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# A-level Chemistry Revision Squad! watch

1. (Original post by MedSchoolDreams)
Hi, How do you balance two half equations when you have water and H+'s on both sides of the two equations after you have balanced their electrons
For example:
1. 5V3+ + 15H2O ------> 5VO3- + 30H+ + 10e-
2. 2MnO4- + 16H+ + 10e- -------> 2Mn2+ + 8H20

Hope you can help
Are you sure uou wrote it correctly, because the number of electrons are not balanced

For the first equation, it +15 on the first side and +20 on the second

Sorry, I submitted an image before but it's not deleting so otherwise it would be correct if the number of electrons were correct and I thought you wrote 5V3 + and not 5V 3+ so excuse this mistake
Attached Images

2. (Original post by NimbleNeil)
Pleasure to come across you again on the Chem forums ;D

Another thing that is irritating me is A, the " pre-exponential factor", again from what i've been able to garner from various videos and websites is that it becomes the Y intercept when you put the whole equation into the y = mx + c form, but I still have no clue as to what it is and how it relates to the rate or activation energy.
A is taken to be constant (actually varies very slightly with temperature but for small changes can be assumed to be constant) and basically takes into account things like the frequency and orientation of collisions.
3. (Original post by NimbleNeil)
Pleasure to come across you again on the Chem forums ;D

I have some quite a long winded question about the Arrhenius equation. I need to be able to work out the Ea of a reaction using the rate of reaction, and the temperature, forming a graph of X axis 1/T ( kelvin) and Y axis ln 1/t ( time, seconds). I think time is used as it is proportional to the rate constant, k.
When you solve for the gradient of the graph, the standard dy / dx, do you need to also solve for the units of the gradient?
My own workings got me s K^-1, but on other websites I have seen units of gradient as being K, i'm just a bit confused about this part of the course generally.
Another thing that is irritating me is A, the " pre-exponential factor", again from what i've been able to garner from various videos and websites is that it becomes the Y intercept when you put the whole equation into the y = mx + c form, but I still have no clue as to what it is and how it relates to the rate or activation energy.

Spec as changed from when I did my alevels but ill explain.

The Arrhenius plot is basically: LnK = LnA - Ea/RT.

Therefore plotting LnK vs 1/T yields a straight line of gradient -Ea/R and an intercept of LnA (applied to y=mx+c)

Things to consider when plotting:
1. T must be in K (obviously)
2. Use a wide range of Temps - otherwise your Ea value will be pretty off.

The main thing to consider is: LnA is rarely the intercept. If you're not plotting from 0 temp (which you probably wont be as 1/T of 293K for example is 0.003413) then you have to use the equation to calculate A.

For example; Lets say you plot your graph, you calculate the gradient (difference in Y/difference in X) to be -4101.

-Ea/R = -4101 (Units = K as its difference in Y/X and Y = no units and X=K-1 so gradient units = K. This makes sense as to calculate Ea we times the graident by -R (units= JK-1mol-1) to give our Ea in J mol-1)
Hence Ea = 34 Kjmol-1

To calculate A from this we put the Ea back into the equation.
Ln(K) = Ln(A) - Ea/RT

^ To do this, you pick any point. Any temp with its corresponding K.
(Remember your value will be Ln(A) so just e^your value to get A)

A basically tells us the value of K at T. Its the measure of species coming together and reacting.

A = Tells us how frequently reactant molecules can collide in the correct orientation.

The -Ea/RT term tells us the fraction of collisions which have the correct activation energy to form products.

Hence the A term and the -Ea/RT (basically the whole equation) tells us how reactants can come together to form products.
As for a collision to be successful (R->P) the molecules must have the correct energy (Ea) and be in the correct orientation.

Hope this helped.

4. (Original post by RMNDK)
Uh oh, we can't see the image.
Try reposting again.
5. (Original post by Zain-A)

The question asks to name this compound, it says the answer is 2-ethoxybutane, but should it not be 2, methyl-ethoxypropane?

Also just noticed this, how tf does the CH2 carbon have 5 bonds?!? Is this question completely wrong or am I missing something major here?

That is seriously weird.
You're right , but to be pedantic with numbers and ordering, I would name that 1-ethoxy-2-methylpropane

This question is completely wrong.
I think they've swapped the CH2 and the CH around. Either way that is definitely NOT butane.

