VCE Chemistry Question Thread

[sweetcheeks]

Hey guys! A bit confused on this question about thermochemical equations, enthalpy etc. (Question 3 from Chapter 2.2 Heinemann Chemistry 2)

The combustion of octane to form carbon dioxide and liquid water can be
written as:
C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(l) ΔH = –5450 kJ mol–1
The combustion of octane to form carbon dioxide and steam can be
written as:
C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(g)


How would the energy released by the combustion of 1 mole of octane to
form steam compare with the energy released by 1 mole of octane to form
liquid water?

Before I answer it, I would like to hear your thoughts. Think about the meaning of ∆H.

[studyingg]

Thank you studyingg for the help  . So in the picture, you gave, since green is what is getting absored, the sample will appear red right? Also, would the observed sample colour depend on the wavelength of colour that passes through the exit slit?

i'm not tooo sure as I haven't looked into that in detail, but from what I know, yes, I think you're correct

[turtlebanana]

Before I answer it, I would like to hear your thoughts. Think about the meaning of ∆H.

∆H tells us how much energy is being absorbed or released during a chemical reaction. So one equation has the product water in liquid state but the other has it in an aqueous state. Now here i'm not sure whether i'm overthinking it or not. If we were to change water from a liquid to a gas that would require energy? Since changing from liquid to a gas is an endothermic process, this means that ∆H is positive, so therefore the ∆H of this equation would be higher than that of the combustion of octane to form carbon dioxide and water? Arghh

[studyingg]

∆H tells us how much energy is being absorbed or released during a chemical reaction. So one equation has the product water in liquid state but the other has it in an aqueous state. Now here i'm not sure whether i'm overthinking it or not. If we were to change water from a liquid to a gas that would require energy? Since changing from liquid to a gas is an endothermic process, this means that ∆H is positive, so therefore the ∆H of this equation would be higher than that of the combustion of octane to form carbon dioxide and water? Arghh

I think you have a good understanding of what ∆H tells u, and from the surface, I guess the answer is somewhat counterintuitive, so dw if you didn't get it straight away. Remember the fact that combustion is an exothermic reaction, where energy is released from the system into the surroundings therefore ∆H is negative, while the process of heating water (until it is converted into steam) is an endothermic reaction where the system absorbs heat energy from the surroundings (the heat source) meaning that ∆H is positive. This means that the change in enthalpy will be lower because although the system lost energy during combustion, it still absorbed energy. This means that the difference between the final energy and the initial energy will be lower (in comparisson to the other situation). For this, think about the energy profile diagram the and the distance between the reactants and the products. As steam will have more energy than water in a liquid state (because it absorbed it during heating) it will be closer to the reactants than the latter.  I think the reason you got confused is because of the negative/positive sign, but for this question you are supposed to just focus on the magnitude of ∆H

[dream chaser]

i'm not tooo sure as I haven't looked into that in detail, but from what I know, yes, I think you're correct

Okay, thanks for the help. Much appreciated. 

[jollyboat]

Hey,

I'm going into units 3/4 next year, and I was wondering if anyone has a list of ions that we should memorise? I never got them completely down last year.

I was also wondering whether we have to memorise a solubility table? I know the SNAPE rule but not the exceptions to it...

Thanks

[Lear]

Going to be completely honest... I didn’t memorise solubility because you just don’t need it. Gravimetric analysis was removed years ago.

As for ions I think you should know
- Charges of elements depending on group
- NO3, SO4, OH

Those are really the main ones. You can figure out most of them from the periodic table in combination with the electrochemical series.

[addict]

Hey,

I'm going into units 3/4 next year, and I was wondering if anyone has a list of ions that we should memorise? I never got them completely down last year.

I was also wondering whether we have to memorise a solubility table? I know the SNAPE rule but not the exceptions to it...

