HSC Chemistry Question Thread

[bluecookie]

Hello, I'm just a bit confused about the ionisation energy of alkali and alkaline metals. My handout sheets say alkaline metals are higher in ionisation energy, yet all elements to the left of the 4th row all have positive oxidation states, meaning they lose electrons to complete the octet rule. When I googled it, it said ionisation energy was the ability of a particle to strip electrons from another atom. None of them strip electrons from other atoms to form compounds but alkaline metals loses more electrons in the process than alkali metals to form compounds and satisfy the octet rule, so how do alkaline metals have a higher ionisation energy than alkali metals?

[RuiAce]

Hello, I'm just a bit confused about the ionisation energy of alkali and alkaline metals. My handout sheets say alkaline metals are higher in ionisation energy, yet all elements to the left of the 4th row all have positive oxidation states, meaning they lose electrons to complete the octet rule. When I googled it, it said ionisation energy was the ability of a particle to strip electrons from another atom. None of them strip electrons from other atoms to form compounds but alkaline metals loses more electrons in the process than alkali metals to form compounds and satisfy the octet rule, so how do alkaline metals have a higher ionisation energy than alkali metals?

I'm actually getting confused at what your point is.

The ionisation energy is the energy required, for another atom, to come take a specified electron FROM the atom in question. The first ionisation energy of potassium is the amount of energy required to take that single valence electron FROM sodium. In other words, it is the amount of energy required to make sodium go from K, to K⁺

Similarly, the first ionisation energy of calcium is the energy required, for another atom, to come take one of the two valence electrons FROM magnesium. It is the amount of energy required to make Mg become Mg⁺
As is the second ionisation energy, the amount of energy required to make Mg⁺ become Mg²⁺

So of course alkaline earth metals such as calcium have a higher ionisation energy than alkali earth metals such as potassium. Alkali metals are more reactive than alkaline earth metals because firstly, they only have one valence electron, not two. Secondly, remember that the electrons are held in orbit of the nucleus, because of the protons. Because something like potassium has less protons (19) in the nucleus than calcium (20), the electrons are held more tightly in alkaline earth metals. So it should be HARDER for that electron to come off an alkaline earth metal, than an alkali metal. Hence why the first ionisation energy would be higher.


Also, I'm not sure how the oxidation states matter here. The oxidation state of any PURE element is ALWAYS 0. You're probably thinking about the more common oxidation states of the ions formed.
That does not matter here though. I don't see how oxidation states relate to ionisation energies.

[bluecookie]

I'm actually getting confused at what your point is.

The ionisation energy is the energy required, for another atom, to come take a specified electron FROM the atom in question. The first ionisation energy of potassium is the amount of energy required to take that single valence electron FROM sodium. In other words, it is the amount of energy required to make sodium go from K, to K⁺

Similarly, the first ionisation energy of calcium is the energy required, for another atom, to come take one of the two valence electrons FROM magnesium. It is the amount of energy required to make Mg become Mg⁺
As is the second ionisation energy, the amount of energy required to make Mg⁺ become Mg²⁺

So of course alkaline earth metals such as calcium have a higher ionisation energy than alkali earth metals such as potassium. Alkali metals are more reactive than alkaline earth metals because firstly, they only have one valence electron, not two. Secondly, remember that the electrons are held in orbit of the nucleus, because of the protons. Because something like potassium has less protons (19) in the nucleus than calcium (20), the electrons are held more tightly in alkaline earth metals. So it should be HARDER for that electron to come off an alkaline earth metal, than an alkali metal. Hence why the first ionisation energy would be higher.


Also, I'm not sure how the oxidation states matter here. The oxidation state of any PURE element is ALWAYS 0. You're probably thinking about the more common oxidation states of the ions formed.
That does not matter here though. I don't see how oxidation states relate to ionisation energies.

Ohh, thank you

I thought the ionisation energy was the likelihood that the atom in question could take an electron from another atom. I think I got it the opposite way around! It all makes sense now, thanks for clarifying

[bsdfjnlkasn]

Hey there,

I'm getting a tad confused with galvanic cells and the list of standard potentials given on the back of the periodic table.

So I understand all the values are given as though the element is undergoing reduction and that you just reverse them when figuring out the cathode's half equation and the associated electrical potential/voltage.

