VCE Chemistry Question Thread

[Sundal]

I don't really understand this question:

Draw an energy profile diagram for this reaction: 2C4H10(g) + 13O2(g) --> 8CO2(g) + 10H2O(l);  Delta H = –5772 kJ mol–1.
Now, describe how your diagram would change for the equation: C4H10(g) +  O2(g) -->  4CO2(g) + 5H2O(l).

The answer is: Activation energy and Delta H values are halved.

I don't understand why the energy of the reactants and the energy of the products are also not halved?

[Sundal]

For this question: Explain why surface properties are important to the operation of catalysts.

The answer is: Reactions involving a heterogeneous catalyst take place at the surface of the catalyst. Reactants form bonds with the catalyst, lowering the activation energy of reactions and allowing them to proceed more rapidly.

Would it then be correct to say that surface properties are important in the operation of catalysts only when it is a heterogeneous catalyst?

Cheers!

[thushan]

Not necessarily a heterogeneous catalyst, but any solid catalyst. Only the outside surface of solid crystals can actually participate in catalysis.

[angelrox00]

Hello everyone, when drawing calibration graphs, and after you have plotted all the different known values, do you draw the line of best fit or join up the dots? Up until now, I thought it was a line of best fit, but in a practice exam I did they joined up the values? (and obtained a non-linear graph)


Thanks very mch
~May god bless
Angel Raju

[sunshine98]

So, I have a question :
Calculate the heat energy for P4(s)-->P4(g)
If P4(s)+5O2(g)-->P4O10(s)           DELTA H= -3005kjmol^-1
AND
 P4(g)+5O2(g)-->P4O10(s)             DELTA H= -3018kjmol^-1
btw don't have the answers for this , just wanted to know if my answer and method was right
So I was thinking because the products are the same there going to have the same enthalpy, and because the reactants have both O2 the effect of oxygen can sort of be disregarded , so subtract the two Delta Hs because you will be basically subtracting the difference between the reactants (that is the P4's )and hence my answer was -13kjmol^-1. Is this correct?
Thanks 

[knightrider]

What does it mean when it says polar molecules ?

How can you tell if a molecule is polar or non polar ?

[grannysmith]

Hello everyone, when drawing calibration graphs, and after you have plotted all the different known values, do you draw the line of best fit or join up the dots? Up until now, I thought it was a line of best fit, but in a practice exam I did they joined up the values? (and obtained a non-linear graph)


Thanks very mch
~May god bless
Angel Raju

A line of best fit is strictly linear; anything else is a curve of best fit.

So, I have a question :
Calculate the heat energy for P4(s)-->P4(g)
If P4(s)+5O2(g)-->P4O10(s)           DELTA H= -3005kjmol^-1
AND
 P4(g)+5O2(g)-->P4O10(s)             DELTA H= -3018kjmol^-1
btw don't have the answers for this , just wanted to know if my answer and method was right
So I was thinking because the products are the same there going to have the same enthalpy, and because the reactants have both O2 the effect of oxygen can sort of be disregarded , so subtract the two Delta Hs because you will be basically subtracting the difference between the reactants (that is the P4's )and hence my answer was -13kjmol^-1. Is this correct?
Thanks 

Yes, however -3005 - (-3018) = +13. Intuitively, solid-->gas requires an input of energy.

What does it mean when it says polar molecules ?

How can you tell if a molecule is polar or non polar ?

A polar molecule is one in which there is an unequal distribution of (negative) charge. The geometry of a molecule, as well as the electronegativity differences of the atoms largely contribute to the polarity of a molecule.

However, we commonly talk about polar molecules as those which are able to be dissolved in a particular solvent, like water. This can be determined by size, shape, and functional groups.

This is a highly simplistic description so I'd encourage further reading here: http://www.chemguide.co.uk/atoms/bonding/electroneg.html

[nerdgasm]

So, I have a question :
Calculate the heat energy for P4(s)-->P4(g)
If P4(s)+5O2(g)-->P4O10(s)           DELTA H= -3005kjmol^-1
AND
 P4(g)+5O2(g)-->P4O10(s)             DELTA H= -3018kjmol^-1
btw don't have the answers for this , just wanted to know if my answer and method was right
So I was thinking because the products are the same there going to have the same enthalpy, and because the reactants have both O2 the effect of oxygen can sort of be disregarded , so subtract the two Delta Hs because you will be basically subtracting the difference between the reactants (that is the P4's )and hence my answer was -13kjmol^-1. Is this correct?
Thanks 

I like your thinking, but unfortunately I think the sign of your answer is incorrect. Using the 'standard' way of working this out, we keep the first equation the same, and reverse the second equation to get:

P4(s) + 5O2(g) ---> P4O10(s); Delta H = -3005kJ/mol (Equation 1)
P4O10(s) --> P4(g) + 5O2(g) ; Delta H = +3018kJ/mol (Reversed Equation 2)

Then if we 'add' the two equations together (or alternatively, consider performing the first reaction to get some P4O10(s), then perform the second reaction to change that P4O10(s) to give P4(g) + 5O2(g)), we get P4(s) + 5O2(g) ---> P4(g) + 5O2(g); Delta H = +13kJ/mol. As 5O2(g) appears on both sides, this means we've effectively used a certain amount of O2(g) to make the exact same amount of O2(g) - so this would have the same enthalpy and as you've mentioned, can be disregarded. Hence, we end up with P4(s) ---> P4(g); Delta H = +13kJ/mol.

