2 can be blown off, which causes the equilibrium to shift to the left; hydrogen ions combine with bicarbonate ions to make carbonic acid, which turns into CO2 gas that is expelled from the lungs; this makes the plasma H+ conc go low, which stabilizes the pH and keeps it from going too low.
Name three characteristics of an irreversible reaction
why don’t reversible reactions go to completion?
when does dynamic equilibrium occur
the reaction goes in one direction only, goes to completion, and the max amt of product that can be formed is det’d by the limiting reagent
reversible reactions don’t go to completion because the products can react together to reform the reactants
dynamic equilibrium occurs when the forward and reverse reactions are equal
for a reversible reaction at a given temp, when will the reaction reach equilibrium
it will reach equilibrium when the entropy of the system is at a max and the gibbs free energy of the system is at a minimum
for the generic reaction aA +bB<> cC +dD what does the law of mass action state?
the law of mass action for a generic reaction says that if a system is at equilibrium at a constant temperature, then the ratio is constant:
Keq= [C]^c[D]^d/[A]^a[B]^b
for the following reaction: 2A <> B + C, what are the rates of the forward and reverse reactions
because the reaction happens in one step, the rates of the forward and reverse reactions are given by
ratef= kf [A]^2 and rater=kr [B][C]
because the rates are equal, what can we do for the rate expressions
because kf and kr are both constants, what new constant can we define(also how can we define it for gases)
what is the ratio of kf to kr?
because the rates are equal, we can set the rate expressions for the forward and reverse reactions equal to each other
kf[A]^2=kf[B][C] > kf/kr= [B][C]/[A]^2
because kf and kr are both constants, we can define a new constant kc, where kc is called the equilibrium constant and the subscript c stands for concentration; for gases its kp where p is pressure
the rats of the forward and reverse reaction rate constants are not usually equal to each other, because the concs of the reactants and prods are usually not equal; the ratio of kf to kr is kc–> kc=keq=kf/kr
how do you find the equilibrium constant for a reaction when it occurs in more than one step
you find this by multiplying the equilibrium constants for each step of the reaction; then divide the concs of the prods by the concs of the reactants, with each conc raised to the stoichiometric coefficient for the respective species
how are the forward and reverse rate constants for the nth step designated
So for example, taking the previous answer into account, for this reaction aA + bB<>cC + dD which happens in 3 steps, each with a forward and reverse rate, then what is kc
these rates are designated kn and k-n
kc= k1k2k3/k-1k-2k-3 = [C]^c[D]^d/[A]^a[B]^b
in equilibrium expressions what can we say about the exponents? in rate laws?
they are equal to the coefficients in the balanced equation; in rate laws the exponents must be det’d experimentally and often DONT equal the stoichiometric coefficients
what is the formula for the reaction ? how does it compare to keq?
Qc= [C]^c[D]^d/[A]^a[B]^b; it looks identical to the keq equation
what 3 things can we say about a reaction when the Q < Keq?
the reaction hasnt reached equiibrium yet; there is a greater conc of reactants and smaller conc of prods than at equilibrium;; the forward rate of reaction is increased to restore equilibrium
what 3 things can we say about a reaction when Q=Keq?
the reactino is in dynamic equilibrium; the reactants and prods are present in equilibrium proportions; forward and reverse rates of rxn are equal
what 3 things can we say about a reaction when Q>Keq?
the forward reaction has EXCEEDED equilibrium; there is a greater conc of prods and smaller conc of reactants than at equilibrium; the reverse rate of rx is increased to restore equilibrium
For a reaction in which QKeq, G>0, what can be said about the dir of the reaction
for QKeq, G>0, the rxn proceeds in the reverse direction
once a rxn has reached equilibrium, any further movement in any direction will be what
it will be nonspon
does the conc of pure liquids and solids appear in the equilibrium constant expression? why or why not?
whats the difference between activities and concs
they dont because its based on the activities of compounds not concs; the activities of pure liquids and solids is 1
there is a negligible difference between conc and activity, for MCAT purposes
Keq is a quality of a reaction at a given what?
if Keq is large, what does this say about the equilibrium position
what is the equil constant for rxn that happens in one direction? reverse direction?
@ a given temp; equilibrium constant is temp-dependent
the larger the Keq, the farther to the right the equilibrium position will be
the equil constant for a rxn that happens in one direction is Keq, for the reverse direction is 1/Keq
how are the concs of prods and reactants related in a Keq formula?
if conc of prods is greater than the conc of reactants, what does this say about Keq? vice versa?
what other value can Keq appear as? describe the conditions of the exponents?