My guess is that this is a printing error, or typing error. The methyl group should actually go on CH. In that case, yes it would be 2-ethoxybutane.
6. (Original post by PlayerBB)
Are you sure uou wrote it correctly, because the number of electrons are not balanced

For the first equation, it +15 on the first side and +20 on the second

Sorry, I submitted an image before but it's not deleting so otherwise it would be correct if the number of electrons were correct and I thought you wrote 5V3 + and not 5V 3+ so excuse this mistake

Yeah it was a made up equation, so that's probably why the electrons are wrong, but you have given me the answer about what to do with the H+ and the H20's so thankyou
7. (Original post by MedSchoolDreams)
Yeah it was a made up equation, so that's probably why the electrons are wrong, but you have given me the answer about what to do with the H+ and the H20's so thankyou
Glad I helped and No problem

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8. (Original post by RMNDK)
Careful..
i.e. should be Na2SO3

Also, how does SO2 act as a base and an acid? It's not amphoteric; only an acidic oxide.
Are you sure you weren't thinking of SiO2..?
yeah my bad! its can only acidic sorry i was thining of silicon dioxide haha
9. Hi guys,

I have made this document with all the reactions of aqueous transition metal ions as part of my revision and I thought I'd share. I am going to use this and get my mum to read out the sub headings and I will recite the equation and the colour of the resulting complex.

Hope this helps some people
Attached Files
10. The reactions of aqueous ions in solution.docx (19.2 KB, 135 views)
11. Why does propan-2-ol have greater enthalpy than butan-2-ol? Does entropy not increase as the molecule size increases?
12. (Original post by RMNDK)

That is seriously weird.
You're right , but to be pedantic with numbers and ordering, I would name that 1-ethoxy-2-methylpropane

This question is completely wrong.
I think they've swapped the CH2 and the CH around. Either way that is definitely NOT butane.

My guess is that this is a printing error, or typing error. The methyl group should actually go on CH. In that case, yes it would be 2-ethoxybutane.
Thanks for the help, one last thing, what does the 1-ethoxy or 2-ethoxy mean? I thought you just write methoxy or ethoxy etc etc and don't use numbers?
13. (Original post by Zain-A)
Thanks for the help, one last thing, what does the 1-ethoxy or 2-ethoxy mean? I thought you just write methoxy or ethoxy etc etc and don't use numbers?
The numbers indicate the carbon atom the group is bonded to.

An ether functional group is just like any other group, like a halogen group or an alkyl group. So treat it it's numbering just like you number halides, or methyl groups or whatever.

It's not like carboxylic acids which don't need the number.

For instance, going back to this

if I put the [O - C2H5] on the middle carbon of the propane (the one with the CH2 that really should be CH), that now becomes 2-ethoxy-2-methylpropane.

So I need to specify the location of the alkoxy group if it potentially can be repositioned elsewhere.
14. (Original post by MedSchoolDreams)
Hi guys,

I have made this document with all the reactions of aqueous transition metal ions as part of my revision and I thought I'd share. I am going to use this and get my mum to read out the sub headings and I will recite the equation and the colour of the resulting complex.

Hope this helps some people
There are some mistakes in your equation of Co2+ with NH3.
You also might want to give the colour of [Co(H2O)6]2+ before it is oxidised by oxygen.

How come you haven't included the equation of Fe3+ with excess OH- ?
The precipitate will dissolve in concentrated NaOH.
15. Can someone help me with this question, i got 0.025 which is D but the answer in the marking scheme is A.thanks
Attached Images

16. and this question too
Attached Images

17. (Original post by djmans)
Can someone help me with this question, i got 0.025 which is D but the answer in the marking scheme is A.thanks
Hi First you need to find the moles of Na2CO3: so we use the equation mass/Mr the Mass is 10.6 and Mr is 106. We divide 10.6 by 106 to get us 0.1 moles.Now that we have the moles we use the volume of the solution to help us find out the concentration: the equation is moles/volume so we divide 0.1 by 0.25 (we need to convert to dm3) to get us a concentration of 0.4.Hope that helps
18. (Original post by haj101)
Hi First you need to find the moles of Na2CO3: so we use the equation mass/Mr the Mass is 10.6 and Mr is 106. We divide 10.6 by 106 to get us 0.1 moles.Now that we have the moles we use the volume of the solution to help us find out the concentration: the equation is moles/volume so we divide 0.1 by 0.25 (we need to convert to dm3) to get us a concentration of 0.4.Hope that helps
thanks i found the molar mass but messed up the next step.
19. (Original post by djmans)
and this question too
82+419+122+(-349) + (-711)= -437
20. (Original post by haj101)
82+419+122+(-349) + (-711)= -437
and also how to identify whether a compound is ionic or covalent or dative
eg: Strontium chloride.etc
21. (Original post by djmans)
and also how to identify whether a compound is ionic or covalent or dative
eg: Strontium chloride.etc
You need to be able to identify the metal and non metal elements in the periodic table,
Ionic- non-metal and metal
Covalent- both non metals
Dative- if the shared pair of electrons come from the same atom

Strontium chloride- strontium is metal and chlorine is non-metal therefore it is ionic

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