VCE Chemistry Question Thread

[HighTide]

@HopefulLawStudent

Gas liquid chromatography involves heating a sample to turn it into a gas. What happens when you heat a protein?

HPLC involves separation of compounds, without heating them and is used as an alternative for analysis of things that cannot be analysed using GLC.

Could you explain why Thin Layer is not an appropriate answer?

HPLC is just a more appropriate answer. It's used for high molecular mass compounds. So, if you consider proteins, there are many amino acids, so there's a high molar mass. TLC can be used for proteins, but the question asks most appropriate, and so it'd be HPLC.

[HopefulLawStudent]

Wait... we can use TLC for proteins? Why/How?

[vox nihili]

Wait... we can use TLC for proteins? Why/How?

Theoretically you could, but it'd give you absolutely useless information to be perfectly honest.

[HopefulLawStudent]

Gotcha. Thanks all!

[larissaaa_]

2 QUESTIONS:

Q1. 1.364 mol of H₂, 0.682 mol of S₂ and 0.682 mol of H₂S were mixed in a 2.00L container at 550 degrees celcius. At equilibrium the concentration of H₂S was measured as 1.02M. What is the value of the equilibrium constant at this temperature? (we had a full on discussion in class about this bc my teacher got the answer wrong then even he got confused and now I have no idea what I'm doing).

Q2. Consider the reaction at equilibrium: N₂O₄(g) <------> 2NO₂(g); delta H = +57 Kj mol^-1 and K = 3.62 x 10²M at 327 degrees celcius. Calculate the equilibrium constant at 327 degrees celcius for the following reaction: NO₂(g) <--------> 1/2N₂O₄(g).

THANK YOU IN ADVANCE

[sweetiepi]

2 QUESTIONS:

Q1. 1.364 mol of H₂, 0.682 mol of S₂ and 0.682 mol of H₂S were mixed in a 2.00L container at 550 degrees celcius. At equilibrium the concentration of H₂S was measured as 1.02M. What is the value of the equilibrium constant at this temperature? (we had a full on discussion in class about this bc my teacher got the answer wrong then even he got confused and now I have no idea what I'm doing).

Q2. Consider the reaction at equilibrium: N₂O₄(g) <------> 2NO₂(g); delta H = +57 Kj mol^-1 and K = 3.62 x 10²M at 327 degrees celcius. Calculate the equilibrium constant at 327 degrees celcius for the following reaction: NO₂(g) <--------> 1/2N₂O₄(g).

THANK YOU IN ADVANCE

Correct me if I'm wrong but for Q2, The K_c for the reverse reaction is calculated by 1/K _c . It is then halved, so it would be K_c value you just calculated, divided by 2. This means K_c= 1/3.62*10² = 2.76*10^-3. And then divide this by 2: 2.76*10^-3 /2 = 1.38*10^-3

(I'm in the middle of solving Q1 too )

Edit:: should defs be sqrt(2). Therefore, listen to Swag, as the answer should definitely be 5.27*10^-2

[Swagadaktal]

Correct me if I'm wrong but for Q2, The K_c for the reverse reaction is calculated by 1/K _c . It is then halved, so it would be K_c value you just calculated, divided by 2. This means K_c= 1/3.62*10² = 2.76*10^-3. And then divide this by 2: 2.76*10^-3 /2 = 1.38*10^-3

(I'm in the middle of solving Q1 too )

I think if you work it out mathematically
Kc of the first reaction = (1/Kc 2nd reaction)^2

So to get Kc second reaction you divide the initial by 1 (coz its reversal) and square root...
Gathering a Kc of 5.26*10^-2

This is the time where someone from the legit chem squad should step in

[sweetiepi]

I think if you work it out mathematically
Kc of the first reaction = (1/Kc 2nd reaction)^2

So to get Kc second reaction you divide the initial by 1 (coz its reversal) and square root...
Gathering a Kc of 5.26*10^-2

This is the time where someone from the legit chem squad should step in

Correct me if I'm wrong but for Q2, The K_c for the reverse reaction is calculated by 1/K _c . It is then halved, so it would be K_c value you just calculated, divided by 2. This means K_c= 1/3.62*10² = 2.76*10^-3. And then divide this by 2: 2.76*10^-3 /2 = 1.38*10^-3

(I'm in the middle of solving Q1 too )

Damn it! Forgot about the sqrt(2) mbmb.