But, when I was writing my study notes I came across something which really confused me and so I was wondering if someone could verify/explain the following dot points to me.

•   The electrode oxidising will always have the lowest SRP
o   Where a non-metal and a metal are present, the metal oxidises
o   Where two non-metals are present, the least reactive oxidises

Thank you!!

[RuiAce]

Hey there,

I'm getting a tad confused with galvanic cells and the list of standard potentials given on the back of the periodic table.

So I understand all the values are given as though the element is undergoing reduction and that you just reverse them when figuring out the cathode's half equation and the associated electrical potential/voltage.

But, when I was writing my study notes I came across something which really confused me and so I was wondering if someone could verify/explain the following dot points to me.

•   The electrode oxidising will always have the lowest SRP
o   Where a non-metal and a metal are present, the metal oxidises
o   Where two non-metals are present, the least reactive oxidises

Thank you!!

The second dot point can be verified by using the table on the data sheet, or checked by some general chemistry knowledge. The table approach is that all the gases are towards the bottom, below all the metals. You won't find a gas somewhere above a metal on there, so you can deduce that the metal is always oxidized and the gas is always reduced.

The general knowledge approach is that, gases would rather gain electrons to have a full outer shell than lose electrons. Remember that oxidation is loss, so the metal would have a greater tendency to lose than the gas.

The third dot point isn't necessary at all. However, you can check the table and note that chlorine is below bromine, so if you form a cell between chlorine and bromine (+ required ions obviously), clearly the bromine will be oxidised. Indeed, bromine is also more reactive than chlorine.

This is also true in general for metals. Zinc is more reactive than copper, and you can check that between these two zinc will be the one getting oxidised.


As for the first one, well this goes completely back to what standard reduction potentials are. The data sheet displays each half equation for reductions, and the corresponding reduction potentials (E^o values). As a rule, we KNOW that the top one would always be the one undergoing oxidation. E.g. between sodium and magnesium, the sodium (found further up) will be undergoing oxidation.

Which makes sense. A more reactive metal will lose its electron more readily. The voltage produced by the cell should be higher if that's the one losing the electron, and giving it to the less reactive metal. From your studies in school, you should be well aware that the metal found closer to the top of the metal undergoes oxidation (and its half equation needs to be flipped).

[bsdfjnlkasn]

Hey there,

I just had a query about the first syllabus dot point for the Production of Materials module which follows: "Construct word and balanced formulae equations of chemical reactions as they are encountered"

Would it really be necessary to include word equations in year 12? Like how often does that come up in the HSC? You would hope the markers don't need the provided reactions spelled out for them and neither do you want us students wasting time writing out chlorine instead of Cl for example. I also don't really want to waste room on my notes including stuff that won't/is unlikely to be assessed so I would really appreciate having that minor issue clarified.

Thank you!!

[RuiAce]

Hey there,

I just had a query about the first syllabus dot point for the Production of Materials module which follows: "Construct word and balanced formulae equations of chemical reactions as they are encountered"

Would it really be necessary to include word equations in year 12? Like how often does that come up in the HSC? You would hope the markers don't need the provided reactions spelled out for them and neither do you want us students wasting time writing out chlorine instead of Cl for example. I also don't really want to waste room on my notes including stuff that won't/is unlikely to be assessed so I would really appreciate having that minor issue clarified.

Thank you!!

Quite rarely. An example of where it appears is actually in the industrial chemistry option, and the reason is because remembering the chemical formulae is a bit too difficult.

But there is no reason for it to be unexaminable. Just because it's rare and usually dominated by the apparently superior balanced formulae equations, doesn't mean it's an impracticality. Plus it is extremely useful to describe general cases such as alkene + water -> alkanol (although you wouldn't if you don't have to).

Spelling, however, they are usually lenient in the HSC. They're not marking you on spelling (that is ridiculous); they are marking you on understanding of the content.

To be honest, IF you know the balanced chemical formulae equations, you should be able to derive the word equations on the spot. There's no good reason to make notes for word equations.

[bsdfjnlkasn]

Hey there,

What basic carbon-chain structures does cellulose have which makes it able to be used in the production of petrochemicals? And how are they helpful?