Thanks

You should know solubility rules, though it is not super important. Usually, SNAPE does the trick for VCE level. However, I like the 'ccops chasing cash'n gia' mnemonic a lot better. Just keep in mind that chlorates, chromates and sulfides don't really come up in VCE where solubility is concerned, and that acetate is just ethanoate.

https://academics.utep.edu/Portals/1788/chemistry/revCASHnGIA_SWRM1.pdf

[Yertle the Turtle]

I would just mention a few ions that Lear didn't mention:
NH₄⁺ can be useful, particularly for when you get to acid/base reactions
CH₃COO^-, useful for the same time.
Also don't forget that you need to be confident of the charges on the ions. SO₄^2- is quite important for when you get to galvanic cells, as car batteries are a popular example, and OH^- and H₃O⁺ are both important to know for acid/base section. Balancing of charges in ionic compounds are really important for the galvanic cells section, so you do really need to be confident on them.

[turtlebanana]

I think you have a good understanding of what ∆H tells u, and from the surface, I guess the answer is somewhat counterintuitive, so dw if you didn't get it straight away. Remember the fact that combustion is an exothermic reaction, where energy is released from the system into the surroundings therefore ∆H is negative, while the process of heating water (until it is converted into steam) is an endothermic reaction where the system absorbs heat energy from the surroundings (the heat source) meaning that ∆H is positive. This means that the change in enthalpy will be lower because although the system lost energy during combustion, it still absorbed energy. This means that the difference between the final energy and the initial energy will be lower (in comparisson to the other situation). For this, think about the energy profile diagram the and the distance between the reactants and the products. As steam will have more energy than water in a liquid state (because it absorbed it during heating) it will be closer to the reactants than the latter.  I think the reason you got confused is because of the negative/positive sign, but for this question you are supposed to just focus on the magnitude of ∆H

Thank you so much for the detailed explanation. Much appreciated


Hey guys! So although i've always kinda known what combustion was, and the general equation for it, i've started to try to actually understand it a bit more instead of just knowing it on the surface. My question is why in a combustion reaction, does a fuel need to burn in the presence of oxygen? What is the significance of oxygen here and what affect will it have if there is/isn't enough for the reaction? (Just what in general is the role of oxygen?)

Mod edit: merged posts

[Yertle the Turtle]

Hey guys! So although i've always kinda known what combustion was, and the general equation for it, i've started to try to actually understand it a bit more instead of just knowing it on the surface. My question is why in a combustion reaction, does a fuel need to burn in the presence of oxygen? What is the significance of oxygen here and what affect will it have if there is/isn't enough for the reaction? (Just what in general is the role of oxygen?)

The oxygen is a necessary part of the combustion reaction. Without any of it combustion will not occur, but with less of it things do start to change. I'll use methane as an example:
CH_{4 (g)} + 2O_{2 (g)} --> CO_{2 (g)} + 2H₂O_(g)
This is called the complete combustion of methane, and it occurs when the reagent in excess is the oxygen (in other words there is excess oxygen that can be used) However, if there wasn't enough oxygen, one of the following equations begins to occur:
2CH_{4 (g)} + 3O_{2 (g)} --> 2CO_(g) + 4H₂O_(g)
CH_{4 (g)} + O_{2 (g)} --> C_(s) + 2H₂O_(g)
Now the second equation may look like there is more oxygen involved, but when you consider it as a ratio to the number of moles of CH4, there is actually less oxygen, and this means that instead of CO2, there is actually carbon monoxide and soot being created. This always occurs in the middle of a fire, and this is why ash and dark smoke is formed. Thus the amount of oxygen available actually impacts the combustion reaction that occurs.
Hope this helps

[dream chaser]

Hi Guys,

Are these diagrams correct(in the attachments of this post). The one without the colour filter is just a solution on its own and how we perceive the colour of the solution to be. The other one is a solution used for colorimetry, hence the coloured filter is used. I'm not sure if they are correct though. Could someone check and let me know. Thanks.

