Chemistry: Post your doubts here!

[Rizwan Javed]

octo nov 2015 mark scheme 21??????

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_ms_21.pdf

 

[The Sarcastic Retard]

LOL! I

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_ms_21.pdf

didnt got it.. THanks thoough

 

[techgeek]

this one Rizwan Javed

 

[Rizwan Javed]

this one Rizwan Javed
View attachment 60507

You are given this situation:
H2 + I2 ⇌ 2HI
In the question, you are told that there are 0.02 mole of H2 and 0.02mol of I2. Consider that the no. of moles of I2 that react with H2 in the forward reaction are y. So if y moles of H2 are reacted with I2, the no. of moles H2 remaining at the equilibrium will be give by (0.02 - y) moles.
Now consider it in terms of mole ratios; 1 mole of H2 reacts with 1 mole of I2 (as given in the equation), so y moles of H2 with react with y moles of I2. In this way, the no. of moles of I2 remaining at equilibrium will be: (0.02 - y) moles.
Do a similar thing to find the no. of moles of HI at equilibrium. 1 moles of H2/I2 produces 2 moles of HI, so y moles of H2/I2 will produce 2y moles of HI. So no. of moles of HI present at equilibrium will be (0+2y) moles or 2y moles. This whole thing can be represented like this:

Now use the Kc expression you might have expressed in the previous parts, putting it equal to the Kc value given in the question. Then simply use algebra, to find the value of y. Once the value is found, then you can calculate the the conc. of substance at equilibrium easily.

Get it?

 

[Rizwan Javed]

this one Rizwan Javed
View attachment 60507

This was a simple question, let's do a bit tricky question now Consider the question below:

You know that the number of moles of Nitrogen, N2, at equilibrium are 2.32mol which are 0.32 moles greater than the number of moles initially. This tells us that, some amount of NH3 has broken down to give this increase in mole of N2 at equilibrium. So by using the mole ratios, find the number of moles of NH3 which have decomposed to give 0.32 mole of N2.
n(N2) : n(NH3) = 1 : 2
^ they are in this ratio, so the moles of NH3 broken are:
0.32 * 2 = 0.64mol

Subtract ^ this amount of mole from the initial number of moles NH3, to find the number of NH3 at equilibrium. 2.40-0.64 = 1.76mol

Now in a similar way, find the number of moles of H2 formed by the decomposition of NH3.
n(H2) : n(NH3) = 3 : 2
so moles of H2 formed = 0.64 * 3/2 = 0.96mol <----- add this amount to the initial number of moles of H2, to find the the no. of moles of H2 at equilibrium: 6 + 0.96 = 6.96mol

Now you have this information:


Since the volume is 1dm^3, the conc. of the substances at equilibrium will be equal to the moles of the substances.
Finally use the Kc expression, to get your final answer:


So the answer is A.

 

[techgeek]

This was a simple question, let's do a bit tricky question now Consider the question below:

You know that the number of moles of Nitrogen, N2, at equilibrium are 2.32mol which are 0.32 moles greater than the number of moles initially. This tells us that, some amount of NH3 has broken down to give this increase in mole of N2 at equilibrium. So by using the mole ratios, find the number of moles of NH3 which have decomposed to give 0.32 mole of N2.
n(N2) : n(NH3) = 1 : 2
^ they are in this ratio, so the moles of NH3 broken are:
0.32 * 2 = 0.64mol

Subtract ^ this amount of mole from the initial number of moles NH3, to find the number of NH3 at equilibrium. 2.40-0.64 = 1.76mol

Now in a similar way, find the number of moles of H2 formed by the decomposition of NH3.
n(H2) : n(NH3) = 3 : 2
so moles of H2 formed = 0.64 * 3/2 = 0.96mol <----- add this amount to the initial number of moles of H2, to find the the no. of moles of H2 at equilibrium: 6 + 0.96 = 6.96mol

Now you have this information:
View attachment 60509

Since the volume is 1dm^3, the conc. of the substances at equilibrium will be equal to the moles of the substances.
Finally use the Kc expression, to get your final answer:

So the answer is A.

