VCE Chemistry Question Thread

[#J.Procrastinator]

With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (aq) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?

[psyxwar]

With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (s) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?

You don't include glucose or water in the formulation of K (edit: glucose is aqueous for photosynthesis so you do include it)

This is because the equilibrium constant actually uses the activity of the species rather than their molarity, and the activity of pure solids and liquids is defined to be one.

Further reading if you're keen, though it's not really relevant to VCE:

http://en.wikipedia.org/wiki/Thermodynamic_activity

Molarity approximates activity so we use it for simplicity's sake.

[Rishi97]

With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (s) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?

As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks

[psyxwar]

As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks

Oh yeah my bad, yeah glucose is aqueous!

[#J.Procrastinator]

You don't include glucose or water in the formulation of K (edit: glucose is aqueous for photosynthesis so you do include it)

This is because the equilibrium constant actually uses the activity of the species rather than their molarity, and the activity of pure solids and liquids is defined to be one.

Further reading if you're keen, though it's not really relevant to VCE:

http://en.wikipedia.org/wiki/Thermodynamic_activity

Molarity approximates activity so we use it for simplicity's sake.

As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks

Oh thank you guys! Yes, whoops, glucose is in it's aqueuos state !

[alchemy]

Question attached.

Another question: if I was asked "what features might an infrared spectrum of propanoic acid have?" Without being provided any data, how would I answer it? The answers in the book actually refer to numerical values and I don't know how they got them.

[Rishi97]

Question attached.

Another question: if I was asked "what features might an infrared spectrum of propanoic acid have?" Without being provided any data, how would I answer it? The answers in the book actually refer to numerical values and I don't know how they got them.

Hi

The numbers are from the data book. Check the data book and based on your knowledge of what propanoic acid looks like, you will be able to find the wavelengths of the features such as a COOH group etc.

[alchemy]

Hi

The numbers are from the data book. Check the data book and based on your knowledge of what propanoic acid looks like, you will be able to find the wavelengths of the features such as a COOH group etc.

Oh ok. What about the question I attached? Were you able to get the answer to that as well?

[keltingmeith]

Yep, that's all in there.

Have you got a copy of the databook, alchemy? If not, I suggest printing off a copy from a VCAA exam, and keeping hold of it whenever you do any question from now on. If you come across something you don't have enough information for, see if there's something in the data book that can help you. Learning that book inside and out now can help you become very quick at pulling information out of it later in your exam. Last year, my teacher gave us all a copy, and we had to use it in all SACs, practice exams, practice questions, etc. and believe you and me - it certainly helped in the long run.

[TimewaveZero]

Hey guys, just wondering if anyone knows of a quick way to deduce weather a fatty acid is saturated/unsaturated without drawing it by looking at the number of carbon atoms compared to the number of hydrogen atoms?

I have come across a number of questions where your are required to decide which of two or more fatty acids would create a saturated /unsaturated fat.

For example;

C17H35-C=OOH

or

C16H31-C=OOH

Also, I just came across a question asking how many double bonds in a particular fatty acid. How can one devise this from a formula without drawing the entire molecule out?

[soNasty]

A saturated fatty acid has the formula

every double bond is 2H less than the saturated fatty acid

So for example, if we have
the saturated version of that would be   <--- 35 came from doubling 17, and adding 1.
Since its 31H rather than 35H, there are 2 double bonds.

[TimewaveZero]

Thankyou!! Much appreciation.

[ETTH96]

Hey guys, can someone help me out with this?

300ml of 1.5 M HCl is mixed with 300 ml of 1.0 M NaOH in a thermally insulated container. the initial temperature of each solution before mixing was 18.
They react according to the equation:
H+ (aq) + OH- (aq) --> H2O(l)    Enthalpy=-56 kJmol-1

if 2.5 kJ is required to raise the temperature of 600ml of water by 1.0^oC, then the final temperature, in degrees celsius, of the solution after mixing would be about:
a) 10
b) 25
c) 28
d) 40

[Yacoubb]

Hey guys, can someone help me out with this?

300ml of 1.5 M HCl is mixed with 300 ml of 1.0 M NaOH in a thermally insulated container. the initial temperature of each solution before mixing was 18.
They react according to the equation:
H+ (aq) + OH- (aq) --> H2O(l)    Enthalpy=-56 kJmol-1

if 2.5 kJ is required to raise the temperature of 600ml of water by 1.0^oC, then the final temperature, in degrees celsius, of the solution after mixing would be about:
a) 10
b) 25
c) 28
d) 40

Is the answer B? If it is, then here's the explanation.

Firstly, the energy released in this reaction:
n(NaOH) = 0.3 * 1 = 0.3 mol
0.3mol * 56kJ/mol = 16.8 kJ
So, 16.8kJ of energy will be released in this reaction (theoretically of course).

16800 J= 4.184 * 600 * change in temp
change in temp = 6.7 = 7

Change in Temp = Max Temp - Min Temp
7 = x - 18
x = 18 + 7
x = 25 degrees C, hence B.

[hyunah]

hello
how would i do this question 14c) and 14d) i know 14 a and b are increase
for 14c) if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
for 14d) if 14c) is in fact ph decrease, wouldnt that mean the water becomes acidic, but then i know that h30+ concerntration = oh- concerntration therefore it has to be neutral, but then the ph decreased?

so confused?