Post by: /0 on November 11, 2009, 03:37:32 pm
GOGOGOGOGOGOGOGO!!!
Here's some stuff to kick us off:
Nuclear Fusion
Energy stored in plants:
Combustion of coal in a power station
Oxidation of sulfur impurities in the coal
(courtesy Kilbaha '09)
- Catalysts speed up reactions by providing an alternative pathway of lower activation energy. Reactants adsorb to the surface of catalysts where they meet and react. Catalysts do not affect equilibrium yield.
- Only temperature can affect the equilibrium constant!
- The conditions for the electrochemical series are 1M, 25 degrees celcius, and 1 atm (e.g. Brine > 1M, so Cl(g)/Cl- drops below the water reaction)
- You can only recharge cells if the products of the discharging reaction do not leave the surface of the electrodes
- If you are electrolysing a solution containing Cl- ions, even though Cl(g)/Cl- is higher than the water reaction, thre is chance that Cl(g) will be given off if the voltage is high enough.
- In a secondary cell, no matter whether you're charging or discharging, the negative electrode stays negative, and the positive electrode stays positive. However, anode and cathode swap.
- If asked to derive
- If asked to derive
- If you dilute a strong acid its pH will change more than for a weak acid. This is due to it already being fully ionised. The weak acid will ionise more if you dilute, due to LCP.
- If you need to draw a half-cell and one of the reactants is a gas, you need to draw a gas electrode (there's a pic in your heinemann textbook)
- In oxygen-hydrogen fuel cells, electrodes must be catalytic and porous. In general they must also be inert and conducting.
- You must use Brine in a membrane cell (producing Sodium hydroxide). It is concentrated NaCl, which allows the Cl(g)/Cl- reaction to go below the water reaction on the electrochemical series.
- The Down Cell (producing Sodium) must use molten NaCl
- You must use Cryolite in a Hall-Heroult Cell (producing aluminium) (you don't need to remember the formula of cryolite), in order to lower the temperature required for the reaction.
- The electrochemical series cannot predict the rate of reaction. Don't even try it.
- In a galvanic cell the ions in the salt bridge must be soluble, but also inert.
- Inert electrodes include Graphite and Platinum, with the former being more expensive but also having some catalytic properties.
-
It can be negative!
- An equation
- In
respectively. i.e. the number in front of the chemical will govern how much the graph of that chemical increases/decreases due to LCP.
- Faraday's Laws
1.
2. In order to produce 1 mole of metal, a whole number of mole of electrons must be consumed.
- Biochemical Fuels
Fermentation of glucose to ethanol:
Post by: Over9000 on November 11, 2009, 03:39:41 pm
Alright yall let's make a thread of STUFF WE GOTSTA KNOWJust remember the velocity of a reaction (courtesy kilbaha)
GOGOGOGOGOGOGOGO!!!
Here's some stuff to kick us off:
Nuclear Fusion
Energy stored in plants:
Combustion of coal in a power station
Oxidation of sulfur impurities in the coal
(courtesy Kilbaha '09)
- Catalysts speed up reactions by providing an alternative pathway of lower activation energy. Reactants adsorb to the surface of catalysts where they meet and react. Catalysts do not affect equilibrium yield.
- Only temperature can affect the equilibrium constant!
Post by: StringFever on November 11, 2009, 03:40:28 pm
Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron :)
Post by: Over9000 on November 11, 2009, 03:43:48 pm
Why u say between those elements its endothermic, theyre the only elements the sun can consist of, nothing heavier exists within a sun.
Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron :)
Post by: Over9000 on November 11, 2009, 03:46:16 pm
Post by: TheJosh on November 11, 2009, 03:50:34 pm
LCP isoffset only.PARTIAL
what do you mean by offset??
Post by: Over9000 on November 11, 2009, 03:51:11 pm
YehLCP isoffset only.PARTIAL
what do you mean by offset??
Post by: TheJosh on November 11, 2009, 03:52:08 pm
YehLCP isoffset only.PARTIAL
what do you mean by offset??
hmmm... that doesn't really answer the question..
Post by: Over9000 on November 11, 2009, 03:53:42 pm
If a change occurs in a reaction to increase the pressure, say decreasing volume, lcp will partially offset with increase in pressure. That is to say there is still an overall net decrease in pressure in the end.YehLCP isoffset only.PARTIAL
what do you mean by offset??
hmmm... that doesn't really answer the question..
