3f Flashcards
get ur pen and paper girl bc we’re doing practice questions
yay
when a questions says its an aqueous conditions what does that mean
the conc of water remains constant
bc ur in water
so even if the conc changes by a tiny amount,, its not gonna make a noticable difference
okay so the overall equation is:
Co(CN)5 OH2^-2 + I- —-> 6(CN)5 I^3- + H2O
the proposed model is:
1. Co(CN)5 OH2^2- k1-> Co(CN)5^2- + H2O
- Co(CN)5^2- + H2O k-1-> Co(CN)5 OH2^2-
- Co(CN)5^2- + I- k2–> Co(CN)5 I^3-
write an equation for a rate we can acc find using experimental conditons and explain why we cant use the first rate equation.
rate = d[Co(CN)5[I^3-] /dT
or
rate = k [Co(CN)5 ^2-] [I-]
but the Co is an intermediate so we cant measure its concentration,, so we need to find an equation that is going to work out for us.
so we apply the steady state concentration which tells us that the rate of change for the conc of the intermediate is constant,, aka 0
we write that as: d[Co(CN)5 ^2-] / dt = 0
and then we need to look back at the proposed model and write down everything that makes it,, and subtract everything that destroys it.
so
d[Co(CN)5 ^2-] /dt = k1[Co(CN)5OH2^2-] - k-1[Co(CN)5^2-][H2O] -[Co(CN)5^2-][I-]
but the conc of water remains constant so we can just ignore that bit.
and then all of that equals 0 bc steady state approx.
and then we try so isolate the conc of the intermediate on one side which gives us
[Co(CN)5^2-] = k1(Co(CN)5 OH^2-) / (k-1 + k2 [I-])
then we sub this into the rate equation we made at first,, the one that describes the rate of the product based on k and the conc of reactants
and we get,, rate =
k2 k1 (Co(CN)5 OH^2- ) ( I-) // K-1 + K2[I-]
we cant cancel out [I] bc its part of an addition in the denominator.
then for fun if we use the linderman approach we can see how the rate equation changes when we get diff conc of stuff
aka the high and low conc for [I] and how if its high conc we can forget k-1 and then cancel out the [I] but if its low [I] then u only have k-1 as the denominatorrrrr
okay example 4:
overall equation: Cl2 = CO -> COCl2
model:
Cl2 <–> 2Cl
Cl + Cl2 –> Cl3
Cl3 + CO –> COCl2 + 1/2Cl2
find the rate of the formation of COCl2
note down anything u realise and u go through it
okay so rate of COCl2 =
d[COCl2]/ dt = k3[Cl3][CO]
but CL3 is an intermediate meaning we cant find its conc so we need to figure out something else
we yse steady state approx.
so d[Cl3]/dt = 0
so we need to see what makes it and what destroys it
0 = k2[Cl][Cl2] - k3[Cl3][CO]
k2[cl][cl2] = k3[cl3][co]
[cl3] = k2[cl][cl2]/k3[co]
then put this back into the equation and cancel what u can cancel out to give u rate = k2[cl][cl2] but if we look close we have cl as one of the concentrations howver cl is also an intermediate bc u form 2Cl and one of these gets used up,, and we cant use steady state approx for this bc ur going from 2Cl to 1Cl
so we use the psuedo equilibrium reaction : and bc its an equilibrium it needs an equilibrium constant which we can make
Keq = [Cl]^2 / [Cl2]
so [Cl] = root (Keq)[Cl2] (as in theyre multiplied together) and then u sub this into the other equation we made up top.
so give rate = k2(root.ke[Cl2]) [Cl2]
and then when u have the square root of 2 things multiplied together u can do X^1/2 multiplied by Y^1/2
so u can get
rate = k2 Ke^1/2 [Cl2]^3/2
will he tell us whether to use steady state or pre equilibrium
yesss hell tell us which ones to use
example 5
overall equation:
2NO + O2 —> 2NO2
model:
2NO <–> N2O2
N2O2 + O2 –> 2NO2
find the rate of [NO2]
d[NO2] /dt = 2k2 [N2O2][O2]
u do 2k2 bc u get 2NO2
but N2O2 is an intermediate so we need to find the rate equation for this and then to sub it in
and then phil said to use pre equilibrium
so we find the equilibrium constant for the N2O2
Keq = [N2O2] / [NO]^2
and so that means [N2O2] = Keq [NO]^2
then put this into the older equation for d[NO2]
2k2 Keq [NO]^2[O2]
precursor kinetics etc etc etc
