VCE Chemistry Question Thread

[Bri MT]

What is the difference between the q in the formula q = mass x heat of combustion and q = mass x specific heat capacity x change in temp? Would you get the same answer with both formulas. My understanding is that the q in q = mcdeltaT is the energy transferred to the water/absorbed by water and q in q = m x Hc is the energy released by the combustion of the fuel.

q represents the energy change yeah - if something is absorbing energy (being heated) then q is how much energy is going in. That energy is then spread out amongst the substance (hence m being part of the equation) and depending on the substance it'll take a different amount of energy to change 1 gram by one degree (this is what c tells us), and the delta T then refers to the temperature change. So we often rearrange is like q/(mc) = [delta]T - the amount of energy going in divided by how much there is and how hard it is to heat up


If you look at q= mH it's job is a bit different. H says, ok, I've got a substance that produces   this much energy per unit. So then if you have m telling you how many units, you can go do     energy/(number of units)  x number of units . The number of units cancels out, and you're left with the amount of energy. Keep in mind that this m is different to the m in q=mc[delta]T. The m in q=mH belong to the thing you're using to get energy (eg fuel you're burning). The m in q=mc[delta]T belongs to the thing that has its temperature being changed


Hope this helps

[huity]

This has been hinted at by miniturtle and Yertle the Turtle, so to make it super clear:
q=m*c*delta T is mostly used for water (where water's c value is given in the data book). However, don't forget that you can actually use q=m*c*delta T for substances other than water, especially if the substance's c value is given. It's rare, but I've seen it appear in practise exams (I believe from VCAA) 

[JR_StudyEd]

Is Cambridge Checkpoints helpful for Chemistry?

[huity]

Is Cambridge Checkpoints helpful for Chemistry?

Some of my friends swore by Checkpoints because it meant that they could do VCAA questions in preparation for SACs. Plus, it's sorted out neatly by topic! Having said that though, some also went through every single VCAA exam and picked out the relevant questions as SAC preparation. Honestly up to you, but the key here is that both methods focus on completing VCAA questions throughout the year (not only in the frantic lead-up to exams) 

[Yertle the Turtle]

Is Cambridge Checkpoints helpful for Chemistry?

I don't know about other people, but I certainly found it useful, though I didn't use it enough. The layout is the main drawback, but otherwise it's good.

[dream chaser]

Hi Guys,

Need clarification with this question. Wouldn't the reactant that has been oxidised be NH4+(aq) and the reactant being reduced is CR2072-(aq)? The answer in the book just says that nitrogen has been oxidised and chromium has been reduced.

Question is in the attachment. All help will be much appreciated. Thanks

[peter.g15]

Hi Guys,

Need clarification with this question. Wouldn't the reactant that has been oxidised be NH4+(aq) and the reactant being reduced is CR2072-(aq)? The answer in the book just says that nitrogen has been oxidised and chromium has been reduced.

Question is in the attachment. All help will be much appreciated. Thanks

I think that it's important to make a distinction between the states in this question since the question states that it is solid ammonium dichromate being decompsed (not aqueous) - just pay attention since examiners/teachers cannot give marks if states are incorrect.

For the oxidation numbers and what has been oxidised and reduced. You should try to label the oxidation numbers of each element within the substance rather than whole ions. For example, breaking up ammonium into its components - hydrogen and nitrogen, and calculating their individual oxidation numbers. In this case, the oxidation number of nitrogen is -3 and the oxidation number of chromium is +6. Therefore, when comparing it to the oxidation numbers in the products, nitrogen has been oxidised (lost electrons and become more positive) whereas chromium has been reduced (gained electrons and become more negative).

Hope that helped!

[dream chaser]

I think that it's important to make a distinction between the states in this question since the question states that it is solid ammonium dichromate being decompsed (not aqueous) - just pay attention since examiners/teachers cannot give marks if states are incorrect.

For the oxidation numbers and what has been oxidised and reduced. You should try to label the oxidation numbers of each element within the substance rather than whole ions. For example, breaking up ammonium into its components - hydrogen and nitrogen, and calculating their individual oxidation numbers. In this case, the oxidation number of nitrogen is -3 and the oxidation number of chromium is +6. Therefore, when comparing it to the oxidation numbers in the products, nitrogen has been oxidised (lost electrons and become more positive) whereas chromium has been reduced (gained electrons and become more negative).

Hope that helped!

Thanks peter.g15 for your reply. So basically you are saying that since ammonium dichromate is solid, we look at the individual elements inside the substance that have been oxidised and reduced. If ammonium dichromate was aqueous, then how I answered would have been correct?

Also, Is the difference between the heat of combustion of a substance and the molar heat of combustion of a substance the fact that the molar heat of combustion of a substance having a positive value whilst the heat of combustion of a substance having a negative value?

