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

1. 3 Energy is given out when one mole of gaseous strontium ions is hydrated.Sr2+(g) + aq Sr2+(aq) This reaction is less exothermic than the corresponding reaction for magnesium ions,Mg2+(g), because
1. A the sum of the first two ionization energies of magnesium is more than that ofstrontium.
2. B the lattice energies of magnesium compounds are more exothermic than thelattice energies of corresponding strontium compounds.
3. C the solubility of magnesium hydroxide is less than the solubility ofstrontium hydroxide.
4. D the ionic radius of Mg2+ is less than the ionic radius of Sr2+.
2. (Original post by Don Pedro K.)
Yeah I think it might just be for gases but:

Attachment 548483Attachment 548483548485Attachment 548483548485548500

The (gaseous) is in brackets, which makes me question if it really does only apply to gases.
It's just for gases. Remembering that Kc would always remain constant, if we say that Kc=[A][B]/ [C][D] then if we imagine that we vary the volume so then we can also write that as (a/x)(b/x)/(c/x)(d/x) where a b c d are mole numbers and x is the volume so will cancel and therefore it won't move at all
3. (Original post by samb1234)
Try and set up a hess cycle, and see if that can help us out
I got +157 (thanks btw, I didn't know you had to use hess's cycle to do this
4. (Original post by dental17)
3 Energy is given out when one mole of gaseous strontium ions is hydrated.Sr2+(g) + aq Sr2+(aq) This reaction is less exothermic than the corresponding reaction for magnesium ions,Mg2+(g), because
1. A the sum of the first two ionization energies of magnesium is more than that ofstrontium.
2. B the lattice energies of magnesium compounds are more exothermic than thelattice energies of corresponding strontium compounds.
3. C the solubility of magnesium hydroxide is less than the solubility ofstrontium hydroxide.
4. D the ionic radius of Mg2+ is less than the ionic radius of Sr2+.
5. (Original post by samb1234)
It's just for gases. Remembering that Kc would always remain constant, if we say that Kc=[A][B]/ [C][D] then if we imagine that we vary the volume so then we can also write that as (a/x)(b/x)/(c/x)(d/x) where a b c d are mole numbers and x is the volume so will cancel and therefore it won't move at all
What if don't have (2) reactants and (2) products on either side?
6. (Original post by samb1234)
Try and set up a hess cycle, and see if that can help us out
Dude, could you explain 12(b) on both the IAL and the UK 2015 paper? I think I understand what's going on in the IAL - equilibrium is shifting to the left due to an increase in H+ ions due to the addition of HCl.
7. (Original post by Dinaa)
I got +157 (thanks btw, I didn't know you had to use hess's cycle to do this
Check your signs, it should be -157 but yeah that is the general method (there may be some formula you can remember for specific situations, but imo you're better off just learning how to use hess cycles as then you can build equations for any situation not just the ones you happen to know the results for)
8. (Original post by Ayman!)
Dude, could you explain 12(b) on both the IAL and the UK 2015 paper? I think I understand what's going on in the IAL - equilibrium is shifting to the left due to an increase in H+ ions due to the addition of HCl.
Yeah your reasoning is right for the IAL one. For the UK one, the NaOH will react with the H+ ions removing them from solution so eq will shift to the right to compensate for that. B adds CN- and 3 and 4 add H+ so a
9. (Original post by Don Pedro K.)
What if don't have (2) reactants and (2) products on either side?
Doesn't matter

10. Is this drawn correctly? samb1234
11. (Original post by Dinaa)

Is this drawn correctly? samb1234
assuming the ones on the left are in aq, then yes apart from the arrow should be from mgl2 to the left
12. (Original post by samb1234)
assuming the ones on the left are in aq, then yes apart from the arrow should be from mgl2 to the left
Ooo, thank you!
13. (Original post by Don Pedro K.)
What if don't have (2) reactants and (2) products on either side?
Scratch that it does matter actually, sorry. Easiest way to see the effects is to remember that Kc remains constant and work it out from there
14. Can someone explain the different types of chromatography to me?
15. (Original post by dental17)
3 Energy is given out when one mole of gaseous strontium ions is hydrated.Sr2+(g) + aq Sr2+(aq) This reaction is less exothermic than the corresponding reaction for magnesium ions,Mg2+(g), because
1. A the sum of the first two ionization energies of magnesium is more than that ofstrontium.
2. B the lattice energies of magnesium compounds are more exothermic than thelattice energies of corresponding strontium compounds.
3. C the solubility of magnesium hydroxide is less than the solubility ofstrontium hydroxide.
4. D the ionic radius of Mg2+ is less than the ionic radius of Sr2+.
Hi, I think the answer is D because a larger radius means lattice enthalpy is less negative. Larger radius means it is easier to bond with due to weaker attraction.
16. (Original post by Dinaa)
Hi, I think the answer is D because a larger radius means lattice enthalpy is less negative. Larger radius means it is easier to bond with due to weaker attraction.
(Original post by dental17)
3 Energy is given out when one mole of gaseous strontium ions is hydrated.Sr2+(g) + aq Sr2+(aq) This reaction is less exothermic than the corresponding reaction for magnesium ions,Mg2+(g), because
1. A the sum of the first two ionization energies of magnesium is more than that ofstrontium.
2. B the lattice energies of magnesium compounds are more exothermic than thelattice energies of corresponding strontium compounds.
3. C the solubility of magnesium hydroxide is less than the solubility ofstrontium hydroxide.
4. D the ionic radius of Mg2+ is less than the ionic radius of Sr2+.
It's D because the smaller the ionic radius, the greater the charge density so the stronger the attraction to the lone pairs on oxygens so hence more energy is released
17. (Original post by samb1234)
Yeah your reasoning is right for the IAL one. For the UK one, the NaOH will react with the H+ ions removing them from solution so eq will shift to the right to compensate for that. B adds CN- and 3 and 4 add H+ so a
Thank you - my train of thought is alright then

Just did the UK paper and it seemed similar at first glance, but took quite a different approach with the questions.

I can't intuitively understand 14(b)(ii) though - is it to do with the Maxwell-Boltzmann distribution and how a greater temperature would have more successful collisions?
18. (Original post by Ayman!)
Thank you - my train of thought is alright then

Just did the UK paper and it seemed similar at first glance, but took quite a different approach with the questions.

I can't intuitively understand 14(b)(ii) though - is it to do with the Maxwell-Boltzmann distribution and how a greater temperature would have more successful collisions?
Essentially yeah - we know that an increased temperature is going to increase the rate as more successful collisions, so as you increase the conc of A more and more so if we wrote the rate eq wrt A and use Arrhenius for k then rate= (Ae^-Ea/RT)[A] so we are not just varying the conc of A hence why it is slightly above the straight line predicted
19. can anyone send me the january 2016 a level paper today? it would be really useful
20. What is does the pKin value for an indicator represent? Is it the pH when it changes colour?

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