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Aqa chem 4/ chem 5 june 2016 thread

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The equation to calculate Gibbs free energy, ΔG, looks like this:
ΔG = ΔH - TΔS
It is the *changes* in enthalpy and entropy which give us Gibbs free energy, ΔG, and from which we can determine if a reaction is spontaneous or nonspontaneous at a particular temperature.

We need to compute the enthalpy change, ΔH. ΔHrx = ∑ΔHf(prod) - ∑ΔHf(react)
ΔHrx = 2mol(-46.1 kJ/mol) = -92.2 kJ .... the sum of the heats of formation of the reactants are zero because they are in the elemental state.

ΔSrx = ∑ΔSf(prod) - ∑ΔSf(react) = 2 mol(0.1923 kJ/Kmol) - (3 mol(0.1306 kJ/Kmol) + 1mol(0.1915 kJ/Kmol)) = -0.1987 kJ/K

ΔG = ΔH - TΔS
ΔG = -92.2 kJ - 573K(-0.1987 kJ/k)
ΔG = +21.7 kJ
Off Yahoo answers is this wrong or right I think its right?

Reply 2781

Suits101

Original post by palacefloor

For when i drew the compIex i didn't draw arrows to show co-ordinate bonds? I aIso didn't put brackets over the whoIe compIex and just put in brackets (+2 on overaII compIex) on top of the drawing. WouId i Iose marks for this??

Maybe 1 for co-ordinate bonds

Original post by koolgurl14

Eh what i thought electronic configuration will be 3d5 4s2 for the Co 2+ due it being more stable ?

Google it:

Co: [Ar] (or electron configuration) 4s2 3d7
Co2+: [Ar] 3d7 (4s electrons are lost first)

Reply 2782

Parallex

Original post by RME11

"Standard enthalpy change of reaction, ΔH°_r - The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state."

^ The question asked for the standard enthapy change of the reaction given and you were given the correct ratio for the reaction stated, which was 2 SO2 + O2 -> 2SO3 and the standard enthalpy of formation values for these substances so you were expected to use D.H = D.H(Products) - D.H(Reactants) to work out the enthalpy change for the reaction given. The D.H calculated would give you the standard enthalpy change of such a reaction under standard conditions, which is what the question asked for.

Then you worked out the entropy change for the reaction given and put the figures into the gibbs free energy equation to work out the free energy change
for the reaction given.

tl;dr: there was no reason to divide by two, if you didn't you are correct.

Exactly this. It 100% said for the reaction given.

Reply 2783

Cadherin

Hey guys, will they accept [Ar] 3d7 4s2 and [Ar] 3d7 for the electron configurations or did you have to do the full ones??

Reply 2784

Applicant75235

Original post by EverydayHell

Thanks you might be right there the question threw me off a bit because it the said the oxide dissolved in water, but only the mark scheme will tell.

I honeslty don't know you might be right I'm just saying that what i wrote. That first question was basically just an engish test with them trying to trick us

Reply 2785

Cadherin

Original post by RME11

Yay!

Reply 2786

sean.17

Original post by Aethrell

You could draw a skeletal formula for the [Co(H2NCh2Ch2Nh2)3]2+, right?

yes

Reply 2787

Ohnis

Honourable mention to @Suits101 for telling me about the chelate effect :3

Reply 2788

Bailey2j09

Original post by Aethrell

You could draw a skeletal formula for the [Co(H2NCh2Ch2Nh2)3]2+, right?

what would that even look like? - not saying its wrong just dont know how you would do it

Reply 2789

_newbe

Original post by JCleggy

Chemguide had this helpful definition:

Standard enthalpy change of reaction, ΔH°_r
'The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state.'

Alevelchemistry notes say this:
'The standard enthalpy change for a reaction is the heat energy change measured under standard conditions: 100 kPa and a stated temperature (usually 298K).'

and then goes on to further clarify:

'Given a reaction: A + 3B -> 2C + 4D

The standard enthalpy change for this reaction is taken to be the enthalpy change under standard conditions when one mole of A reacts with three moles of B to give two moles of C and four moles of D.'

From this I'm now certain that dividing by 2 was not required for the marks in this question.

(It is also mentioned in these notes that it is convention to leave the answer in kJmol^-1 which means those who did divide by two to get these units have a strong case for also gaining full marks.)

you would be right if it said that but it said formation there was delta Hf not delta Hr

Reply 2790

EverydayHell

Original post by Cadherin

Hey guys, will they accept [Ar] 3d7 4s2 and [Ar] 3d7 for the electron configurations or did you have to do the full ones??

It didn't ask for the full config so you should be ok. :biggrin:

Reply 2791

26december

I didn't put a 2+ or put brackets around the complex when drawing the compound with ethane1-2diamine - do you think I'll lose marks?

Posted from TSR Mobile

Reply 2792

EverydayHell

Original post by 26december

I didn't put a 2+ or put brackets around the complex when drawing the compound with ethane1-2diamine - do you think I'll lose marks?

Posted from TSR Mobile

You do usually lose a mark so no charge I think.

Reply 2793

Cadherin

Original post by Aerosmith

Hilarious how people just cannot accept that the question required you to have an appreciation of the fact that 1 mol is formed when standard 😂😂

You're wrong.

The question asked 'for this reaction' - this is the standard reaction that occurs when OXYGEN is in its standard state (O2) is the one that was given.

Halving the values would be meaningless and would not be for the standard reaction when reagents and products are in their standard states.

Reply 2794

Cadherin

Original post by EverydayHell

It didn't ask for the full config so you should be ok. :biggrin:

I did the shorthand cos effort :P

Reply 2795

emma_1111

Can anyone link me to chem4 unofficial mark scheme? Thank you!!

Reply 2796

Suits101

Original post by Ohnis

Honourable mention to @Suits101 for telling me about the chelate effect :3

It came up!!! :smile:

Reply 2797

EverydayHell

Original post by koolgurl14

I honeslty don't know you might be right I'm just saying that what i wrote. That first question was basically just an engish test with them trying to trick us

Hahaha, that's AQA!

Reply 2798

sean.17

Original post by Ginpls

Q8. First equation was just reacting it with 6 water moleculesP was Cr(H2O)3(OH)3Reagent was anything with OH- ions

NH3 can be used as a reagent here too! :smile:

Reagent for cr3+ to cr2+ is Zn and H+

Hydrogen fuel cell half eqn was

O2 + 4e- + 2H2O ---> 4OH- postive electrode

H2 + 2OH- ---> 2H2O + 2e- negative electrode

Overall: o2 + 2H2 ---> 2H2O

No effect if pressure increases as you are only increasing the pressure of oxygen so the same number of electron are released so same number of redox reactions??? (Not sure though)

Graph was a straight horizontal line

No effect on emf for increase in surface area of pt

Environmental advantage was h2o doesnt contribute to global warming as much as co2 and/or acid rain

For the reagents of Cr3+ to 2+, could you put HCl instead of H+?

Reply 2799

Spectral

Original post by _newbe

you would be right if it said that but it said formation there was delta Hf not delta Hr

The reaction in the contact process is not delta Hf, SO2 and O2 are not 'elements'
It is delta Hr, there is no need to divide by 2, in fact technically it is wrong to divide by 2

As a simple analogy:
For this reaction A + B -> C I'm sure you'd be fine with this being standard
About this one: A + B -> C + 2D Well C is 1 mol, about D?

That's where the argument of 1 mol breaks down. In fact, standard enthalpy of reaction specifically does not specify 1 mol for that very reason - because it's incorrect