Chemistry: Post your doubts here!

[techgeek]

Only two papers in every Oct/Nov session only are same. Otherwise, every other paper is different.

what do you mean by only two papers are same?

 

[Rizwan Javed]

what do you mean by only two papers are same?

There are 3 variants of Chemistry papers. Out of those 3, 2 are same in every Oct/Nov session.

 

[techgeek]

There are 3 variants of Chemistry papers. Out of those 3, 2 are same in every Oct/Nov session.

oh really? I didn't know that, thanx
yaeh, therefore yesterday's paper was the same I had solved already but today's turned out to be different

 

[fatimarehman]

Can someone help me with chemistry paper 5 Oct Nov 2014 var 52 Q1 part d subpart I and ii. How to compare the concentrations or molecules?

 

[ahmedish]

If E cell is positive then the reaction will be proceeding in forward direction. This means more products will be formed.

E cell is always positive though

 

[fatimarehman]

Can someone explain me chemistry paper 5 Oct Nov 2014 var 52 part d sub part I and ii.

 

[qwertypoiu]

Help part b ivView attachment 60404

J is formed by replacing H from the tertiary carbon in the middle. This happens at rate=21.
K is formed by replacing H from any one of the primary carbons on the side. It happens at rate =1, but since there are 9 such H atoms, 9*1=9
So relative amounts will be 21:9

 

[The Sarcastic Retard]

E cell is always positive though

Not always...

 

[ahmedish]

Not always...

I realised.
Can you explain why having a positive E cell mean there are more products?

 

[ahmedish]

Not always...

wait i dont get how E cell can be negative LOL
How can E cell be negative?

 

[TariqBhai]

June 13 P41, 1) c) ii)
mark scheme ays its forward (more product) because E is more than 0 (its positive)
but isnt that always the case ?? Even if both half cells is negative its going to be (-0.12)-(-0.67)= positive??
so why are there more products?

wait i dont get how E cell can be negative LOL
How can E cell be negative?

I realised.
Can you explain why having a positive E cell mean there are more products?

Ecell will be positive for reactions which are feasible/spontaneous in the forward direction. If Ecell is negative then the reaction is feasible only for the backward direction.

Since you calculated that Ecell was positive then the reaction:
2Fe + 2I <=> 2Fe+ I 2
Will only occur in the forward direction: 2Fe + 2I => 2Fe+ I 2

If Ecell was negative then the reaction would only occur in backward direction: 2Fe+ I 2 => 2Fe + 2I

MS is talking about Ecell and not E.

 

[The Sarcastic Retard]

I realised.
Can you explain why having a positive E cell mean there are more products?

wait i dont get how E cell can be negative LOL
How can E cell be negative?

Ecell will be positive for reactions which are feasible/spontaneous in the forward direction. If Ecell is negative then the reaction is feasible only for the backward direction.

Since you calculated that Ecell was positive then the reaction:
2Fe + 2I <=> 2Fe+ I 2
Will only occur in the forward direction: 2Fe + 2I => 2Fe+ I 2

If Ecell was negative then the reaction would only occur in backward direction: 2Fe+ I 2 => 2Fe + 2I

MS is talking about Ecell and not E.

As TariqBhai explained, if Ecell value is positive means reaction will be occurring. As reaction is occurring that simply means reaction is moving in forward direction. This means more product. And the reaction which arent feasible are the ones that have negative Ecell and the reaction will move backwards.

 

[MafaldaC]

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

 

[The Sarcastic Retard]

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

I couldnt find any ms. But here is what u can get idea similar to ms. http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_er.pdf
scrol down till u get paper 21 er.

 

[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