4.4 How far? — entropy
Posted by: intechemistry on: September 23, 2010
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35 Responses to "4.4 How far? — entropy"
November 4, 2010 at 3:19 pm
Waalaikumassalam wbt.
If exo, delta(H)/T will be positive. If more exo, then delta(H)/T will be more positive.
Because deltaS(tot) = deltaS(sys) – (deltaH)/T
then if (deltaH)/T becomes more positive, when we minus that bigger positive number {as the formula above tells us to do} then deltaS(tot) becomes more negative. Hence less feasible, i.e. the reaction, X(s) —> X(aq), is less ‘capable’ of happening. It doesn’t progress towards the products. For species with an even larger deltaH, the reaction is even less likely to progress
I’m assuming there isn’t a significant difference in deltaS(sys) between the different species
Does that help?
January 23, 2011 at 9:52 pm
salam sir….
may i ask, which one has the greater entropy? fluorine gas or bromine gas?
April 18, 2011 at 9:28 pm
Salam sir:) May I know what does univalent ion means? I came across this when answering Q13(d) in GF about entropy, in which it says in the answer: Since the reaction [NH4NO3 (s) + aq --> NH4+ (aq) + NO3-] involves univalent ions dissolving, so the water is only slightly ordered. So, what does univalent means? Tq in advance:)
April 19, 2011 at 9:56 am
Univalent in this context means ions with a single charge.
Dissolving of a divalent ion e.g. the Ca in CaCl2 may release more species (3, one Ca2+ and 2 Cl-) into the solution and be more disordered
May 11, 2011 at 8:23 pm
Salam sir…
[Co(H2O)6]2+ + (EDTA)4- ——> [Co(EDTA)]2- + 6H2O is thermodynamically feasible.
Can I say that this is because the degree of complexity of [Co(EDTA)]2- is much more complex than that of [Co(H2O)6]2+, so the absolute entropy [Co(edta)]2- will be more +ve than [Co(H2O)]2+?
Or should I say that this is because the number of moles of reactants involved is only 2 moles, much more smaller than that of products formed which is 7 moles. So the entropy of the system will be large and positive.
This is if I am to predict whether the reaction is feasible or not without the use of data booklet.
Thank you.
May 12, 2011 at 1:03 am
@6, amirul arif, May 11, 2011 at 8:23 pm
What you must say really depends on the Q asked. Sometimes they impose directions, or give you clues, so that you answer talks about some specific reason or part of the reaction.
I wouldn’t be happy with your first answer, but much more happy with your second 
What say in the first answer is right, [Co(edta)]2- will be more complex however as it does contain more atoms and more electrons, but both complexes are quite ‘complex’ already so the individual difference in complexity between them is probably not such a big factor.
I prefer your second answer. As the total number of molecules in the products is significantly greater than the total number of molecules in the reactants, going from reacts -> prods, entropy will increase as more molecules become present, hence increasing the feasibility of the process. That is a quick intelligent guess at the answer without the data book, but I don’t think the data book will give values for complex ions, so I doubt you will have to use the data book for that anyway.
(Note we are assuming delta H is negligible)
June 4, 2011 at 1:00 pm
Hi sir,
There was some confusion over which definition to use for enthalpy of solution and hydration. So, i’d just like to confirm, which one should we stick to for the next 3 weeks?
June 6, 2011 at 5:57 am
Hi Xin Ling.
The answer was given here:
http://intechemistry.wordpress.com/2011/03/28/more-questions-to-edexcel/
Use the ‘to infinite dilution‘ definition
which is equivalent to: water is added to one mole of the compound (or gaseous ions for hydration enthalpy) until no further heat changes occur.
June 6, 2011 at 9:31 pm
Go back to the old definitions? alright, noted and thanks!:)
June 12, 2011 at 1:16 pm
Yeah. Golden Oldies (like me :p)
Rod Beavon said so and Rod’s da boss.
June 15, 2011 at 12:26 am
Sir, again, sorry for confirming at this hour. But i figure i should ask anyway 
The absolute entropy given in the data booklet (for elements; pgs 2-4), its for single atoms.
So, if we have O2(g), we have to multiply the S value by 2 i.e. 2(+102.5 J/K/mol).
If we have 2 O2(g), we have to multiply the S value by 4 i.e. 4(+102.5 J/K/mol).
For organic compounds (pg20), its stated that the absolute S for diatomic gaseous elements is for 1/2 of the element. There are no such gases (eg O2) in this particular table, aren’t they?
As for the rest, regardless if its delta H or absolute S values (organic & inorganic compounds), just multiply the number of moles accordingly as stated in the equation.
