STPM化学专区

Allison

我也一样。。。现在还在上网
完蛋了

darksider

原帖由 Allison 于 2009/8/24 04:10 PM 发表
erm。。。
如果是concentrated NaCl，那代表Na & Cl ion 比 H2O多
所以Cl ion 会discharged
aqueous 的话， H2O 比较多所以OH ion discharged
你是lower的？

Correction
我想这样的解释会好一点：

In Anode, two possible ions can be oxidised , ie , cl- and h2o
Cl2(g) + 2 e− ⇄ 2 Cl−    E°red= +1.36V  - (1)
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O E°red= +1.23V - (2)

Overvoltage needed for the formation of oxygen is higher than that of chlorine gas because activation energy needed for transfer of electrons at the electrode-solution surface is high which corresponds to slow rate (transfer of electron from anode to cathode). Thus, extra voltage must be applied so that the reaction can proceed in a satisfactory rate. Since rate of formation of chlorine gas is faster than that of oxygen gas due to lower activation energy,Cl- is oxidised and chlorine gas is produced. (Reaction which requires lesser overvoltage will occur instead of the one requiring higher overvoltage)

Hopefully I didn't confuse any of you.

Additional notes  (not related)
----------------------
Let say the electrolyes used are all aqueous

i) In anode , possible half reactions :
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O  E°red= +1.23V --- (1)
O2(g) + 2 H2O + 4 e− ⇄ 4 OH−(aq) E°red= +0.40V --- (2)
Cl2(g) + 2 e− ⇄ 2 Cl−    E°red=  +1.36V  - (3) ----->>> just an example, replace it with any other anion's half eq

According to MPM,
if the electrolye used is neutral, equation (2) can be eliminated from the possible half equations and we consider only (1) and (3) instead. This is because in neutral solution, most of the water molecules remain undissociated and [OH−]is very small , ie , 1x10−7mol dm-3. So we compare only equation (1) and (3) and the one with lower E°red will be discharged.

If the electrolyte used is alkaline, then equation (1) should be ignored because the [OH−] cannot be ignored now. Hence, we compare equation (2) and equation (3)


ii) In cathode, possible half reactions :
2 H2O + 2 e− ⇄ H2(g) + 2 OH− E°red= -0.8277V -----(1)
2 H+ + 2 e− ⇄ H2(g) E°red= 0.00V ----(2)
Cu2+ + 2 e− ⇄ Cu(s) E°red= +0.34V ----(3) ---- just an example, replace it with any other cations will do


If the electrolye used is neutral, equation (2) should be ignored because large amount of h2o remains undissociated and [H+] is relatively small , ie , 1x10−7mol dm-3. Hence, we compare only equations (1) and (3) if the electrolye is neutral.


If the electrolye used is acidic, equation (1) can be ignored since  [H+] is relatively large now. Hence, we compare only equations (2) and (3).


According to MPM, most candidates do not realise the presence of h2o molecules in aqueous electrolye.


Again, I hope I didn't confuse you. Good luck!


Btw, for most textbooks I've used, I found out Chemistry by John McMurry and Robert C.Fay is the best. None of the malaysian text books could be compared to it. However it is expensive but you know in internet we have such thing called download, so

[ 本帖最后由 darksider 于 27-8-2009 07:41 PM 编辑 ]

darksider

If a solution is aqueous, it has water molecules.

If the solution is dilute, it means that the water molecules are present in large amount.

If the solution is concentrated, it means that the water molecules are present in relatively small amount.

Dicardo

so if it only written 'aqueous' , we can assume it is dilute ?

darksider

原帖由 Dicardo 于 2009/8/27 11:32 AM 发表
so if it only written 'aqueous' , we can assume it is dilute ?

Yes, most probably. If it never mentions concentrated salt, then it should be a dilute one.

darksider

做了些改正，有错的地方就提醒我吧！

@影子@

帮我解决这问题

A glass bulb contains 4.0g of gas at a pressure of 100kPa. If the pressure in the bulb is lowered to 0.1kPa by removing the gas in it,what is the number of molecules that will be left in the bulb?(Mr of gas=40 ; Avogadro constant = 6 X 10^23)

darksider

原帖由 @影子@ 于 2009/8/27 08:56 PM 发表
帮我解决这问题

A glass bulb contains 4.0g of gas at a pressure of 100kPa. If the pressure in the bulb is lowered to 0.1kPa by removing the gas in it,what is the number of molecules that will be lef ...

Take note that the volume of glass bulb is constant.
Initially, at pressure of 100kPa, there is 4.0g of the gas.
Hence, at 0.1kPa, there is x g of gas.

