Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

OP

Original post by Ayman!

June 2015 IAL was definitely challenging - tested many interesting concepts. I really enjoyed the acid-base and rates questions. Only scraped 76/90 though. :colondollar:

Yeah it's quite interesting to see how even though the questions are based on the same concepts as the uk one they still managed to make it quite different

Reply 542

imnoteinstein

2

Original post by Ayman!

June 2015 IAL was definitely challenging - tested many interesting concepts. I really enjoyed the acid-base and rates questions. Only scraped 76/90 though. :colondollar:

isnt it almost exactly like the normal june '15 paper tho? O:

Reply 543

Ayman!

15

Original post by samb1234

Yeah it's quite interesting to see how even though the questions are based on the same concepts as the uk one they still managed to make it quite different

Original post by imnoteinstein

isnt it almost exactly like the normal june '15 paper tho? O:

I haven't done that one yet - will get on it in a few!

Reply 544

Peppercrunch

14

Why is the answer D in this?
I thought it'd be C as the reaction is endothermic, so by decreasing the temperature, the equilibrium would shift to the left, and by increasing the concentration - more particles = more collisions.

Which of the following changes will lead to the greatest increase in the rate of thefollowing endothermic reaction?
N2(g) + O2(g) → 2NO(g) ∆H +ve

Temperature - Initial concentration of N2 and O2
A decrease by 15% - decrease by 15%
B increase by 15% - stay the same
C decrease by 15% - increase by 15%
D increase by 15% - increase by 15%

(edited 7 years ago)

Reply 545

Funky_Giraffe

10

Original post by Peppercrunch

Why is the answer D in this?
I thought it'd be C as the reaction is endothermic, so by decreasing the temperature, the equilibrium would shift to the left, and by increasing the concentration - more particles = more collisions.

Which of the following changes will lead to the greatest increase in the rate of thefollowing endothermic reaction?
N2(g) + O2(g) → 2NO(g) ∆H +ve

Temperature - Initial concentration of N2 and O2
A decrease by 15% - decrease by 15%
B increase by 15% - stay the same
C decrease by 15% - increase by 15%
D increase by 15% - increase by 15%

Your reasoning for increasing the concentration is correct, however decreasing the temperature is not going to increase the rate. I think you are confusing the equilibrium yield of N2 and O2 with the rate of reaction.

Reply 546

imnoteinstein

2

Original post by Don Pedro K.

Okay firstly work out the number of moles of HCl to begin with. This is
(volume * concentration / 1000) = (10*1.0)/1000 = 0.01 moles of HCl.

Since HCl is a strong acid, it will dissociate completely, meaning that the concentration of HCl = H+ ion concentration.

What will the concentration of [H+] be in 990cm^3 of water?

Rearranging the formula above, you get:

Concentration = (moles * 1000) / volume.
= (0.01 * 1000) / 990 = 0.01010101.....

pH = -log([H+]) = -log(0.0101010101...)

Therefore the pH = 1.99... = 2 when you round it up. So the answer should be C!

https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202014%20(IAL)%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf
hey for Q19b, why do we start off with 2NO2? Why cant it be NO2? The equations would still work out fine
( NO2 --> 0.5N2 + O2 and
CO +0.5N2 + O2 --> NO + CO2) - is it bc its second order?

Reply 547

Original post by imnoteinstein

https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202014%20(IAL)%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf
hey for Q19b, why do we start off with 2NO2? Why cant it be NO2? The equations would still work out fine
( NO2 --> 0.5N2 + O2 and
CO +0.5N2 + O2 --> NO + CO2) - is it bc its second order?

Yeah since it's second order with respect to nitrogen dioxide, the rate equation will be: rate = k[NO2]². Since the rate equation shows the species and the number of molecules of that species involved in the rate determining step, the rate determining step will start with 2NO₂ :smile:

Reply 548

Original post by Peppercrunch

Why is the answer D in this?
I thought it'd be C as the reaction is endothermic, so by decreasing the temperature, the equilibrium would shift to the left, and by increasing the concentration - more particles = more collisions.

Which of the following changes will lead to the greatest increase in the rate of thefollowing endothermic reaction?
N2(g) + O2(g) → 2NO(g) ∆H +ve

Temperature - Initial concentration of N2 and O2
A decrease by 15% - decrease by 15%
B increase by 15% - stay the same
C decrease by 15% - increase by 15%
D increase by 15% - increase by 15%

The forward reaction is endothermic. Increasing the temperature would therefore, by Le Chatalier's Principle, shift the equilibrium to the right (favouring the endothermic forward reaction), causing the system to absorb more heat. So for temperature, increasing by 15% would increase the rate of the endothermic reaction.

As for initial concentration, you seem to already know why it has to increase by 15%, so yeah :smile:

EDIT: I just saw that it's not an equilibrium lol. In this case, you therefore need to think about the kinetic energy of the reactants - increasing temperature will increase the kinetic energy within the reactant molecules, giving a greater proportion of molecules the activation energy required to react as they collide more powerfully and more often :smile:

(edited 7 years ago)

Reply 549

gabby07

11

What kind of scores are you guys getting in the papers for unit 4 & 5?

Reply 550

drj24

8

Original post by gabby07

What kind of scores are you guys getting in the papers for unit 4 & 5?

Unit 4 generally around 72, haven't done many unit 5 yet

Reply 551

@Ayman! @samb1234, and everyone else, does changing the volume of the container for an equilibrium ever not have an effect on position of equilibrium lol not Kc you silly guy Don Pedro K.? Or will it always shift the equilibrium to the side of fewer molecules when volume is decreased/side of more molecules when volume is increased?

(edited 7 years ago)

Reply 552

OP

Original post by Don Pedro K.

@Ayman! @samb1234, and everyone else, does changing the volume of the container for an equilibrium ever not have an effect on Kc? Or will it always shift the equilibrium to the side of fewer molecules when volume is decreased/side of more molecules when volume is increased?

Kc only changes with temp.

Reply 553

Original post by samb1234

Kc only changes with temp.

Sorry, I meant the position of the equilibrium XD

Reply 554

OP

Original post by Don Pedro K.

Sorry, I meant the position of the equilibrium XD

Well then no, unless I was changing the concs/number of moles as well to compensate

Reply 555

Original post by samb1234

Well then no, unless I was changing the concs/number of moles as well to compensate

So if you had an aqueous equilibrium, with 3 moles of reactants and 5 moles of products,

increasing volume would shift equilibrium to the right and,

decreasing volume would shift equilibrium to the left?

Reply 556

OP