[larissaaa_]

I think if you work it out mathematically
Kc of the first reaction = (1/Kc 2nd reaction)^2

So to get Kc second reaction you divide the initial by 1 (coz its reversal) and square root...
Gathering a Kc of 5.26*10^-2

This is the time where someone from the legit chem squad should step in

Wait woah this will probably be really stupid but where did the square roots come from

Correct me if I'm wrong but for Q2, The K_c for the reverse reaction is calculated by 1/K _c . It is then halved, so it would be K_c value you just calculated, divided by 2. This means K_c= 1/3.62*10² = 2.76*10^-3. And then divide this by 2: 2.76*10^-3 /2 = 1.38*10^-3

(I'm in the middle of solving Q1 too )

Edit:: should defs be sqrt(2). Therefore, listen to Swag, as the answer should definitely be 5.27*10^-2

[zsteve]

larissaaa_, basically here if you half the coefficients of a chemical reaction, the value of K will be square-rooted, because all the indices of the concentration terms will be halved (hence square root)

[sweetiepi]

Wait woah this will probably be really stupid but where did the square roots come from

It comes from when you halve the molar ratio of the equation itself. (Whereas if you double it, you square the K_c) Hope this helps

(Also I'm not having much luck with Q1, but I believe that it may have something to do with the ICE tables method I'll let the legit chem team answer it )

(so basically what zsteve said )

[keltingmeith]

Wait woah this will probably be really stupid but where did the square roots come from

Let's go first principles, for funsies! For the chemical reaction N2O4 -----> 2NO2, the equilibrium constant K is given by:



For the chemical reaction NO2 ------> (1/2)N2O4, the equilibrium constant K is given by:



So, if we do some algebraic manipulations, we'll get the following:



Hopefully that clears things up a bit! This area of study requires you understand some basic exponential laws to really piece this stuff together, so it might seem a bit complex if you're not doing methods. If that's the case, it might help to learn the basics from Khan Academy!

[larissaaa_]

It comes from when you halve the molar ratio of the equation itself. (Whereas if you double it, you square the K_c) Hope this helps

(Also I'm not having much luck with Q1, but I believe that it may have something to do with the ICE tables method I'll let the legit chem team answer it )

larissaaa_, basically here if you half the coefficients of a chemical reaction, the value of K will be square-rooted, because all the indices of the concentration terms will be halved (hence square root)

(so basically what zsteve said )

Let's go first principles, for funsies! For the chemical reaction N2O4 -----> 2NO2, the equilibrium constant K is given by:



For the chemical reaction NO2 ------> (1/2)N2O4, the equilibrium constant K is given by:



So, if we do some algebraic manipulations, we'll get the following:



Hopefully that clears things up a bit! This area of study requires you understand some basic exponential laws to really piece this stuff together, so it might seem a bit complex if you're not doing methods. If that's the case, it might help to learn the basics from Khan Academy!

Got it thank you all so much cleared it up heaps!

[HopefulLawStudent]

A pure substance was suspected of being the high molecular weight anabolic steroid Stanozolol. A sample of this substance was dissolved in a suitable pure solvent and injected into a gas chromatograph. How many peaks would you expect to see in the resulting chromatogram? Justify your result. [2 mark]

According to the answers (that only tell me what the answer isn't): If you responded with ‘only one peak’ and then asserted that there was only one substance present, award yourself one mark.

a) Why is the answer not "only one peak"?

[Hopelesshopefull]

A pure substance was suspected of being the high molecular weight anabolic steroid Stanozolol. A sample of this substance was dissolved in a suitable pure solvent and injected into a gas chromatograph. How many peaks would you expect to see in the resulting chromatogram? Justify your result. [2 mark]

According to the answers (that only tell me what the answer isn't): If you responded with ‘only one peak’ and then asserted that there was only one substance present, award yourself one mark.

a) Why is the answer not "only one peak"?

There would be a peak for the solvent and one for the substance.
Hope this helps