Keq=[prods]/[reactants]
if the conc of the prods is greater than the conc of the reactants than Keq is a top heavy fraction and must be greater than 1; vice versa, Keq becomes a bottom heavy fraction and must be less than 1; Keq can also appear as a single value using exponents»>
A reaction that strongly favors prods will have a large, positive exponent, and the larger the reactant, the less reactant that will be present at equil.
what does a large positive exponent indicate about a rxn? large negative exponent?
it indicates that the reaction has almost gone to completion
it indicates that a reaction strongly favors reactants at equil(only a little reactant is converted to product)
in this example consider the reaction, A<> B+C with Keq= 10^-12 and a starting conc of [A]=1M, what does Keq look like? if x amount of [A] has reacted, how much [C] and [B] will be produced at equil and what will the equil conc be?
what doe it look when these value are plugged into the Keq expression? solving this equation would be difficult so what can we do to help? give a specific example, like what if Keq was 10^-12?
Keq= [C][B]/[A]
a amount of C and x amount of B have been produced at equil, and the equil conc of A will be [1-x] M.
Keq=10^-12 = (x)(x)/(1-x)=x^2/1-x
[DONT USE ANY ICEBOX ON THE TEST BUT PRACTICE IT THOUGH :d]
to help with the above hard equation, we will instead acknowledge that the value for Keq is a large negative exponent so we use the “x is negligible” shortcut.
if Keq was 10^-12, we can assume x is negligible and round the denom to the starting conc, so eqn looks like this» 10^-12= x^2/1, with 1 being the starting conc
what does le chatelier’s princple say about stress? what happens to the equil? what are two reasons that a rxn can be thrown out of equiL?
it says that stress is applied to a system then shifts to relieve that stress
regardless of the form that stress takes, the reaction is temporarily moved out of equil»two reasons a rxn can leave equil is becuz: 1)the concs or partial pressures of the system are no longer in the equilibrium ratio OR because 2) the equilibrium ratio itself has changed as a result of a change in the temp of the system
what is the relation btween the Keq and the Qc when you change a conc of a reactant or product?
regarding le chatelier’s principle, what dir will the system always react in?
when you change the conc of a reactant or product Qc doesn’t equal the Keq
the system will always react in the direction away from the s=added species or toward the removed species
when a system is compressed what happens to its volume and pressure?
according to the ideal gas law, what kind of relationship does pressure have with volume?
according to le chatelier’s principle, what happens when you increase the pressure of a system?
the volume goes down and the pressure goes up
the two have a direct relationship according to the ideal gas law
when you increase the pressure in the system, it will respond by decreasing the total number of gas moles, thereby BRINGING DOWN the PRESSURE.
watch the le chateliers principle youtube video again
what changes when you raise the temp of a rxn, Keq or Qc
in regards to le chatelier’s principle, if a reaction is endothermic(delta H>0), how does heat function? if exothermic, how does heat function?
Keq changes
Heat acts like a reactant; acts like a product
REMEMBER: WHATEVER CHANGE YOU MAKE TO THE SYSTEM, THE SYSTEM IS GUNNA WANNA UNDO THAT CHANGE)
At lower temps with smaller heat transfer, what type of product is formed?
At higher temps with a larger heat transfer, what type of product is formed?
which pathway requires more energy, thermo or kinetic? and which prods form faster? what product has more free energy?
a kinetic product; a thermodynamic product
the free energy that must be added is higher for the thermo pathway than for the kinetic pathway
kinetic prods form faster and are called fast products
the kinetic prod has more free energy
Take the example of 2-methylcyclohexanone. in the thermo pathway, there is a double bond between the C-1 and the methyl group. why does it require more energy to form the transition state of this reaction? how does the base reach the carbon with the methyl group?
whay is the product of this reaction more stable and less likely to react than the kinetic product?
this reaction requires more energy to form transition state of this reaction becuz the base must overcome the steric hindrance created by the methyl group
the base squeezes in to reach the carbon with the methyl group
the product from this reaction is more stable and less likely to react any further becuz the double bond is more substituted than the one in the other pathway.
Take the example of 2-methylcyclohexanone. in the kinetic pathway, there is a double bond between C-1 and C-6. Why is this pathway preferred when little heat is available?
why can the base more easily reach C-6 and to do what? and what forms?
why is this product not as stable as the thermo product? becuz its less stable, what can happen to it?
this pathway is preferred when little heat is available becuz less energy is needed to reach the transition state
the base can more easily reach C-6 for the purpose of removing a proton and the resulting enolate can form.
this product is not as stable becuz it has a less sub’d double bond; becuz of this lack of stability, this product(ring) can be attacked.
