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Question 1

what do we have when we have a consecutive reaction

Answer 1

we have an intermediate

Question 2

when we have consecutive reactions what do we normally assume

Answer 2

we assume theres a pre equilibrium to simplify our model of kinetics

Question 3

okay lets say we have an overall equation of:
2Br- + H2O2 + 2H+ —> Br2 + 2H2O

and we find the rate using isolation method or initial rates and the rate equation is

rate. = k [Br-][H2O2][H+]

what do we need to think of

Answer 3

if the rate is all to the order of 1’s for each conc

there needs to be 1 of each moelcule involved at some stage

Question 4

we get a proposed model of the reaction and we have

H+ + H2O2 -><- H2O+-OH
where –> is k1 and <– is k(-1)

H2O2+-OH + Br –> HOBr + H2O
where —> is k2

then u have
HOBr + H+ + Br —> Br2 + H2O
where –> is k3

the 2nd equation is the slowest,, what does this mean

Answer 4

we care about the 1st and 2nd reaction

not the 3rd bc the 3rd is after the RDS so its irrelavent.

we look at the overall equation to see which product we have

then we take the conc of that product and do rate=d[pro]/dt
where its gonna be positive bc its the conc of the products

the rate will also equal what

rate = k2 [H2O2+-OH][Br] bc these are the reactants of the RDS!!!!!!!!! and this is what dictates the rate of the reactio n

Question 5

when we have a rate constant of reactions and they wanna know what the rate of a reaction is ,, whatdo we do

Answer 5

we see what k is the smallest as this will be the RDS

then we take the reactants of this and we multiply it by k ,, aka rate consatnt of the RDs

then we put it to the order of how many of them there are.

so if theres only 1 stoichiometrically,, its to the power of 1.

Question 6

in an elementary step what does the rate depend on

Answer 6

the rate dependwson

[reactants]k

where k is rate constant

Question 7

when the RDS reactant is an intermediate what happens

Answer 7

we cant measure the conc of this

we we need it in the form of smt we can take the conc of

so we assume a pre equilibrium and we look at the equation where the ‘equilibrium’ arrows are

to define a Keq aka an equilibrium constant

Question 8

whats the Keq of a reaction

Answer 8

[products] /// [reactants]

Question 9

if Keq = [H2O+-OH]/ [H+][H2O2]

and we dont know the conc of H2O2+-OH but we kinda need it for an equation,, what do we do

Answer 9

we rearrange so that we get

[h2o2+-oh] = Keq x [H][h2o2]

then we can sub this into the

k2 [ h2o2+-oh][Br] equation to find rate

now the equation incolves concentrations we know we can measure

therefore the proposed model fits with experiemnetla kinetics ]

but that doesnt mean the model is correct.

Question 10

what can we use if theres no pre equilibrium

Answer 10

we take steady state equilibrium aka SSA

Question 11

what is steady state equilibrium all about

Answer 11

its about how the intermediate conc is 0

bc its made and used up at the same rate

so d(intermediate) / dt = 0

we also assume the conc of the intermediate is much smaller than reactants

Question 12

descobe the graph we use to represent SSA

Answer 12

we have conc against time

conc of reactant decreases

conc of product increases

conc of intermediate increases,, stays constant and then dips back down

so it looks like a little lid of a cardboard box

bc d(intermediate)\ dt = 0

this is the linderman mechanism

Question 13

okay so lets say we have a problem that asks us why does the order of a reaction change from 1st order to 2nd order when u decrease pressure

when the overall reaction is:
a —-> p
and the mechanism is:
a + a –> <– a* + a
a* —> p

where the arrows are k1, k-1, k2

Answer 13

rate of reaction is obvs the rate of change of product conc and it will be positive

rate = d[p]/dt
= rate = k2 [a*] bc its the reactant

but we cant take the coc of a* bc its an excited state

so we need to switch up
so we find rate of a*

rate = d[a]/dt
= k1[a]^2 - k-1[a][a]-k2[a*]

bc we need to think about what reactions make and destroy a*. we add the ones that make it and we subtract the ones that destroy it.

but obvs we assume d[a*]/dt is 0

so that all equals 0

then u rearrange to find {a*] so u can use the answer to sub in stuff u can find the conc of

and then u put them into the equation and cancel out any stuff.
remmeber that when we say the pressure of a is large that means ur forming it,, meaninf that k value is larger

and when we say the pressure of that species is low,, we mean that the k value is now smaller

so we can choose to ignore the small ones bc they dont make much diff to the equation. then we cancel out and we should see how at less pressure we have ^2 and how at more pressure we have ^1