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

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Stuck on 7b in the june 2010 paper (the last q)

paper-
http://www.ocr.org.uk/download/pp_10_jun/ocr_57551_pp_10_jun_gce_f325.pdf

mark scheme-
http://www.ocr.org.uk/download/ms_10/ocr_56094_ms_10_gce_f325.pdf

I just dont understand why you multiply 40 to the moles of h202 and the mr of H202 34, where does it come from??

thanks!

Reply 481

SmartFool

NEED SOME HELP OF SOMEONE PLEASE!
On page 117 in the chemistry textbook there is a q on orders and the rate equation, someone explain to me how you work out the answers!

rate = k[p]2[Q]
How will rate change if

a) Conc of P is doubled
b)Conc of Q is tripled
c)Conc of P and Q are both tripled?

Its probably an easy set of Q'S but it is really baffling me how you work them out e.g the answer to C is 27 but how the hell do you get that answer?! help please

Also on page 131 there is a question on KC but Why is reaction B Endothermic when KC IS DECREASING in the table when temp is increased and not increasing?

If someone could whip out their textbook and just explain those to me i will be really grateful!

Reply 482

haydyb123

[QUOTE="SmartFool;31976685"]NEED SOME HELP OF SOMEONE PLEASE!
On page 117 in the chemistry textbook there is a q on orders and the rate equation, someone explain to me how you work out the answers!

rate = k[p]2[Q]
How will rate change if

a) Conc of P is doubled You can tell from the rate equation that P is 2nd order, therefore if it's doubled the rate will increases by 4 (2squared)
b)Conc of Q is tripled From the rate equation, Q is first order, therefore by tripling the Conc of Q, you will triple the rate as a first order reactant is proportional to the rate
c)Conc of P and Q are both tripled? And here is the more difficult question, you know that increases the conc. of the 2nd order reactant by 3 will increase the rate 9 times, add that to tripling the conc. of Q and this increases 9 by 3 = 27 times, so you see from that all you have to do for any of those types of questions is to remember to X your two numbers together

Its probably an easy set of Q'S but it is really baffling me how you work them out e.g the answer to C is 27 but how the hell do you get that answer?! help please

Also on page 131 there is a question on KC but Why is reaction B Endothermic when KC IS DECREASING in the table when temp is increased and not increasing?
You know that increasing the temperate has increased the value of Kc, higher temperatures favour the endothermic reaction. When Kc increased the concentration of the products is larger, and the concentration of the reactants is smaller, therefore you can tell that increasing temperature has increased the number of products, and because you mean that high temperatures favour the endothermic reaction, reaction B must be endothermic

If someone could whip out their textbook and just explain those to me i will be really grateful![/Q]

hope this has helped :smile:

Reply 483

wilsea05

Original post by cws121

right so you need the grams per dm3 of H2O2.
so, you start by working out the moles of KMnO4 by doing 0.02345x0.02=0.000469 (n=c*v n=moles, c=concentration, v=volume in dm3)
from the equation at the start of the question you know that 2MnO4- reacts with 5H2O2 so you need to multiply 0.000469 by 2.5 (5/2) =0.0011725 moles of H2O2 in 25cm3
by originally there was 250cm3, so you need to multiply that number by 10 = 0.011725 (as its asking for g dm-3 of the UNDILUTED solution of H2O2)

0.011725 per 25cm3 H2O2 (undiluted) so you rearrange n=cxv to get n/v=c
so 0.011725/0.025 (this is the same as multiplying it by 40) = 0.469 moles per dm3
now you do n=m/M (n=number of moles, m=mass in grams, M = molecular mass)
so rearrange it to get m=nxM so you do 0.469x34 (16x2+1x2) = 15.946 g dm-3

Reply 484

SmartFool

Original post by haydyb123

hope this has helped :smile:

Hey, thanks for that first bit on rates and orders i get it now! : :biggrin:

But im still not understanding the KC question because in the table reaction B at 500k is 0.045 and then at 800k kc is 0.0028. So looking at these values Kc has decreased when temperature has risen from 500k to 800k. So hasent raising the temperature decreased the value of KC?
:confused:

Sorry to bother you, It's probably really obvious but i just don't get it!

Reply 485

Killmepls

Original post by cws121

I did a different method, I don't know why they did x40 honestly but there is more than one way of doing it. Tbh a lot of the calculations we do in chem you can do in more than one way e.g. for buffers, ph of alkali, kw..
I just checked the markscheme look in the right box it has allow for the other method to the right of where it says that line you where talking about.

Reply 486

Killmepls

Original post by SmartFool

Hey, thanks for that first bit on rates and orders i get it now! : :biggrin:

Sorry to bother you, It's probably really obvious but i just don't get it!

lol look at the numbers closer, one is - power one is plus :P

Reply 487

haydyb123

Original post by Killmepls

lol look at the numbers closer, one is - power one is plus :P

Haha, I had to look twice as well, no worries honestly! :biggrin:

Reply 488

SmartFool

Original post by Killmepls

lol look at the numbers closer, one is - power one is plus :P

omg

i have spent ages on that trying to work it out FFS they could have made it clearer in the book!
LOL thanks though it all makes sense now :smile:

Reply 489

INeedToRevise

Can someone please explain ionic size and charge density etc. like when it increases/decreases, effect on lattice enthalpy :smile:

e.g. How is the ionic radius of Iron (III) smaller than calcium ion, 2+?

Reply 490

sc0307

Original post by INeedToRevise

Can someone please explain ionic size and charge density etc. like when it increases/decreases, effect on lattice enthalpy :smile:

e.g. How is the ionic radius of Iron (III) smaller than calcium ion, 2+?

The lattice enthalpy of an ionic compound is affected by two factors:

1. Ionic Radius
2. Ionic Charge

The bigger the charge, the more EXOTHERMIC, no good just saying bigger or smaller, the lattice enthalpy will be as the ionic bonds will be stronger.

The smaller the radius of the ion, the more EXOTHERMIC the lattice enthalpy will be as the ions can pack closer together increasing the strength of the ionic bonds.

For Fe(III), it's charge of 3+ is stronger than Ca2+ so i guess that makes the ionic radius smaller than the calcium ions. Iron is also further across the same period as calcium giving iron a smaller radius as well.

Reply 491

k9markiii

Original post by Jtking3000

i take what the specimen paper mark scheme says with a pinch of salt, found so many mistakes in there ¬¬

same thanks. That's 3 more marks.

Reply 492

k9markiii

Original post by sportycricketer

Yeah I got double that answer :s-smilie:

is that gonna be wrong then?
cos I did
2 x (Na2O) - [(4xNa) + (1xO2)]

I need about 105 marks to get an A this summer or 59% I guess with the current grade boundaries.

Reply 493

sportycricketer

Original post by k9markiii

I need about 105 marks to get an A this summer or 59% I guess with the current grade boundaries.

Pretty similar to me then :biggrin:

Reply 494

INeedToRevise

Original post by sc0307

Thanks

Don't know why I got negged for asking a question :confused:

Reply 495

haydyb123

Original post by INeedToRevise

Thanks

Don't know why I got negged for asking a question :confused:

Trust me, you get negged for nothing on this thread.

Reply 496

sportycricketer

Original post by haydyb123

Trust me, you get negged for nothing on this thread.

Probably because everyone's gone mad with the revision :P lol

Reply 497

haydyb123

Original post by sportycricketer

Probably because everyone's gone mad with the revision :P lol

Haha quite possibly, I think I can feel myself going insane actually, all I want to do is play cricket, not sit in my room and read about Stalin's foreign policy in the origins of the Cold War.