HSC Chemistry Question Thread

[amelia20181]

If it's a condensation polymer it most likely means that a water molecule comes off when the monomers join. By looking at the polymer you can see that 4 monomers are joined together (2 of each type and they alternate). If you circle around the first monomer from the first oxygen to the second oxygen on either side of the ring (counting left to right), you can see it repeats again after the second monomer, so where the monomers join must be at the second oxygen (right side of ring). The answer looks like D because when the monomers join you can see it will leave one oxygen when a water molecule forms.

Hopefully that made sense and helped you

But doesn't B also leave one oxygen when a water molecule forms 

[dermite]

Hi there, im struggling with how to do part (ii)
if anyone can help me with this Question on the Equilibrium Constant, it would be much appreciated. 

also, what's the general method of approaching equilibrium constant questions?

[radnan11]

Question regarding Industrial Chem:
Is ammonia considered a raw material for the Solvay Process? Reading from two textbooks with one saying yes,the other saying "sometimes, depending on the situation"

[KT Nyunt]

Hi there, im struggling with how to do part (ii)
if anyone can help me with this Question on the Equilibrium Constant, it would be much appreciated. 

also, what's the general method of approaching equilibrium constant questions?

Ok so the method I like to use to approach equilibrium constant questions is through a RICE table: Ratio (mole ratio) , Initial Concentration, Change in concentration and Equilibrium concentration, as shown in the photo attached. By filling out the information they give us (fyi we have to convert everything into concentrations first) we can find the change in concentration of H_2(g). This change would be proportional to the change in concentration for the other reagents, in their molar ratios. From this, we get all the change in concentrations from the initial concentrations in CO and CH3OH as well, allowing us to calculate the final equilibrium concentration.

Then we use the equilibrium concentrations and plug that into the K expression.

Unfortunately, I'm a bit unsure about the answer I got. I'm a bit confused since the question says the volume was increased (ie. pressure was decreased). According to Le Chatelier's principle, this would cause the equilibrium to shift to the side where there are more gas moles. This would be the reactants side. But my calculations show that K>1, meaning the equilibrium has shift to the right.

Can someone please explain if there's something I'm not getting  

[dermite]

Ok so the method I like to use to approach equilibrium constant questions is through a RICE table: Ratio (mole ratio) , Initial Concentration, Change in concentration and Equilibrium concentration, as shown in the photo attached. By filling out the information they give us (fyi we have to convert everything into concentrations first) we can find the change in concentration of H_2(g). This change would be proportional to the change in concentration for the other reagents, in their molar ratios. From this, we get all the change in concentrations from the initial concentrations in CO and CH3OH as well, allowing us to calculate the final equilibrium concentration.

Then we use the equilibrium concentrations and plug that into the K expression.

Unfortunately, I'm a bit unsure about the answer I got. I'm a bit confused since the question says the volume was increased (ie. pressure was decreased). According to Le Chatelier's principle, this would cause the equilibrium to shift to the side where there are more gas moles. This would be the reactants side. But my calculations show that K>1, meaning the equilibrium has shift to the right.

Can someone please explain if there's something I'm not getting

you subtracted 0.32 instead of 0.36. Is there a reason for this?

[KT Nyunt]

you subtracted 0.32 instead of 0.36. Is there a reason for this?

I subtracted 0.32 because that was difference in the concentration between the initial concentration of the H2 and the final concentration I calculated. But, I did misread the question and thought 0.36 were the final amount of moles in the vessel at equilibrium. Therefore, the final equilibrium concentration I calculated was incorrect. 
I'm still unsure on how to fix this though, no matter what I try I'm getting K>1.

I just retried the RICE table after I realised my mistake but got a K value of 15 (2sf). Do you have the answers on you? Maybe we can work backwards.