Post by: TheJosh on November 11, 2009, 03:57:45 pm
Post by: NE2000 on November 11, 2009, 04:06:11 pm
Post by: arthurk on November 11, 2009, 04:07:11 pm
Nuclear fusion reaction was only in previous study designbut it appeared in kilbaha and that makes us worry i guess
Post by: BiG DaN on November 11, 2009, 04:07:37 pm
Post by: arthurk on November 11, 2009, 04:15:18 pm
any1 feel like teaching me unit 4 in 16 hours 53 minutes?So you get one oxidant and a reductant and they make the good stuff
but itll only work if they are good together^^
Post by: hyperblade01 on November 11, 2009, 04:21:26 pm
- In a secondary cell, no matter whether you're charging or discharging, the negative electrode stays negative, and the positive electrode stays positive. However, anode and cathode swap.
Could you re-word that/explain again :S
- If asked to derive, find
where X is anything. Do not use Faraday's constant, it is itself derived from
Do you happen to remember an exam from 08-09 that has a question involving finding Faraday's constant? I know there's one but otherwise I'll just go searching through each slowly..
And to add to this thread: Don't forget that
Post by: /0 on November 11, 2009, 04:27:18 pm
Code: [Select]
___(V)__
| |
| |
\_/ \_/
Just say the right electrode is negative. It will be the anode. Then the left will be positive and cathode.
Code: [Select]
___+Battery-__
| |
| |
\_/ \_/
Now when charging, the right electrode is still negative, but now it's the cathode.
The left electrode is still positive but it's the anode.
Post by: dekoyl on November 11, 2009, 04:27:50 pm
Post by: Mao on November 11, 2009, 04:28:27 pm
Why u say between those elements its endothermic, theyre the only elements the sun can consist of, nothing heavier exists within a sun.
Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron :)
False. In a dying star, lots of things can happen. (Stars are universal element factories, they convert hydrogen to helium to lithium and keep going until Iron, in their healthy lifespan. Just before they completely die, they spit out the heavier more exotic elements such as lead, uranium, etc, basically everything in the periodic table).
Post by: StringFever on November 11, 2009, 04:31:59 pm
(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!
Post by: hyperblade01 on November 11, 2009, 04:38:18 pm
Thanks /0 for the diagrams, the original wording kinda confused me (probably 'cuz of my downtime or something)
Question about the star thing, why does it stop at Iron? irrelevant to the upcoming exam I know, but it sounds interesting :)
Post by: StringFever on November 11, 2009, 04:39:18 pm
Lol at the efficieny-style question :P
Thanks /0 for the diagrams, the original wording kinda confused me (probably 'cuz of my downtime or something)
Question about the star thing, why does it stop at Iron? irrelevant to the upcoming exam I know, but it sounds interesting :)
I don't remember :(
And yeah, positive reinforcement for the efficiency questions :)
Post by: /0 on November 11, 2009, 04:40:30 pm
http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg
Post by: dekoyl on November 11, 2009, 04:43:15 pm
Efficiency-style questions:Sorry StringFever, could you (or someone else) explain that again?
(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!
I remember a VCAA paper that had 60% efficiency for some hydrogen cell. And, instead of multiplying by 0.6, you had to divide by 0.6. Why is this so?
It took me a while to get and I just wanna go over it again =\
Post by: Over9000 on November 11, 2009, 04:43:48 pm
Thats weird I thought that it goes up to iron only when its dying and while its healthy it simply converted hydrogen to helium (I better go bak to the drawing board).Why u say between those elements its endothermic, theyre the only elements the sun can consist of, nothing heavier exists within a sun.
Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron :)
False. In a dying star, lots of things can happen. (Stars are universal element factories, they convert hydrogen to helium to lithium and keep going until Iron, in their healthy lifespan. Just before they completely die, they spit out the heavier more exotic elements such as lead, uranium, etc, basically everything in the periodic table).
I thought once the sun got up to iron, the star was choked under the hihg weight of iron and then proceeded to die (but maybe that is the phase before becoming a red giant).
Post by: StringFever on November 11, 2009, 04:45:38 pm
Efficiency-style questions:Sorry StringFever, could you (or someone else) explain that again?
(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!
I remember a VCAA paper that had 60% efficiency for some hydrogen cell. And, instead of multiplying by 0.6, you had to divide by 0.6. Why is this so?
It took me a while to get and I just wanna go over it again =\
This is assuming that you had to find the hydrogen used in the cell, right?
Because if we think about it, the cell was 60% efficient right - therefore, only 60% of the hydrogen was used to produced energy.
For us, we've only got data about the energy produced from the cell, and that's 60% of the hydrogen used.
Since the actual amount of hydrogen is greater, we need to multiply by (100/60) which is the same as dividing by 0.6! :)
Post by: /0 on November 11, 2009, 05:19:21 pm
Post by: kimtywong on November 11, 2009, 05:29:54 pm
is it ALWAYS 100/x, x being efficiency percentile?
Post by: StringFever on November 11, 2009, 05:37:11 pm
Hey u know for redox reactions how for acidic medium you addthen
, does it work the opposite way for alkaline medium? I've done a few examples adding
first and then
I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.
