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# Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

1. (Original post by Reccoshai)
Can some one help me with the question in the text book module 2 examination questions in the OCR book, question 5 (a) how exactly do they calculate this? looked at the marks scheme and i dont even understand that
Edit: My bad, wrong question.
2. (Original post by Ayostunner)
Can someone please explain the concept of the feasibility of a reaction, i cant seem to get my head round it
Do you mean the entropy stuff or the electrode potentials stuff?
3. (Original post by Reccoshai)
Can some one help me with the question in the text book module 2 examination questions in the OCR book, question 5 (a) how exactly do they calculate this? looked at the marks scheme and i dont even understand that
Q=mcT

Q= (100+8.48g)(4.18)(34.5-22)

Mass of water = m
c= specific heat capacity 4.18
T= temperature change

Q= 5668.08 J
Q= 5.66808KJ

so for delta H (enthaply change of soloution) the units are KJ/mol

so you need to divide the KJ/moles of LiCl

moles of LiCl = mass/Molar mass = 8.48/ (6.9+35.5) = 0.2 moles

delta H =5.66808KJ/0.2 moles = 28.34KJ/mole

because the temperature INCREASED in the surronding the system DECREASEd temperatureits an exo reaction so its neg delta H

- 28.345KJ/mole

AM I RIGHT?

oh **** i did a mistake the mass is 8.48 + 100
EDIT: corrected
4. (Original post by susan23)
i love you goldlock!
No worries, glad I helped
5. (Original post by goldlock)
It's like this: 2 moles of Mn 3+ to everything else in equation 2.
4 moles of Mn 3+ to everything else in equation 1.
Equations 1 and 2 are completely independent of each other, so you can't compare the ratio of moles of Mn 3+ between them

Hope that helped
so the first equation is independant from the second. But the second and third equations are linked right?
6. (Original post by susan23)
Q=mcT

Q= (100g)(4.18)(34.5-22)

Mass of water = m
c= specific heat capacity 4.18
T= temperature change

Q= 5225 J
Q= 5.225KJ

so for delta H (enthaply change of soloution) the units are KJ/mol

so you need to divide the KJ/moles of LiCl

moles of LiCl = mass/Molar mass = 8.48/ (6.9+35.5) = 0.2 moles

delta H = 5.225J/0.2 moles = 26.125KJ/mole

because the temperature INCREASED its an endothermic reaction so its positive delta H

+ 26.125KJ/mole

AM I RIGHT?
Thats what i though yet in the book it says its -28.38K per mol

Oh i see where i went wrong i didnt add the 100 to 8.48 like you haha thanks for the help
7. Good luck tomorrow everyone. Let's hope OCR give us all some nice order, buffer and ligand questions..... Xxx
8. (Original post by Schoolio93)
so the first equation is independant from the second. But the second and third equations are linked right?
No sorry I think I explained it poorly - they're all linked but I meant independent in the sense that the ratios are independent from one equation to the next. People were thrown by the fact that out of all of the important reactants in those equations, only the number in front of Mn3+ changed (from 4 to 2) - but that doesn't mean that you change its molar quantity as you look from equation 2 to equation 1. The number of moles of the stuff is the same.

Hope that answered your question
9. HELP!!! WHAT DO THEY MEAN BY SCALING UP TO 1 MOL OF H2O??? for e.g if you had

3x + 4Y -------> 2H20

AND Y IS IN EXCESSS ---- HOW WOULD YOU CALCULATE n TO BE USED TO CALCULATE ENTHALPY CHANGE OF NEUTRALISATION!!!
10. (Original post by Twinkles)
So if the equation was:

Fe2+ + 6H2O --> [Fe(H2O)6]2+
( I don't know whether this is right )

What would the Kstab be?
[[Fe(H20)6]2+] over [Fe2+]

sorry, I don't know why I included H20 the first time. sorry for any confusion caused.
11. Does anybody know the half equations for 4Fe2+ + O2 + 4H+ --> 4Fe3+ + 2H2O ?
12. (Original post by entertheOJI)
[Fe(H20)6]2+ over [Fe2+] [H20]to the power of 6.
Thank you
Now I just don't understand when not to include H20 in Kstab
aah.
13. (Original post by Twinkles)
Thank you
Now I just don't understand when not to include H20 in Kstab
aah.
You never include water molecules in K stab. K stab is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions. So H20 is the solvent, so you never include it.
K stab is products over reactants. Remembering to exclude water molecules and raise concentrations to the power of moles.
14. (Original post by bambinooo)
HELP!!! WHAT DO THEY MEAN BY SCALING UP TO 1 MOL OF H2O??? for e.g if you had

3x + 4Y -------&gt; 2H20

AND Y IS IN EXCESSS ---- HOW WOULD YOU CALCULATE n TO BE USED TO CALCULATE ENTHALPY CHANGE OF NEUTRALISATION!!!
If Y is in excess I take it you mean that 4 moles of Y is in excess, so focus on using the moles of x - if it takes 3 moles of x to make 2 moles of water, then it takes 1.5 moles of x to make 1 moles of water - so you divide your q value by 1.5 instead of dividing it by 3

(Original post by Twinkles)
Thank you
Now I just don't understand when not to include H20 in Kstab
aah.
The concentration of water is taken to be a constant since everything's dissolved in it as it's in large excess, so we don't bother with it
15. (Original post by Twinkles)
Thank you
Now I just don't understand when not to include H20 in Kstab
aah.
I think thats wrong.
You never include h20 in kstab .. just like in ka .
Also its double brackets around the other 2.
16. Are people doing an all nighter here?
17. Help! in June 2010 the last question, why do they times it by 40. Where the hell does 40 comes from HELP! " concentration in gdm^(-3) of original H2O2 = 40 x 1.1725 x 10^(-2) x 34 = 15.9
18. (Original post by intellectual1)
Are people doing an all nighter here?
nope :P lol.. gonna wake up early in the morning and read over the definitions and learn the colours lol
19. (Original post by faez307)
Help! in June 2010 the last question, why do they times it by 40. Where the hell does 40 comes from HELP! &quot; concentration in gdm^(-3) of original H2O2 = 40 x 1.1725 x 10^(-2) x 34 = 15.9
They're actually dividing by 0.025, which is equivalent to multiplying by 40
20. (Original post by intellectual1)
Are people doing an all nighter here?
Maybe me!

Going through Jan 2011 paper as i type this, not a bad paper... I get most of it, and am checking through text and mark scheme for what OCR penalise.

I have a few past papers i'll take tomorrow morning to do in the Library

EXAM IS AT 13.00 yeah? Just checking

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