[dermite]

I subtracted 0.32 because that was difference in the concentration between the initial concentration of the H2 and the final concentration I calculated. But, I did misread the question and thought 0.36 were the final amount of moles in the vessel at equilibrium. Therefore, the final equilibrium concentration I calculated was incorrect. 
I'm still unsure on how to fix this though, no matter what I try I'm getting K>1.

I just retried the RICE table after I realised my mistake but got a K value of 15 (2sf). Do you have the answers on you? Maybe we can work backwards.

here's the answers:
i dont understand why they added instead of subtracted the change.

[KT Nyunt]

here's the answers:
i dont understand why they added instead of subtracted the change.

Ok so they used moles instead of concentrations, which I guess is easier for this question. Then at the end they converted this into concentrations.

So the question says the number of moles of H₂ 'changes' by 0.36 moles. As the question also informs us that the volume is increased, we know from le chatelier's principle that this will cause the equilibrium to shift to the reactants where there are more gas moles.

Hence, the number of moles of H₂ will increase by 0.36 moles. Using molar ratios, we can also deduce that CO will change by 0.18 moles, but since it is also a reactant it will increase by 0.18. Whereas, CH₃OH is a product, so it will decrease by 0.18

 :)I hope that makes sense...

I may still be confused as to why K is >1 but at this point I'm just accepting this lol.

[amelia20181]

do you need an equation or diagram for a question that is 3 marks or more

how do you memorise lots of equations

[KT Nyunt]

do you need an equation or diagram for a question that is 3 marks or more

how do you memorise lots of equations

Generally it is the 4+ mark questions where you'd consider including a diagram or equation, unless the question specifies otherwise. As a general rule, if it is a 4+ mark question and you know an equation that is related to the question, always include that equation.

As for memorising lots of equations... Either memorise it using flash cards and writing it out a couple times or just remember what reactants and products are involved and then just remember you might need to balance it.

Hope this helps! 

[amelia20181]

is a 6 mark question usually a band 6 type of question

[varun.amin]

is a 6 mark question usually a band 6 type of question

Usually, it's the extended response questions that seperate Band 6 students from Band 5 students. Both know their content and can pick up marks in MCQ and Shorter response questions (1-4 marks). The level of detail you put into an extended response and how well you answer it can be the difference between a 3/6 and 6/6. Most Band 6 students I've talked to have practiced the extended response questions quite a few times and have developed a refined way to answer them. From my own experience, they tend to use the same topics in extended response questions, so if you practice them enough you'll nail them.

[amelia20181]

should you use dot points to answer a 7 mark question


what is the chemical equation for the reaction when octyl ethanoate is made by reacting 1-octanol with ethanoic acid using the catalyst, concentrated sulfuric acid in a refluxing mixture

and what property of esters allows them to be used in industry as solvents or thinners



whats the equilibrium reaction for the incomplete ionization of the weak acid acetic acid


is sulfate or hydrogen carbonate amphoteric

[flacko]

Hey
I have been confused about bond breaking and endo/exothermic reactions.
Everywhere I look it states that when bonds break energy is required and when bonds form energy is released. If for example the energy required to break a bond is lower than the energy released by making a bond the reaction is exothermic as heat/energy is released. I understand this, however, doesn't this defy the law of conservation of energy? The elements require a certain amount of energy to break their bonds and release even more energy when bonds are formed. Where is this extra energy coming from? Thanks

[fun_jirachi]

Hey
I have been confused about bond breaking and endo/exothermic reactions.
Everywhere I look it states that when bonds break energy is required and when bonds form energy is released. If for example the energy required to break a bond is lower than the energy released by making a bond the reaction is exothermic as heat/energy is released. I understand this, however, doesn't this defy the law of conservation of energy? The elements require a certain amount of energy to break their bonds and release even more energy when bonds are formed. Where is this extra energy coming from? Thanks

someone correct me if i'm wrong, im using my prelim knowledge here
from my understanding to break bonds you need an energy source ie. activation energy, and the energy released is the energy used to break the bonds + the energy stored in the bonds, idk