Thanks,
Xin Ling
June 15, 2011 at 3:01 am
Yes. This is one of the BEWARE points about the data book.
You would indeed have to multiply the ‘atom entropy’ by 2, if you were talking about 1 O2(g).
Yes, for 2 O2(g), you would need to multiply the +102.5 by 4 as 2 O2 has 4 oxygen atoms in it.
You are right. P20-23 has no diatomic gaseous elements in it.
“As for the rest, regardless if its delta H or absolute S values (organic & inorganic compounds), just multiply the number of moles accordingly as stated in the equation.” – Yes. The deltaH values given are all molar quantities, so must multiply up the moles there. The entropy values you would also multiply up the numbers based on the number of moles and for diatomic elements, multiply up by the number of atoms of that elements as in the examples above.
You’re OK with with everything here.
July 19, 2011 at 10:09 pm
Salam sir, I’m from group ALM11 and I would like to ask a question related to entropy..
I know that the standard enthalpy of formation of an element is zero, but why it is not applicable for standard entropy values of the same element? And can you explain on how we can obtain the standard entropy values of substances? I just don’t understand where they get the values from…
I’m looking forward to your reply, sir…
July 24, 2011 at 2:33 am
Salam Shahrizan.
Entropy for a species, elements included, are calculated using the Boltzmann (entropy) equation. S = k ln(W)
k = Boltzmann constant, 1.3807*10^(-23)
W is the number of ways of distributing the energy inside that species. The ‘Pearson’ book tells you how to calculate W
i.e. W= [ (N+q-1)! divided by (N - 1)! q! ]
where
N = No of species (we deal with mole quantities at A-level)
q = # of quanta to be arranged amongst those species.
! = factorial function (e.g. 4! = 4x3x2x1)
That’s how we get S values. Once we have S values, we can get delta(S) values by calculating the difference between entropy of products and reactants: delta(S)system = (sum of entropies of products) – (sum of entropies of reactants)
Why aren’t enetopies of an element zero? Well, they would be if we measured the entropy at 0 Kelvin and the element was in a perfect crystal. But the entropy values we use (see your data book) are (usually) standard entropies i.e. at 25oC, i.e. 298 K, a temperature at which all elements have some thermal vibration i.e. heat energy within itself which can be distributed in a number of ways, so the entropy could not be zero.
Standard enthalpy of formation for elements is zero because of the definition. The definition declared: The enthalpy change when 1 mole of a compound is formed from its elements in their standard states at 298K and 1 atm. If we wanted to form oxygen, the starting material would be oxygen, and hence there will no enthalpy change in going from O2(g) –> O2(g) is zero.
The standard enthalpy is not defined that way.
Does that help?
July 25, 2011 at 12:26 am
Ooh, now i understand the difference..
Thank you sir, for this brilliant explanation…(^_^)
August 14, 2011 at 11:33 am
Greetings Sir. Doubts on Enthalpy changes and dissolving,
1) Which definition is preferably better for ‘standard enthalpy of hydration’?
-Enthalpy change when the moles of a gaseous ion is added to water to infinite dilution’ OR ………..
-Enthalpy change when 1 mole of gaseous ions is added to EXCESS water .
and can you explain a bit more on the significance of the term ‘excess’ water in the second version of the def.? is it referrin to as ‘infinite dil.”?
2) Why and how does the degree of covalency in ionic bonding affects the lattice energy of a compound ? ..and it will also be indeed helpful if I can get a complete definition for enthalpy of lattice energy here;
Thank you in advance Sir.;)
August 15, 2011 at 9:27 am
1) Actually if you put this one, you should be OK…
“One mole of a gaseous ion is added to water to infinite dilution’
I know it’s a bit different from what I said in class, but that’s what I’d now like to see. It’s more clear.
Excess water is a bit vague, I mean excess of what? hence it’s dangerous to use. Stick with the infinite dilution thingy (or until there is no further heat change)
2) The more covlanet, the more exo the lattice enthalpy. The LE assumes a 100% ionic model. so if you have covalent bonding there, the effect is equivalent to “more bonding”.
Definition: “Lattice enthalpy when one mole of an ionic solid is formed from it’s constituent gaseous ions”
You may need to add the bit about standard conditions if the Q is talking about std. conds.
February 5, 2012 at 3:59 pm
salam sir,
In GF A2 pg.40, one of the factor affecting lattice energy is the extent of covalency..what does it mean by ‘extent of covalency’ and how it affects the solubility..
February 5, 2012 at 4:23 pm
Salam.
