unit 2 help - chem
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
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
(edited 7 years ago)
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!
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!
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!
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 xfor 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
(edited 7 years ago)
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
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
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.
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 ........
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!
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!
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
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