brightsky's Chem Thread

[lzxnl]

1)

okay I've come across this question in a random practice exam:

NH4Cl(s) <--> NH3(g) + HCl(g), dH = +176 kJ mol-1

when solid NH4Cl is added to the equilibrium, what will happen to the forward and reverse rates?

now I know LCP might say that if you add stuff to the LHS, the equilibrium position will be shifted to the right, hence the rate of forward reaction will increase instantaneously and then decrease a little and the reverse reaction increase, before equilibrium is regained. but does adding solid NH4Cl really affect the equilibrium at all? if we consider the Kc of the reaction, NH4Cl is a solid, so the concentration of NH4Cl really isn't relevant. so Kc = [NH3][HCl]. now if we add NH4Cl, Qc still remains the same as Kc. in other words, it appears that adding NH4Cl doesn't really affect the equilibrium in any way, since the concentration of the solid more or less remains constant, if it is even right to speak of concentration of solid.

what are your thoughts? I may be missing something critical...in which case I apologise in advance for the stupid question :p

2) also just a quickie: if the question does not explicitly tell us to use LCP, can we argue using formulas such as the van't hoff equation? I don't understand how one can really use LCP to explain why, for an endothermic reaction, an increase in temp results in an increase in Kc. sure, the equilibrium position will shift to the right, but that doesn't mean Kc increases.

thanks!

1. From my knowledge, solids don't contribute to the equilibrium as such, so what you have is fine.
2. LCP DOES explain, for an endothermic reaction, the equilibrium constant increases. The equilibrium position is shifted to the right through no change in volume or concentration or pressure, so the only quantity that could have changed is the equilibrium constant.
And I wouldn't use Van't Hoff's equation as VCAA doesn't like it when you use more advanced knowledge.

[brightsky]

1. From my knowledge, solids don't contribute to the equilibrium as such, so what you have is fine.
2. LCP DOES explain, for an endothermic reaction, the equilibrium constant increases. The equilibrium position is shifted to the right through no change in volume or concentration or pressure, so the only quantity that could have changed is the equilibrium constant.
And I wouldn't use Van't Hoff's equation as VCAA doesn't like it when you use more advanced knowledge.

2. ah okay I see. smart...smart...
1. but does adding more of the solid speed up the rate in any way? i'm tempted to say yes, but my chemical knowledge pulls me in the other direction. I mean...the concentration of the solid doesn't change, and so all you have is a greater chunk of solid...only factors that affect rate are concentration, temperature, surface area, light and catalyst. I fail to visualise how this reaction actually takes place. you have a solid in a vessel, and some how the solid particles react with each other to produce gases. now if I think about it that way, if there is more solid, then intuitively the reaction rate will increase. i'm not sure...

[lzxnl]

I'm not sure how solids react either, but it seems to me that having more solid could increase the surface area of the solid phase reaction.
BUT, I don't know much about solids in equilibria.

[brightsky]

ah I see.

I have another minor query. how are we supposed to express the redox pair for the reaction H2O + 2e- --> H2 + 2OH-? do we right H2O/H2,OH- or just H2O/H2?

[lzxnl]

Hydroxide isn't the result of an oxidation or reduction; H2O is involved as one hydrogen atom is reduced, while H2 is the result of the reduction. So you'd have H2O/H2

[brightsky]

1) one of the questions from thushan's book: should Ca2+(aq) + 2OH-(aq) be combined and written as Ca(OH)2(s) given low solubility of Ca(OH)2?
2) also, will we be penalised if we write 2-methylbutane on the exam? my teacher discourages trying to be smart while naming organic compounds and so advises that we include the 2- in such cases. but technically, the 2- really isn't required...

[brightsky]

also, how are we supposed to know that MnO2 and Mn2O3 are solids? is there are way of determining state symbols, or does it just come with practice?

[lzxnl]

also, how are we supposed to know that MnO2 and Mn2O3 are solids? is there are way of determining state symbols, or does it just come with practice?

That's a compound with a metal and a non-metal. Generally, these compounds are ionic; assume they're solids.