[RuiAce]

Hey there,

What basic carbon-chain structures does cellulose have which makes it able to be used in the production of petrochemicals? And how are they helpful?

I see carbon atoms everywhere in cellulose. That is it. A basic carbon-chain structure is just a chain of carbons being there. That is not meant to be overanalysed.

Petrochemicals. Things such as ethylene and ethanol, and say octane and polyethylene. They're full of carbon chains. So if we break down cellulose and gather all the carbon chains, we should be able to make other petrochemicals.
(Easiest way to demonstrate how it works - just talk about how you can convert cellulose all the way down to ethanol. Which can be converted to ethylene, and then into a bunch of other stuff.)

How are they helpful? Well right now we produce our petrochemicals from crude oil. What about it? It's a non-renewable resource (will run out).
Cellulose on the other hand? Abundant, and renewable. Meaning we won't easily run out.

[bluecookie]

Potassium reacts with water to form potassium hydroxide and hydrogen gas. When a piece of potassium reacted with 200ml of water, the resulting solution was found to have a concentration of 0.0046 mol L.

What is the mass of the potassium that reacted with the water in the scenario described above?
What volume of hydrogen gas would be produced during the reaction at 25 degrees c and 100 kpa?

[RuiAce]

Potassium reacts with water to form potassium hydroxide and hydrogen gas. When a piece of potassium reacted with 200ml of water, the resulting solution was found to have a concentration of 0.0046 mol L.

What is the mass of the potassium that reacted with the water in the scenario described above?
What volume of hydrogen gas would be produced during the reaction at 25 degrees c and 100 kpa?

Translating the words into a balanced chemical equation we have:
2 K_(s) + H₂O_(l) → 2 KOH_(aq) + H_2(g)

The resulting solution was dissolved in 200 mL of water. Its concentration is given as 0.0046 mol L^-1.
So by using n = CV, we have n_KOH = 0.0046 * 0.2 = 0.00092 mol

The reaction shows that 1 mol of KOH is produced alongside 0.5 mol of H₂
Hence n_H2 = 0.00046 mol

Since we are at 25^oC and 100 kPa, the data sheet tells us that the molar volume is V_M = 24.79 L mol^-1
So by using n = V/V_M we have V = 0.0046 * 24.79 = 0.114034 L

[bsdfjnlkasn]

Hey there,

How often does the HSC ask for literal methods of experiments conducted during the year? Should I spend a long time memorising them (as in the specific measurements and equipment)?

Thank you 

[RuiAce]

Hey there,

How often does the HSC ask for literal methods of experiments conducted during the year? Should I spend a long time memorising them (as in the specific measurements and equipment)?

Thank you 

If the results are quantitative, then they probably will not matter.

However, you must know the experimental method to fine details if you want mark maximisation. As well as some results if they are qualitative.

It is possible to get away with writing a method using a numbered format, I believe (might want to double check with Jake). However, you absolutely need to know your method as it is always possible (sometimes likely) that you will be asked on any random method of the course. They will explicitly say something along the lines of "outline an experiment where you ..."

An experimental method without numeric quantities and not stating what materials are used (e.g. conical flask) can easily show signs of lack of understanding. There's situations where you would prefer a conical flask over a beaker because you're interested in things not spilling out, and measurements may not be required. And you never want to add too much of a substance either (e.g. you need to balance out the amount of alkanol v.s. alkanoic acid for esterification, as per LCP and other factors)

You also need to be able to assess the validity, reliability and accuracy of the experiment you conducted.

All of this is typically a pain to most students. However, understanding the experiments you performed and being able to describe each and every one of them is definitely examinable, and takes considerable time to know fully. Experiments are examined when you least expect it to be.

[bsdfjnlkasn]

Hey there,

When we are calculating the enthalpy of a certain fuel (q=mcAt), why do we use all the water measurements? i.e. it's mass, 4.18 and change in temp?

Thank you 

[RuiAce]

Hey there,

When we are calculating the enthalpy of a certain fuel (q=mcAt), why do we use all the water measurements? i.e. it's mass, 4.18 and change in temp?

Thank you 

q is the energy required to heat the water so that the temperature changes by ΔH. So everything on the RHS is related to the water.