I think the one without the colour filter is wrong. Because I think more that than just red will get transmitted and the colours that get transmitted will combine to form one colour which ultimately becomes red. For instance, if yellow and green transmitted through the test tube, it would combine to make blue, whilst the other colours of light got absorbed.

The diagram without colorimetry used would probably work for opaque(not clear objects) like a leaf. A leaf would however absorb all colours of light and reflect green which makes us see leaves as green colour.

Can someone let me know on this. All replies will be much appreciated. Thanks


Hi Guys,

Can someone please explain to me how AAS works. I'm not to sure whether the light leaving the hollow cathode lamp is of a particular wavelength or of all wavelengths of light? If it is of a particular wavelength, how is it determined which wavelength to use?

Mod edit: merged posts. Please consider editing rather than double posting

[sweetcheeks]

Hi Guys,

Can someone please explain to me how AAS works. I'm not to sure whether the light leaving the hollow cathode lamp is of a particular wavelength or of all wavelengths of light? If it is of a particular wavelength, how is it determined which wavelength to use?

In AAS you are analysing the metal content of a sample. For example, you may want to work out the concentration of copper ions in drinking water. When you run the sample through the flame, the copper ions will become copper atoms in gas phase (and some of these copper atoms will then ionise).

The light source is the hollow cathode lamp. It will contain a copper cathode. The lamp is chosen based on the metal you are analysing (whatever metal you are analysing will be cathode). The copper atoms in the lamp are excited and they will emit light (emission spectrum). As the light passes through the flame, the copper atoms in gas phase will absorb this light, due to the principle of the emission/absorption spectrum.

There will be several wavelengths of light emitted by the cathode lamp (it will match the emission spectrum of the metal atoms).

[dream chaser]

In AAS you are analysing the metal content of a sample. For example, you may want to work out the concentration of copper ions in drinking water. When you run the sample through the flame, the copper ions will become copper atoms in gas phase (and some of these copper atoms will then ionise).

The light source is the hollow cathode lamp. It will contain a copper cathode. The lamp is chosen based on the metal you are analysing (whatever metal you are analysing will be cathode). The copper atoms in the lamp are excited and they will emit light (emission spectrum). As the light passes through the flame, the copper atoms in gas phase will absorb this light, due to the principle of the emission/absorption spectrum.

There will be several wavelengths of light emitted by the cathode lamp (it will match the emission spectrum of the metal atoms).

Okay Thanks. Will the copper atoms in the gas phase emit all the light it absorbed? Also, what would the detector measure?
Could someone explain how HPLC works. I understand most parts of it. It is just the absorption and the purpose of the pump part what I don't get. It says in the text book that "In HPLC, the components are usually detected by passing the eluent stream through a beam of ultraviolet (UV) light. Many organic compounds absorb UV
light. When an organic compound passes in front of the beam of light, a reduced signal is picked up by a detector. The amount of light received by the detector is converted into an absorbance measurement and recorded on a chart that moves slowly at a constant speed or on a computer. The resulting trace is called a chromatogram."

When does "passing the eluent stream through a beam of ultraviolet (UV) light" take place? Before or after the eluent interacts with the stationary phase and the injected sample? What is the purpose of the pump? What is the "waste" in the diagram below? How would the organic compounds absorb the light when it is the eluent stream that the beam of ultraviolet light goes through?

Could someone explain it please. Would really appreciate it. A diagram of HPLC is in the attachment of this post. Thanks

Mod edit: merged posts

[Yertle the Turtle]

Hi dream chaser,
In HPLC, the compound is injected into the solid stationary phase. The compound (mobile phase) then gradually moves down the stationary phase column. While it does this, the different types of molecules react to the stationary phase different amounts, and this relates to the length of time that the mobile phase stays in the column. Thus, the different components of the compound exit the column at different times, and go through the UV machine, which analyses the different compounds, and their retention times, as well as the analysis, create a picture of the different compounds, and thus the different components are analysed. That is the entire function of a HPLC machine. Hope this helps