You are Here: Home

# unit 2 help - chem Watch

1. could someone pls explain the two questions below...

for the first (concerning the oxidising agent) why is the answer not B, the oxidising agent is itself reduced and the H2SO4 is reduced from +4 to +2 so why is it wrong - the correct answer is D

as for the sec, increasing the concentration means increase of the number of reacting particles so the whole graph should be lifted upwards - but the answer is no change wth

pls pls helpppppp
Attached Images

2. Hey nice question!

Ok for the first one:

It's H2SO3, not H2SO4 (maybe you got it wrong because you read this wrong?)
If you mean H2SO3, the oxidation state of S in H2SO3:
-We know the oxidation state of H2SO3 is 0 because its not charged
-We know the oxidation state of H is +1
-We know the oxidation state of O is -2
-Therefore with some maths: 2(1) + [Oxidation state of S] + 3(-2) = 0
-[Oxidation state of S] = 4. It's oxidation state does not change in B, it's +4 in HSO3 and +4 in SO2

Second one:

Ea refers to the activation energy of a reaction, not what you might think as the rate of reaction.
The activation energy of a reaction is something that is specific to a reaction, and changing the conditions won't change it. (Even adding a catalyst, the catalyst would just provide an alternative pathway which this pathway would have a lower Ea - still the original reaction has the same Ea)

Hope I helped a bit!
3. (Original post by Spectral)
Hey nice question!

Ok for the first one:

It's H2SO3, not H2SO4 (maybe you got it wrong because you read this wrong?)
If you mean H2SO3, the oxidation state of S in H2SO3:
-We know the oxidation state of H2SO3 is 0 because its not charged
-We know the oxidation state of H is +1
-We know the oxidation state of O is -2
-Therefore with some maths: 2(1) + [Oxidation state of S] + 3(-2) = 0
-[Oxidation state of S] = 4. It's oxidation state does not change in B, it's +4 in HSO3 and +4 in SO2

Second one:

Ea refers to the activation energy of a reaction, not what you might think as the rate of reaction.
The activation energy of a reaction is something that is specific to a reaction, and changing the conditions won't change it. (Even adding a catalyst, the catalyst would just provide an alternative pathway which this pathway would have a lower Ea - still the original reaction has the same Ea)

Hope I helped a bit!
yup, I misread the q for the first , but thanks a lot x

for the sec it didnt ask about the Ea , it asked abt the diagram in general, whivch is supposed to be lifted upwards since the fraction of molecules increase. and the Ea will remain the same
4. Oh omg, I thought it was an energy-level diagram!
It's a maxwell Boltzmann diagram right? If you could take a picture of the diagram I'll give it a second go
5. (Original post by Spectral)
Oh omg, I thought it was an energy-level diagram!
It's a maxwell Boltzmann diagram right? If you could take a picture of the diagram I'll give it a second go
ohhhh so sorry forgot to mention that, here u go
Attached Images

6. For the 2nd one, there would be no change because you are increasing the concentration of the particles. This does not mean you are increasing the number of particles - number of particles stay the same, and they have the same energy therefore there would be no change to the Mb distribution curve.
7. (Original post by hoafanuk)
For the 2nd one, there would be no change because you are increasing the concentration of the particles. This does not mean you are increasing the number of particles - number of particles stay the same, and they have the same energy therefore there would be no change to the Mb distribution curve.
but increasing the conc increases the rate of the rxn, therefore more particles should have energy higher than Ea for more successful collisions, but this graph is contradicting this ........
8. Ah ok I understand now, thanks.

Yeah, if there's one thing to remember from what I write here, is that changing concentrations of reactants will not affect the shape of the Maxwell-Boltzmann curve.

I think the first thing to realise is that the y-axis is 'fraction of molecules with specific energy'. Adding more molecules (i.e. increasing concentration) won't affect a fraction. This is because increasing concentration of reactants doesn't mean that the reactants gain more energy. The fraction of molecules with a specific energy doesn't change.

If it's still troubling you, or if i didn't answer it properly, please say so!
9. (Original post by pondsteps)
but increasing the conc increases the rate of the rxn, therefore more particles should have energy higher than Ea for more successful collisions, but this graph is contradicting this ........
increasing the concentration does not increase the energy of the reactant particles

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 2, 2016
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.