1) one of the questions from thushan's book: should Ca2+(aq) + 2OH-(aq) be combined and written as Ca(OH)2(s) given low solubility of Ca(OH)2?
2) also, will we be penalised if we write 2-methylbutane on the exam? my teacher discourages trying to be smart while naming organic compounds and so advises that we include the 2- in such cases. but technically, the 2- really isn't required...

Calcium hydroxide is partially soluble in water...depends on how much of it you have.

As for your second point...I wouldn't do that either because VCAA examiners may not immediately pick that up and just mark you wrong. Is it worth it?

[brightsky]

yeah so am I right to think that all compounds containing OH- are more or less insoluble except those with NH4+, Group 1 metals, Ba2+ and Sr2+? are there any other exceptions that i'm not aware of?

with regards to the question about nomenclature...damn. I hate it how the chem examiner is so vague as to what kind of answer is actually required. apparently when drawing the new reaction pathway with the addition of a catalyst on an energy profile diagram, we need to draw it so that the curve ends earlier than the original curve, but the horizontal axis isn't even a time scale...

with regards to metal+nonmetal, wait but surely that principle does not apply in general. stuff like MgCl2 contain a metal and a non-metal but is soluble...

[lzxnl]

yeah so am I right to think that all compounds containing OH- are more or less insoluble except those with NH4+, Group 1 metals, Ba2+ and Sr2+? are there any other exceptions that i'm not aware of?

with regards to the question about nomenclature...damn. I hate it how the chem examiner is so vague as to what kind of answer is actually required. apparently when drawing the new reaction pathway with the addition of a catalyst on an energy profile diagram, we need to draw it so that the curve ends earlier than the original curve, but the horizontal axis isn't even a time scale...

with regards to metal+nonmetal, wait but surely that principle does not apply in general. stuff like MgCl2 contain a metal and a non-metal but is soluble...

Sort of ironically given that hydroxide is the conjugate base of water and can hydrogen bond with water, but yes, very few hydroxides are soluble. And what the heck is ammonium hydroxide? It's just an ammonia solution.

Wait what, curve ends earlier as well!? That's weird. For starters, catalysed reactions may take more steps for all we know.

MgCl2 would also be a solid unless you added it to water. I didn't know the context. That's because chlorides are generally soluble, with the exception of silver, mercury...lead(II) chloride isn't very soluble from memory.

Metal oxides are generally insoluble, barring the exceptions given.

[brightsky]

how detailed do our answers need to be for questions like "what is one advantage of solar power"? would "expensive" be sufficient to gain the mark? or do we need to explain why it is expensive? the knowledge I gained during year 9 geography has already pretty much dissipated...

[lzxnl]

Advantage? Renewable works for me...

Its price isn't an advantage of solar power

[brightsky]

oh woops I meant disadvantage lol. okay so we don't need to elaborate? one word answers are fine?

also, do the units of calibration factor need to be J K-1 or can they be kJ K-1? *pedantic*

[Limista]

oh woops I meant disadvantage lol. okay so we don't need to elaborate? one word answers are fine?

also, do the units of calibration factor need to be J K-1 or can they be kJ K-1? *pedantic*

obviously not nliu, but I think I can give an answer to your questions. 

I wouldn't give one word answers for those industrial chem questions. I'd give atleast a sentence. So for solar energy - they require lower costs to run (or something along these lines). If you say that they are just 'inexpensive', then the examiner might be questioning what you mean. Like, do you mean the panels are inexpensive (which we know is totally not the case. The panels are expensive!), or do you mean that the electricity-generating process is expensive? We know the electricity generating process is not expensive, because solar cells are a direct source of electrical energy. I think they're supposed to be as efficient (if not more) as the galvanic cell - but don't quote me on this.

As for units for calibration factor - it's J/degrees Celsius. I'd just stick with joules per degrees Celsius, because that's the value you're going to get in your calculations & that's the 'accepted' form I guess. Technically, kJ/degrees Celsius is not wrong, because it's basically just in a different form (kind of like metres vs. kilometres).

[Dismounted]

also, do the units of calibration factor need to be J K-1 or can they be kJ K-1? *pedantic*

Depends on the context of the question. If it specifies "give me calibration factor in joules per Kelvin", then yes, obviously. Otherwise, it is up to you. Just don't give answers in some obscure unit.