In some sense, you can 'subtract' equations from each other, as reversing an equation turns it into its 'negative', and adding the negative of something is like subtracting the original 'something'. In this case, we 'added' the negative of Equation 2, to Equation 1. This would be the same as Equation 1 - Equation 2, which again gives you the net enthalpy change of +13kJ/mol.

I suppose one final way to look at it, would be to draw an energy profile of the two (original) reactions on the same axes. Because the products of the two reactions are exactly the same, you can put one level for P4O10(s). Now, P4(s) + 5O2(g) will be 3005kJ/mol above this level (note the reaction is exothermic, so the enthalpy of the product is lower than that of the reactants), and P4(g) + 5O2(g) will be 3018kJ/mol above this level. Hence, the P4(s) + 5O2(g) level is 13kJ/mol below the P4(g) + 5O2(g) level. As 5O2(g) makes the same enthalpy contribution to both levels, we can state that the hypothetical P4(s) level will be 13kJ/mol below the P4(g) level. Hence, if we were to draw a reaction profile of P4(s) --> P4(g), we would have the product 13kJ/mol above the reactants, and hence delta H would be +13kJ/mol.

[angelrox00]

So is a calibration curve always linear? (So you always draw a line of best fit?)

Also, in the attached question, why is the answer not A. As you are always trying to get the functional group on the lowest carbon no, shouldn't it be A? (The answer is D) In VCAA 2008 Exam 1, there was a similar question (Q6) and methyl was prioritised over a chlorine funtional group when numbering the carbon attoms.

[dankfrank420]

So is a calibration curve always linear? (So you always draw a line of best fit?)

Also, in the attached question, why is the answer not A. As you are always trying to get the functional group on the lowest carbon no, shouldn't it be A? (The answer is D) In VCAA 2008 Exam 1, there was a similar question (Q6) and methyl was prioritised over a chlorine funtional group when numbering the carbon attoms.

Perhaps the priority of the -OH group has something to do with it?

[BakedDwarf]

So is a calibration curve always linear? (So you always draw a line of best fit?)

Also, in the attached question, why is the answer not A. As you are always trying to get the functional group on the lowest carbon no, shouldn't it be A? (The answer is D) In VCAA 2008 Exam 1, there was a similar question (Q6) and methyl was prioritised over a chlorine funtional group when numbering the carbon attoms.

Functional groups are prioritised over one another:
The order of functional groups from highest priority to lowest priority is carboxylic acid, alcohol and amine (chloro, or any halogen, is never a main functional group). Hence, if there's a methyl group, it is prioritised over chloro; that is, it receives the lowest carbon number.

However, in the diagram for your question, there is two functional groups, alcohol and chloro, and a methyl group. So, as I said above, the alcohol must receive the lowest carbon number, 4, because it has a higher priority than the methyl and chloro. And then the methyl group receives the lowest carbon number, 3, as chloro is not prioritised over anything. Therefore, chloro receives the next lowest possible carbon number, 7.

I know this is probably confusing because I haven't explained it clearly, but what you should understand is that certain functional groups are prioritised over one another. If you're still unsure, hopefully someone else with clearer knowledge can explain it in a more understandable way.

[angelrox00]

Ok. Thanks guys.

[knightrider]

A polar molecule is one in which there is an unequal distribution of (negative) charge. The geometry of a molecule, as well as the electronegativity differences of the atoms largely contribute to the polarity of a molecule.

However, we commonly talk about polar molecules as those which are able to be dissolved in a particular solvent, like water. This can be determined by size, shape, and functional groups.

This is a highly simplistic description so I'd encourage further reading here: http://www.chemguide.co.uk/atoms/bonding/electroneg.html

Thanks so much grannysmith 

[bavelzzz]

Could someone answer these questions on chromatography?

1. Both paper chromatography and thin layer chromatography are commonly used to separate components in a mixture. Explain why a more polar organic compound might adsorb more strongly in paper chromatography.

2. Explain why gas chromatography is unsuitable for measuring the concentration of glucose in a urine sample.

3. A student runs an unknown saccharide through a high performance liquid chromatograph. The retention time matches the retention time of galactose recorded the previous week. Why cannot it be assumed that the sample is indeed galactose?

[bavelzzz]

Could someone explain to me how a nucleotide is formed...I can't remember (for example) which carbon in a base joins to the sugar molecule etc. Is there a way to figure out where one molecule joins to another?