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

OP

Original post by Don Pedro K.

Yeah I think it might just be for gases but:

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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]/ [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

Reply 562

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 :colondollar:

Reply 563

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+.

Help please! Also could someone please help me with the solubility chapter and the main thi

Is the answer D?

Reply 564

Don Pedro K.

9

Original post by samb1234

It's just for gases. Remembering that Kc would always remain constant, if we say that Kc=[A]/ [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?

Reply 565

Ayman!

15

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.

Reply 566

OP

Original post by Dinaa

I got +157 (thanks btw, I didn't know you had to use hess's cycle to do this :colondollar:

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)

Reply 567

OP

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

Reply 568

OP

Original post by Don Pedro K.

What if don't have (2) reactants and (2) products on either side?

Doesn't matter

Reply 569

Is this drawn correctly? @samb1234

Reply 570

OP

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

Reply 571

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! :bigsmile:

Reply 572

OP

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

Reply 573

Whizbox

7

Can someone explain the different types of chromatography to me?

Reply 574

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+.

Help please! Also could someone please help me with the solubility chapter and the main thi

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.

Reply 575

OP

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+.

Help please! Also could someone please help me with the solubility chapter and the main thi

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

Reply 576

Ayman!

15

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 :biggrin:

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?

Reply 577

OP