colloid practicequestions Flashcards
(70 cards)
Describe how consideration of the pairwise interaction energies in a liquid can
explain the phenomenon of surface tension
- surface tension occurs due to the surface particles lacking neighbours o n the air side so the bulk pulls the surface particles closer to itself creating surface tension. increasing thr pairwise interaction enrgy which is the enrgy that neighbouring particles experience, the surface particles would be pulled in lcoser to the bulk therefore increasing the surface tension.
this is seen in the equation: ST = Zb Nb W.bb
What happens to the surface tension of a surfactant solution as the
concentration of surfactant increases? Explain why these changes occur.
- surfactant is added to the surface and reduces surface tensiondue to disruption strong interactions such as hydrogen bonding!
- further increasing conc of sf continues to decrease the surface tension.
- the cmc is then reached where the surface is saturated with surfactant and the sf starts to form micelles in the bulk.
once micelles form,, the st doesnt decreease asny more!! as sf forms micelles and not adsorb onto surface
What are the different thermodynamic contributions to surfactant micellisation?
gibbs is negative aka the reaction is spontaneous.
enthalpy is usually small but positive as ur not making or breaking bonds,,its all about intermolecular interactions.
entropy is negative due to the freeing of water which has to surround the individual surfactant molecules and once micelles form these are released into the bulk increasing its entropy!!! also the hydrophobic tails are able to freely move due to interacting less with water when in micelle form….
Explain how the closed association model and the phase seperation models
treat a surfactant solution from two different thermodynamic perspectives. What
common equation is obtained for the free energy of micellisation?
closed association states that there is an equailirbium between micelles and starting surfactant monomer solution!! normally given by nK1 <–> Kn
as there cant be an equilibrium constant for each sf turnig into a micelle.
the phase separation approach talks about how they (micelles and surfactant monomer solution) are separate phases and the micelles only form once the cmc has been reached.
they both say that G =RT ln(k) //cmc
Explain why the surface tension of the solution is around
40 mN m-1 whereas the surface tension of pure water is
approximately 72 mN m-1.
surfactant adsorbs onto the interface an changes its composition.
this reduces the h bonding of the pure water surface and adds vdw character,, which are lower in energy than hydrogen bonds therefore reducing the strength of surface tension.
u need to break imf in order to make surface.
when were given cmc what do we need to make sure we do
we need to make sure we find the mole fraction
turn mM into M by /1000.
then do y/y+55.5
describe what a microemulsion looks like
see through
transparent bc particles are too small to reflect light!!!
Describe the arrangement of molecules in the microemulsion.
Using the modified Gibbs-Helmholtz equation as the basis of
your answer, explain why this arrangement is the most
thermodynamically favourable.
- microemulsion : particles are too small to reflext light
delta H is small bc no bonds being broken or formed
entropy is large bc ur going from bulk oil to tiny oil droplets
change in A is also very large bc the small particles have a larger surface area than bulk oil
surface excess is low bc all surfactant adsorbed onto surface of the interface
delta G overall is negative, making the microemulsion formation spontaneous and favoured
For a hydrocarbon surfactant with cmc = 8 x 10-3 mol dm-3,
estimate the surface tension of a solution of this surfactant at the cmc
hydrocarbon surfactant has a surface tension of around 30/40 mNm-1
so say a value between these.
n an oil-in-water (o/w) microemulsion system, the distribution of small
droplets of oil throughout the whole volume of the system is
thermodynamically more favourable than separation of the system
into bulk oil and water phases with surfactant dissolved in both.
Use the contribution of each term in the Gibbs-Helmholtz equation to
explain why microemulsions are thermodynamically stable, but
emulsions are not.
microemulsions: more favoured aka more stabe more g being negative.
delta H is small for both bc no. bonds are being broken // formed.
delta A for microemulsion is lrger bc smaller and more particles have a larger surface area than just the bulk.
the entropy of microemulsion is also much larger bc u have a lot of particles spread out,, aka the entropy is much larger. CANCELS OUT THE LARGE CHANGE IN AREA
the surface tension of microemulsion with surfactant is smaller bc less H bonds and more VDW tsaking their place. which recuces the surface tension. MAKES THE LARGE CHANGE IN AREA A SMALELR VALUE
Explain how the following equation leads to a formal definition of
surface tension: dG = dA
shows that surface tension is the energy needed to increase / change the area of the surface by a small amount.
yhh bc surface tension is the energy required to chnage the area of the surface
describe by ethnol has a larger surface tension than a fluorinated moelcule
bc ethanol has OH groups which can form hydrogen bonds with the solvent and neighbouring molecules which increase the surface tension.
the fluorinated molecule only has CF bonds which cant form hydrogen bonds only VDW which are weaker,, therefore the surface tension is lower.
whats surface excess
the amount of something that has adsobed onto the surface of the interface per unit area compared to what would be preset if all the bulk decided to become surface particles.