Ionic bonds are treated as though the TOTAL transfer of electrons has occurred. Occasionally some species don’t totally give away their electrons. Why? because ‘when’ these species give away their electrons, they become +’ve, and positive species attract electrons, so the electrons would be attracted to go back to where thy came from!
To ‘get around this problem’ the electrons make a partial return to the species which just ‘lost’ them. Perhaps 5% of the electron density comes back or 7% or 14% etc… the result is that there is a little bit of sharing going on, and are familiar with the concept of electron sharing when with regards to covalent bonds..
So the degree or amount to which they do this ‘partial return’ happens is called the extent of covalency.
[Please note: species like NaBr are still treated as ionic but are said to have partial covalent character.]
If you confiscate your little sisters ice lolly, she threatens to tell your mum. You give her a little back to keep her quiet {p.s. please don’t do this}.
The grater the extent of covalency, the less polar the bond. An ionic bond being the most extreme extent of a polar bond. It will usually cause a decrease in the solubility. The silver halides are a great example of this. AgF, AgCl, AgBr and AgI
Going down the series, the anion is more polarisable. The cation distorts the anions electron cloud toward itself increasing the extent of covalency (or degree of covenant character) in the molecule hence going down the series the solubility decreases. This manifests itself in a difference in values between the theoretical lattice energy (assumes 100% ionic compound) and the experimentally derived lattice enthalpy (where the extent of covalency produces a different value).
February 5, 2012 at 6:35 pm
ok..what i’ve understood from your explanation,substances that are ionic are very soluble in water as they are polar. AgI has the least solubility in water as iodide ion is bigger in size(as going down the group,the atomic radius increase) which cause iodide ion become more polarisable compared to fluoride,chloride and bromide ion. Thus, AgI has the covalent characteristic as the anion cloud is being distorted by cation. Is it true sir? Can i conclude that when ionic substances has the covalent characteristic, they become less soluble in water as the bond become less polar?
February 17, 2012 at 10:01 pm
greetings …
Sir ,can you pls provide a brief explanation as to why when the total entropy change is negative , the reaction is said to favor reactants ??
10q
February 18, 2012 at 12:16 am
delta(S) = R lnK
If delta(S) is negative, K will be very small, K = prods / reacts
therefore if K is small, eqm lies in favour of the reactants.
brief enough? :p
February 18, 2012 at 12:26 am
Errrrm…
3 | ridhwan, January 23, 2011 at 9:52 pm
You need to specify a temp. If at room temp, then Br2 is a liquid.
Data book gives standard entropies:
Cl2(g) = 85.2 x 2 = 170.4 J K-1 mol-1
Br2(l) = 174.9 x 2 = 349.8 J K-1 mol-1
Unusually (relative to what we frequently see in A-level class), the liquid has higher entropy. Such “unusual” cases like this are unlikely to appear on exams, unless for example they tell you what the situation is and you have suggest why. Thy will usually give you two compounds with pretty similar number of electrons so you can apply a simple rule of ‘gases have higher entropy’.
V.v.v.v.v. sorry for not spotting your question, and hence taking over a year to see it. I know it’s too late now, but your Q may help others.
EDIT: Oops, you said fluorine gas.
Data book gives standard entropies:
F2(g) = 79 x 2 = 158 J K-1 mol-1, as eppected smallr than Cl2 (less e- and also gaseous)
March 4, 2012 at 6:35 pm
Hi sir, which one has higher entropy, aqueous state or liquid state? Thank you sir.
March 4, 2012 at 6:40 pm
Shalom Kosher.
It depends. If you have similar sized molecules (same # of electrons) then the liquid state is by far the most likely to have the greatest entropy..
March 4, 2012 at 11:43 pm
”2 | intechemistry
November 4, 2010 at 3:19 pm
Waalaikumassalam wbt.
If exo, delta(H)/T will be positive. If more exo, then delta(H)/T will be more positive.
Because deltaS(tot) = deltaS(sys) – (deltaH)/T
then if (deltaH)/T becomes more positive, when we minus that bigger positive number {as the formula above tells us to do} then deltaS(tot) becomes more negative. Hence less feasible, i.e. the reaction, X(s) —> X(aq), is less ‘capable’ of happening. It doesn’t progress towards the products. For species with an even larger deltaH, the reaction is even less likely to progress”
Sir I read this a few times but I couldn’t get a hang of it. (i noticed this problem when i read the textbook too.)firstly, isnt exo the deltaH/T will be negative?! only when we minus it (in the equation) it will then become a positive. So it would be like – (-H/T) = +H/T then the delta S total will be more positive. thus it should be feasible and bla blaa.
delta H(solution) is always negative when the solid is soluble, isn’t it?However the text says ‘eq driven to the left if H(soln) is exo, i THINNNKK it should be endo,no?