[peter.g15]

Thanks peter.g15 for your reply. So basically you are saying that since ammonium dichromate is solid, we look at the individual elements inside the substance that have been oxidised and reduced. If ammonium dichromate was aqueous, then how I answered would have been correct?

Also, Is the difference between the heat of combustion of a substance and the molar heat of combustion of a substance the fact that the molar heat of combustion of a substance having a positive value whilst the heat of combustion of a substance having a negative value?

Sorry if that was confusing, the states bit was just an unrelated note that is important to take into consideration in all questions since it can cost a mark or two in assessments. For redox reactions, you'll just always need to look at individual elements rather than larger ions such as ammonium or dichromate (their whole charges are still useful for determining the individual charges of elements). Does that make sense?

Not quite... If you look at table 11 on the VCAA data booklet (the one with fuels and their heats of combustions), you'll see that it includes 'heat of combustion' and 'molar heat of combustion'. All the figures are positive in this table and the only difference is the units. Heat of combustion is given as kJ/g while molar heat of combustion is given as 'kJ/mol'. You'll need to make these figures negative (and appropriate to your equations mole ratio) in thermochemical equations. Otherwise, they can remain positive.

Hope that helps and let me know if you need more clarification

[Monkeymafia]

galvanic cells questions:
- how do we know what possible solutions could be used in each half cell?
- how do we know what materials the electrodes should be made of?

Thanks!

[sweetcheeks]

galvanic cells questions:
- how do we know what possible solutions could be used in each half cell?
- how do we know what materials the electrodes should be made of?

Thanks!

There are several considerations to make when determining what solutions to use for a half-cell
-You want a solution that isn't going to interfere with the desired reaction occurring. For example, if you have a zinc-copper galvanic cell, you don't want to use a copper sulfate solution in the zinc electrode, as you will get a spontaneous reaction within that half-cell and you can't utilise the energy.
-You want it to be a strong electrolyte. This means that you need a substance that is highly soluble and highly dissociates (forms ions). You don't want to use an electrolyte that may form insoluble precipitates (it can coat the electrode, making it unable to function and/or it reduces the ions concentration in solution).
-The anion needs to be inert.
-Factors such as cost, toxicity, availability are also considerations to make.

Ideally you use a solution that contains the substance associated with that electrode, along with an inert ion such as sulfate. Most commonly, copper (ii) sulfate, zinc sulfate are used. Nitrates can be used for some reactions, however they are powerful oxidising agents and under the right conditions, they can react and form NOx.

As for the electrode:
-It needs to be capable of conducting electricity
-It should be unreactive to the electrolyte used in the half cell (e.g. you wouldn't put a zinc electrode in a copper half-cell)
-For a zinc half-cell you need to use a zinc electrode, because otherwise you wouldn't have a source of oxidisable zinc (assuming you are using it as the oxidation half-cell)
-For substances such as iodine, which can't be used as electrodes (doesn't conduct electricity), or solutions (e.g. Fe2+/Fe3+ half cell), an inert electrode, such as graphite or platinum can be used.
-Factors such as surface area, cost, overpotential (related to activation energy) are also considerations to make.

[EllingtonFeint]

Hey,
I’m a bit confused about how to tackle this question...
Could somebody maybe give me a few hints about which formula to start with or some kinda explanation please?
Thank you

[EllingtonFeint]

I’m also stuck on the next question.
My answer is wrong.
I’ve attached the question as well as my working out.
Please tell me if you can spot what my errors are.
Thank you.

[sweetcheeks]

Hey,
I’m a bit confused about how to tackle this question...
Could somebody maybe give me a few hints about which formula to start with or some kinda explanation please?
Thank you

For this question, start by working out how many moles of ethane is required to be combusted.

I’m also stuck on the next question.
My answer is wrong.
I’ve attached the question as well as my working out.
Please tell me if you can spot what my errors are.
Thank you.

It appears that you have made a mistake with the stoichiometry/ratio. 2 mole of octane should release 10900 kJ. In your working the 2.19/2 = x/5772 implies that 2 mole of octane only releases 5772 kJ, rather than 10900 kJ. You also show in your equation, as well, that the ∆H = -5772 rather than -10900 (is this a mistake in copying it?).

[EllingtonFeint]

For this question, start by working out how many moles of ethane is required to be combusted.

Rightttt, OKAY Yesss, thank you! That worked 

It appears that you have made a mistake with the stoichiometry/ratio...(is this a mistake in copying it?).

Ohmigosh I feel so utterly dumb. Yes you are absolutely right!  I just copied it wrong!  Ahhh I legit spent like 20 minutes trying figure out what was wrong. Thank you xx