For a given surfactant solution, = 3.2 x 10-5 mol m-2.
i) Explain where this equation applies, and what it means
this is the surface ewxcess and is seen i n’x’ equation.
it tells u the amoun =t of smt that adsorbs onto the surface of the gas liquid interface compared to if all thr bulk partivles became surface particles.
a larger surface excess means a weaker surface tension
Calculate the interfacial area occupied by an individual surfactant
molecule at the specified value of surface excess.
if surface excess value = 3.2x10^-5 molm^-2
surface excess x avogadro
= 1.92704x10^19
then do 1/ans to get it in m2.
and we get 5.19x10^-20 m^2 as our answer.
Microemulsions have an extremely high interfacial area between the two
phases. Explain how this property contributes to their spontaneous
formation when the appropriate combination of components are
combined
- high interfacial area = lots of area for surfactants to adsorb onto = prevents particles from agglomerating into bulk = keeps entropy large and positive,, aiding in keeping G negative!!
- surfactants adsorbing = lower surface tension bc less H bonds and more lower energy VDW. makign it easier to for mmore surface.
-
In a microemulsion, effectively all of the surfactant resides at the
interface between the phases present. Describe how this is different to
an aqueous surfactant solution at concentrations above and below the
cmc. [3]
okay so in a microemulsion,, basically all the surfactant is on the interface,, preventing the particles from agglomerating into bulk which is why microemulsions are so stable dispersion.
in the other one, below the cmc the surfactant will either adsirb onto the interface or be dissolve as a monomer in the bulk.
above the cmc,, the surfactant will no longer adsorb onto the interface but will form micelles in the solution!!!
when we analyse graphs for physisorption for N2O,, where is there 2 N peaks and why do they have different binding energy
bc the 2 N are in different chemical environments.
aka one is closer to OP and one is further. the middle N also has a + charge on itslef
the larger binding energy peak is due to the more positive middle N.
if were looking at physiosorption peaks and theres a diff in binding enrgy from a low to high temp,, what is this due to
this is due to the change in chemival environment when we go from 80k to 280k.
aka change in chemical environement when the reaction is heated.
on the physiosorbption xray electron thing,, how can u draw equations from it:
at 80k u have 2 N peaks and 1 O peak for N2O
at 280k u have 0 N peaks and 1O peak but its got a higher binding energy
- okay so the 80K one makes sense,, bc N2O has 2 nitrogen environments,, bc u have N2O,, the middle N is positive.
when u increase to 280k,, u lose the N2 peaks - we assume they have desorbed.
at 280k the O peak also increases in binding energy, this is bc its been chemiosorbed bc chemisorption has stronger energy bc ur acc forming bonds,, not just VDW. thats how we know.
okay so at 80k we have
N2O(g)—> N2O(a)
at 280k we have
N2O(a) –> N2(g) + O(a). bc O stays adsorbed bc of the peak and N peaks are gone meaning theyre no longer asorbed.
when we have the xray graph type thing what does a peak mean
a peak means that smt is adorbed there!!! diff peaks can be due to different chemical environments meaning they will adsorb with diferent binding energy,, the more positive it is,, the more binding energy it will have.
in the xps thing,, when a peak changes binding enery at a higher temp,, what does this mean
it means that its been chemisoorbed,, or that theres a shift in chemical environment
in the xps thing,, what does it mean when the peaks disappear at a larger temperature
it means that its been desorbed back into a gas.
when they give u pressure and volume and ask u to find some volume of the monomlayer using the langmuir isotherm, what new equation that we just learnt do we use and how do we figure out how to use it
P/V = 1/KVm + P / Vm
do this for each set of info
then do the biggest LHS - LHS
then do the same thing to the rhs.
then u have a simple triangle to find Vm!!