Post by: NE2000 on November 11, 2009, 05:41:58 pm
Hey u know for redox reactions how for acidic medium you add
I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.
This = the chemguide way :)
Post by: d0minicz on November 11, 2009, 05:42:06 pm
can you give an example please?Hey u know for redox reactions how for acidic medium you add
I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.
Post by: StringFever on November 11, 2009, 05:42:56 pm
can you give an example please?Hey u know for redox reactions how for acidic medium you add
Give me an example question and I'll do it for you! :)
I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.
OK, let's take Zn to Zn(OH)2.
Step 1: Set up skeleton equation
Zn --> Zn(OH)2
Step 2: Balance like in acidic environment (with H+s and waters)
Zn + 2H2O --> Zn(OH)2
Zn + 2H2O --> Zn(OH)2 + 2H+ + 2e-
Step 3: Balance any H+ with an OH- (ON BOTH SIDES)
Zn + 2H2O + 2OH- --> Zn(OH)2 + 2H+ + 2e- +2OH-
Step 4: H+ and OH- form water
Zn + 2H2O + 2OH- --> Zn(OH)2 + 2H2O + 2e-
Step 5: Clean it up and voila
Zn + 2OH- --> Zn(OH)2 + 2e-
Post by: homghomg1 on November 11, 2009, 05:43:37 pm
Hey u know for redox reactions how for acidic medium you addhalf reactions in alkaline conditions are harder to do
i recommend this
http://www.chemguide.co.uk/inorganic/redox/equations2.html
explains how to do it very well, with some good examples
the main point is that sometimes it is a little difficult to know whether to add OH- or H2O first, but if you do it as if it was under acidic conditions and then cancel out the hydrogen ions by adding OH- to each side, it works out
read some of their examples and you'll see what i mean
Post by: /0 on November 11, 2009, 05:44:03 pm
Post by: StringFever on November 11, 2009, 05:46:33 pm
Hey u know for redox reactions how for acidic medium you addhalf reactions in alkaline conditions are harder to do
i recommend this
http://www.chemguide.co.uk/inorganic/redox/equations2.html
explains how to do it very well, with some good examples
the main point is that sometimes it is a little difficult to know whether to add OH- or H2O first, but if you do it as if it was under acidic conditions and then cancel out the hydrogen ions by adding OH- to each side, it works out
read some of their examples and you'll see what i mean
Like my example above :)
Post by: dekoyl on November 11, 2009, 05:55:02 pm
Code: [Select]
A|___
B|___
C|___
|___:_________
t
At time t, if we add a little concentration of B, B spikes up but as it establishes equilibrium again, its concentration is higher than before you added it. A gradually goes up and establishes equilibrium. But what happens to C? In a trial exam, it shows that C spikes up as well (but not as much as B) and goes back to being the same concentration again. Is this right?Post by: /0 on November 11, 2009, 05:56:46 pm
Post by: dekoyl on November 11, 2009, 05:58:33 pm
LOL learning from trial exams like this will screw me over =\
Post by: kimtywong on November 11, 2009, 06:08:28 pm
Post by: ayu on November 11, 2009, 06:47:16 pm
Post by: dekoyl on November 11, 2009, 08:19:43 pm
- An equationOkay, in relation to this^does not imply that the concentration graphs will start with
NO! You must look at the equilibrium constant!
If we have
I looked at the concentration of the three to determine it (which was the correct answer) but wasn't sure how to use the equi. constant.
Thanks
Edit: oops stuffed the LaTeX up for the equi constant =\
Post by: /0 on November 11, 2009, 08:23:37 pm
"- In
respectively. i.e. the number in front of the chemical will govern how much the graph of that chemical increases/decreases due to LCP."
Which basically means they have to DO something to the equilibrium so you can observe the change.
e.g. If they put in extra reactant A, then graph A will spike up but then graphs A and B will decrease, but B will decrease twice as much as A, so that's how you tell A and B apart. Also, C will increase and that's how you locate C
e.g. If they dilute the solution then all 3 will decrease, but then A and B will increase, with B increasing twice as much as A. C will decrease.
Post by: dekoyl on November 11, 2009, 08:25:14 pm
Post by: Over9000 on November 11, 2009, 09:16:43 pm
Well if the equations happen in reverse, does that mean the electrovoltage becomes the opposite charge (i.e + to -)
Say for example if
would
Post by: /0 on November 11, 2009, 09:18:57 pm
As long as you have the difference in potentials
Post by: Over9000 on November 11, 2009, 09:21:06 pm
Hmm I don't think that applies to the electrochemical series, only enthalpies etc.Thanks, ths electrocehmical stuff is soo hard, I much prefer LPG.
As long as you have the difference in potentialsyou can find out how much voltage is produced by a spontaneous reaction or how much voltage is required to drive a non-spontaneous reaction.