Hi! This is a question regarding the option Industrial Chemistry.
For the Saponification dot point, do I have to memorise the entire structure of a specific fat/oil, or would it be sufficient to put an R at the end of each carbon chain?
They won't expect you to memorise a specific triglyceride, and will give you the formula if they ask for a specific fat. Otherwise, they may ask you to draw the general equation for saponification, which you would put the "R" like what Zainbow said.
I was just wondering if I could have the screenshotted notes explained to me?
I'm not too sure how the hydronium ion is working as an oxidant (nor do I now what that means)
Also how are the reactive metals being oxidised by the sulphuric acid?
Like i'm really not understanding the vocabulary and so am feeling really lost when looking at the given equation and trying to make sense of it all...
An oxidant is a species that causes oxidation of another but gets reduced itself. If you look on the Data Sheet, the H
+ (same as hydronium) has a value of 0.00V. It will oxidise any metal above it from Pb to K.
This is because it must be a spontaneous reaction with a positive Eo value.
For example, let's oxidise zinc:
To oxidise Zn into Zn
2+, it is 0.76V.
To reduce H
+, it is 0.00V.
Remember back in galvanic cells, you add these two together to get 0.76V total, and it is spontaneous. If it were negative, the metal will be unreactive to the hydrogen ions under the same conditions.
The H
+ will get reduced as a result of oxidising a metal by gaining its electrons. Hence why acid + metal releases hydrogen gas, because the hydrogen ions get reduced into hydrogen gas.
Dilute sulfuric acid oxidises metal because it provides the hydorgen ions.
a) Why does temperature only affect the equilibrium constant? If it's a measure of concentration, then why don't changes in ion concentration (which is the most obvious in my mind) and changes in pressure/volume change K? Because when these factors are changed, don't they change the position of the equilibrium and so the relative concentrations of the products/reactants?
I was once confused by this too. Remember that the equilibrium constant is the RATIO of products to reactants. Once you add more moles, the equlibrium will shift, yes. BUT, the equilibrium will shift back into that special ratio, hence the K constant remains unchanged.
Temperature changes only affect the equilibrium because it moves the whole system to one side and keeps it there as long as that temperature is held constant. Once the system stays on one side, the entire RATIO changes because the products may have permanently increased or decreased depending on the heat of reaction.
b) I've just read that an explanation for the exothermic nature of sulphuric acid's ionisation is because the energy released by the bond forming is greater than the total energy required to ionise the molecule and to overcome the hydrogen bonding that holds the water molecules together.
What bond formation is being referred to? How is it different to the ionisation being referred to (and the hydrogen bonding for that matter - why is it even relevant?)
The dilution of sulfuric acid is actually a two-step reaction.
}\rightarrow\ce{H+(aq)}+\ce{HSO4^{-}(aq)})
}+\ce{H2O(l)}\rightarrow\ce{H3O+(aq)})
The dissociation is endothermic, absorbs a bit of heat because bonds are broken.
The actual hydronium ion forming is exothermic, releases much more heat as bonds are formed. Hence, ionisation is exothermic because it produces hydronium ions (that's the relevant part).
c) Lastly, I would like some clarification for these notes:
Are these just general rules/blanket statements that apply to all equilibriums? I've never heard of them before and can't really understand them. Could someone possibly illustrate the point with an example?
When H2O is added the concentration of all species decreases. So the reactants are favoured to increase the concentration of dissolved ions
When H2O is removed, concentration of all species increases. So the products are favoured to minimise total amount of dissolved ions in the system.
This isn't something I've seen, but I wouldn't use it as a general rule. Equilibriums can be liquid or gaseous. A gaseous equilibrium, unless steam is involved, would not change if water/steam were added. If you added water to a liquid equilibrium, that doesn't have water involvement, it wouldn't have much of an effect
An example of an equilibrium that will be affected is the dilution of weak acids:
}+\ce{H2O(l)}⇌\ce{H3O+(aq)}+\ce{CH3COO^{-}(aq)})
Adding water to this will shift the equilibrium to the right.
Hope this helps
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