(my reasoning brain was about to explode after typing out all these. ARGH) thanks to clear things up.
March 5, 2012 at 2:39 pm
I think there’s an error somewhere in the above. It has to do with the minus sign. I think it’s been used twice by the time we use the expression deltaS(sys) – (deltaH)/T
Let’s go through it…
Using the diagram at the top of the page, we can see it’s minus d(H)/T which becomes more +’ve as the enthalpy become more exo.
And because
deltaS(tot) = deltaS(sys) + deltaS(surr), and also deltaS(surr) = -d(H)/T
then
deltaS(tot) = deltaS(sys) + { -d(H)/T }
so if -d(H)/T becomes more +’ve then deltaS(tot) becomes more positive.
This is in line with the simplification we learned, which was usually exothermic reactions are favorable and endo ones are not.
It’s good that you didn’t get it – you could not accept the erroneous info.
~ ~ ~ ~ ~ ~
delta H(solution) is always negative when the solid is soluble, isn’t it? Often, but not always. H(solution) for NaCl(s) at 25oC is about +4 kJ mol-1 and NaCl(s) is soluble at that temp. The deltaS(sys) is very positive (the system experiences a big increase in entropy) so deltaS(surr) can be a bit negative (meaning d(H) can be positive) and still deltaS(tot) can still be +’ve overall, hence spontaneous, hence soluble.
If reaction was MORE endo, then equilibrium would indeed be driven to the left as you correctly said, hence less soluble
and if reaction was MORE exo then equilibrium would be driven to the right (more soluble)
Well done for your perseverance in getting through all of your question
March 5, 2012 at 10:50 pm
thanks! so there is error in the text? since it says reaction more exo then equilibrium driven to the left, more endo then driven to the right. This statement only applicable to a few exceptions like NaCl right?
March 6, 2012 at 1:19 am
Hummmm. I have found a copy of G.F. A2.
If you are talking about the same thing that I’m looking at i.e. Summary at end of page 69, then actually the text is correct.
Because the summary is actually talking about the effect of changing temperature and keeping the magnitude of delta(H) fixed, but considering whether delta(H) is exo or endo, whereas I was taking about the effect of varying delta(H) values and fixed T.
So was it P69 (and 68) you were referring to?
March 6, 2012 at 4:45 pm
P69 summary has no problem. I meant the text at p38. after the equilibrium X(s) = X(aq), it says less soluble if Hsoln is exo and more soluble if it is endo.
March 6, 2012 at 10:38 pm
Ok.
Not sure I can explain it. Let me try and rationalize it instead.
If the eqm shifts to the left, K(eqm) must decrease. dS(tot) is proportional to K(eqm) so dS(tot) must have decreased. If rxn was exo there is only one way dS(tot) could decrease and that is if dS(sys) became more negative than dS(surr) was positive.
This reveals the context in which that equilibrium, X(s) X(aq), is discussed. It is talking about dissolving ionic salts which cause an increase in order in the solvent and is greater than the increase in disorder we get from the ions separating i.e. ionic lattice breaks down. This, we are told, happens with small highly charged ions. So the more exo the process, (balance between hydration enthalpy and lattice enthalpy) it may cause more ordering of the solvent.
But we can look at the other case too…
In cases where the dissolving solid did NOT cause the solvent to be more ordered than the degree to which the entropy increases due to lattice breakdown, then the the more exo the reaction, the equilibrium would shift the other way to the right (more soluble).
So I don’t think the X(s) X(aq) shifting to the left if more exo is a general ‘always true’ statement. I think it depends on the special case given in the preceding lines in the book.
But to be honest, I myself am not 100% certain.
Please note, the above is me simply trying to rationalise what is written. I could be wrong. I cannot imagine that level of detail / specificity / particular case will be required for A-level. George Facer occasionally seems to add material that shows you there are exceptions to any “must be true for absolutely everything” ideas which we may get in our head (human nature).
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November 2, 2010 at 7:37 am
assalamualaikum warahmatullah
sir, there is a statement in g.facer that i’m not really understand
page 38
“the equilibrium:
x(s) x(aq)
is driven to the left(less soluble) if ^Hsol is exothermic and to the right (more soluble) if ^hsol is endothermic”
is it that if more exo then more soluble??? esp to the right..better… why it’s written to the left…
if to the left ye, i agree that hsol exothermic would be better for it..
i tried to consider the s sys.
still couldn’t get it..