100000 Pa = 4g
        100Pa = xg
x= 4x100 / 100000
  = 4/1000 = 0.004g

mol of the gas = 0.004g / 40 g per mol
                         =  0.0001 mol

Therefore, number of molecules of the gas left = 0.0001mol x 6.022x10^-23 per mol
                                                                               = multiply yourself

@影子@

1.In temperature T K,1dm³ of an equilibrium mixture contains 0.60mol N2,0.50 mol O2 and 0.15 mol NO.
(a)Calculate the equilibrium constant Kc at T K for the following equilibrium:
                                   N2(g) + O2(g) ~~~~2NO(g)                          ~~~represent forward n backward reaction
(b)0.20 mol N2 is then added to the equilibrium mixture,whilst maintaining the temperature  at T K. Calculate the value of      (NO)²       at the moment when N2 is introduced into the mixture.
         （N2）（O2）
(c)Calculate the concentration of N2,O2 and NO at the new equilibrium.



2.                             2NO(g) + Cl2 ~~~~ 2NOCl(g)
At temperature T℃，the partial pressures of NO,Cl2 and NOCl in a 1dm³ equilibrium mixture are 60kPa, 50kPa and 80kPa respectively.
(a) Calculate the value of the equilibrium constant,Kp.
(b) When additional Cl2 is added so that its partial pressure is 70atm, the new equilibrium partial pressure of NOCl is 88atm.Determine the partial pressures of NO and Cl2 in the new equilibrium mixture if temperature remains constant.

darksider

http://www.youtube.com/watch?v=CGJjjm7-DJ4

@影子@

什么意思啊？

darksider

去看看
Reaction in Equilibrium
http://www.youtube.com/watch?v=C ... =player_profilepage

Intuition behind formula of Keq
http://www.youtube.com/watch?v=O ... =player_profilepage

Le Chatelier's principle
http://www.youtube.com/watch?v=4 ... =player_profilepage

那些问题都很简单，你去看这些video后明白了才做。

Tips for 1(c)  Calculate concentration at new equilibrium
                      N2        +            O2  <------>  2NO
Initial             6+2                      5                   1.5
Change          -x                       -x                  +2x
equilibrium   8-x                     5-x                  1.5+2x
(new)
Since Keq is only affected by changes in temperature, hence it remains constant when concentration of N2 increases.

Therefore, Keq = 0.075 = (1.5+2x)^2 / (8-x).(5-x)
Find x and you will get the answer but substituiting x into concentraton at new equilibrium.

@影子@

噢，好！
我不会再问！
谢谢你！

idontwant2b

请问
hydrolysis, condensation, hydration, dehydration分别在哪里？

premier66

hydration 是加 H2O ,dehydration 是把H2O 弄走 ， hydrolysis 是把H2O 分解， 有两种 alkaline hydrolysis :normally we use  sodium hydroxide ,NaOH to produce OH ions , another type is acidic hydrolysis :use sulphuric acid ,H2SO4 ,to get H ions.

大概是这样。。。

darksider

原帖由 idontwant2b 于 2009/10/9 08:52 PM 发表
请问
hydrolysis, condensation, hydration, dehydration分别在哪里？

Hydrolysis is a reaction that involves breaking down of water molecules.[Alkaline hydrolysis, acidic hydrolysis etc]

Condensation is a reaction involving formation of water molecules. [Condensation polymerisation etc]

Hydration is reaction involving addition of water molecules(act as ligands) to a molecule or compound.
eg, Cu2+ + 6H2O ----> [Cu(H2O)6]2+

Dehydration is a reaction that involves removal of H and OH atoms from a compound to form H-OH (water)

nkrealman

有什么简单的方法去读thermochemistry这课？
尤其是born-haber cycle.
还有 hybridisation要怎样看？

darksider

原帖由 nkrealman 于 2009/10/29 02:35 AM 发表
有什么简单的方法去读thermochemistry这课？
尤其是born-haber cycle.
还有 hybridisation要怎样看？

thermochemistry..
明白所有enthalpy change...可以去www.khanacademy.org参考。
你需知道怎样
-enthalpy change of reaction = enthalpy change of formation of product - enthalpy change of formation of reactant
-enthalpy change of combustion
-enthalpy change of solution = - Lattice energy + enthalpy change of hydration
-explain the enthalpy change of neutralisation of weak acid/base (energy absorbed to dissociate weak base/acid)
-hess's law ... enthalpy change is a state variable... enthalpy change of a reaction is independent of the route it takes to complete the reaction.
-born haber cycle is used to calculate ionisation energy.
若arrow向上，你用+sign，若arrow向下，你用negative sign.

in born haber cycle，
total energy in the LHS = total energy in the RHS
- assume all arrows are heading upwards in both sides(change sign if necessary). then equate both and you will be able to find the desired value.

-------------------------------------------
hybridisation is the fusion of orbitals of different energy levels to degenerate orbitals in order to faciliatate bonding.
There are three types in our syllabus.... sp, sp2, sp3...

[ 本帖最后由 darksider 于 29-10-2009 08:24 AM 编辑 ]

idontwant2b

请问考试complex的颜色要背吗？
2008的好像要背的耶。

darksider

原帖由 idontwant2b 于 2009/11/8 11:03 PM 发表
请问考试complex的颜色要背吗？
2008的好像要背的耶。

对。。要背。。但是来来去去都是一样。。。