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that was awesome!
thank you so much

 

[Rizwan Javed]

​

that was awesome!
thank you so much

No problem

 

[funky brat]

In the infrared spectroscopy question given at the end of the book (the Roger Norris thing) there's a question which asks to state the three distinct points for indentifying the compound or more generally if theres such a question in which there is only one identifiable peak for say OH group and then other complicated peaks occur in the fingerprint region, what could be the three reasons to state?

 

[Rizwan Javed]

In the infrared spectroscopy question given at the end of the book (the Roger Norris thing) there's a question which asks to state the three distinct points for indentifying the compound or more generally if theres such a question in which there is only one identifiable peak for say OH group and then other complicated peaks occur in the fingerprint region, what could be the three reasons to state?

Can you please post the picture of that question?

 

[ahmedish]

Should I study infrared spectroscopy if im doing A2? can it come in Paper 4 ? (its a new addition to the AS syllabus)

 

[The Sarcastic Retard]

Should I study infrared spectroscopy if im doing A2? can it come in Paper 4 ? (its a new addition to the AS syllabus)

It myt come.

 

[ahmedish]

It myt come.

Man but wtf where do they expect us to learn it.
Its quite big as well.
Did you study it?

 

[princess Anu]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s08_ms_5.pdf

Q1c
Temp rise is included in independent as well as dependent variable :/
WHY

 

[Eugene99]

D. how?

 

[Dark Destination]

View attachment 60524

D. how?

I think because the products of A and C are both carboxylic acids, which have hydrogen bonding between their molecules.
There won't be any reaction for B, cause it's a tertiary alcohol.

The product of D is a ketone, which doesn't have hydrogen bonding between its molecules, leading to a lower boiling point?

I'm sorry if I'm wrong. This is probably the first chemistry question I tried in like months...

 

[Eugene99]

I think because the products of A and C are both carboxylic acids, which have hydrogen bonding between their molecules.
There won't be any reaction for B, cause it's a tertiary alcohol.

The product of D is a ketone, which doesn't have hydrogen bonding between its molecules, leading to a lower boiling point?

I'm sorry if I'm wrong. This is probably the first chemistry question I tried in like months...

you are right about A and C, but actually ketones have higher boiling points, because they have two alkyl groups on either side of their C=O which makes it more stable. which is why I was confused
Hasn't it got something to do with branching?
But branched also have lower boiling points

 

[Rizwan Javed]

I think because the products of A and C are both carboxylic acids, which have hydrogen bonding between their molecules.
There won't be any reaction for B, cause it's a tertiary alcohol.

The product of D is a ketone, which doesn't have hydrogen bonding between its molecules, leading to a lower boiling point?

I'm sorry if I'm wrong. This is probably the first chemistry question I tried in like months...

You are right :3

 

[The Sarcastic Retard]

I think because the products of A and C are both carboxylic acids, which have hydrogen bonding between their molecules.
There won't be any reaction for B, cause it's a tertiary alcohol.

The product of D is a ketone, which doesn't have hydrogen bonding between its molecules, leading to a lower boiling point?

I'm sorry if I'm wrong. This is probably the first chemistry question I tried in like months...

Its correct.

 

[Dark Destination]

you are right about A and C, but actually ketones have higher boiling points, because they have two alkyl groups on either side of their C=O which makes it more stable. which is why I was confused
Hasn't it got something to do with branching?
But branched also have lower boiling points

Yeah, but still no hydrogen bonding which is a strong force of attraction leading to increased boiling points. Absence of it would surely lower the boiling point.

And yes, I think straight chain alkanes have HIGHER boiling points cause of more contact point or something... but the comparison here is clearly because of the intermolecular forces of attraction.

 

[Eugene99]

Yeah, but still no hydrogen bonding which is a strong force of attraction leading to increased boiling points. Absence of it would surely lower the boiling point.

And yes, I think straight chain alkanes have HIGHER boiling points cause of more contact point or something... but the comparison here is clearly because of the intermolecular forces of attraction.

get that..thank you