ALL CIE CHEMISTRY DOUBTS HERE !!

[tmisterr]

using the mole ratio, x mole of HI will produce 0.5 mole of H2 and 0.5 mole of I2 so so at equilibrium there will be b-x mole of HI remaining.
 Now Calculate the partial pressures gas, in total we have b-x+0.5x+0.5x which is b moles of gases. partial pressure HI is therefore b-x/b for H2 is 0.5x/b and for I2 it is again 0.5x/b.

 Replace these values into the Kp equation = p[H2]*p[I2]/p[HI]²

we get (0.5xp/b)²/((b-x)p/b)²
 
from here it is just simple simplification, the p² and b² will cancel out top and bottom and in the end you get 0.25x²/(b-x)² which is = x²/4(b-x)²

[elemis]

+rep tmisterr

Thanks

[elemis]

http://www.freeexampapers.com/download.php?l=A%20Level/Chemistry/CIE/2002%20Nov/9701_w02_qp_1.pdf&t1=2wypmop&t2=31e3abol

Question 17.

For Question 36 Ammonia is a reducing agent because it loses Hydrogen, correct ?

[Deadly_king]

http://www.freeexampapers.com/download.php?l=A%20Level/Chemistry/CIE/2002%20Nov/9701_w02_qp_1.pdf&t1=2wypmop&t2=31e3abol

Question 17.

For Question 36 Ammonia is a reducing agent because it loses Hydrogen, correct ?

Nov 02 p1

17. In this question, all the four answers sounds good from the first view. But there's once best and we need to find out which one.

The question is asking about the enthalpy change of formation, that is the energy required when HCl is formed from its reactants at rtp or it could also be the reverse, that is the conversion of HCl to its reactants.

For this we just need to break the C-Cl bond and H-I bond.

Electronegativity does not affect enthalpy changes. Activation energy will not be the reason since it's only the energy required for the reaction to start. Since we're doing the reverse bond energy of the halogens is not required.

I'll take only HCl for explanation. Same principle is applied for HI.
Option C refers to the energy required to break the bond between Hydrogen and the chlorine atom, that H-Cl . Upon breaking this bonds we obtain the atoms of hydrogen and chlorine which will combine together to form hydrogen gas and chlorine molecule.

In other words we have converted H-Cl to H₂ and Cl₂ and the energy required to do so is the same as the enthalpy change of formation of the molecule except for the sign which will be opposite. This is we're converting H-Cl to H2 and Cl2 while enthalpy change of formation is the conversion of Cl₂ and H₂ to HCl.

Bond energy for H-Cl is greater since Cl is a smaller atom, hence force of attraction will be greater.

So answer is C.

36. You're right.

[elemis]

Dude, the answer is 17 C.

Anyways, how is it C ? What does option C even mean ?

[Deadly_king]

Dude, the answer is 17 C.

Anyways, how is it C ? What does option C even mean ?

Ooh yeah.......i'll modify my post.

[elemis]

Bond energy for H-Cl is greater since Cl is a smaller atom, hence force of attraction will be greater.

H-Cl has a shorter bond length than H-I therefore it is stronger and hence requires greater energy to break the bond, right ?

[Deadly_king]

H-Cl has a shorter bond length than H-I therefore it is stronger and hence requires greater energy to break the bond, right ?

You're absolutely correct br0

[elemis]

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s03_qp_1.pdf

Q33

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf

Q2

[Deadly_king]

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s03_qp_1.pdf

Q33

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf

Q2

Jun 03 p1 No 33

It is given that the conversion of graphite to diamond is endothermic.

1. The enthalpy change of atomisation of diamond is smaller than that of graphite -----> It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.

2. The bond energy of the C-C bonds in graphite is greater than that in diamond -----> It is correct as well since in diamond, each carbon carries 4 bonds while in graphite they carry only 3 bonds.

3. The enthalpy change of combustion of diamond is greater than that of graphite -----> It is correct too dor the same reason as the first one.

So answer is A.


Nov 03 p1 No 2
First you need to find the % of phosphorus in P₂O₅ ----> 2(31)/142 x 100 = 43.7 %

It is given that P₂O₅ is 30% of the fertiliser. Hence phosphorus will be 43.7% of this 30%.

% by mass of phosphorus = (43.7/100 x 30/100)/1 x 100 = 13.1%

Therefore answer is B.

[elemis]

Jun 03 p1 No 33

It is given that the conversion of graphite to diamond is endothermic.

1. The enthalpy change of atomisation of diamond is smaller than that of graphite -----> It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.

2. The bond energy of the C-C bonds in graphite is greater than that in diamond -----> It is correct as well since in diamond, each carbon carries 4 bonds while in graphite they carry only 3 bonds.

3. The enthalpy change of combustion of diamond is greater than that of graphite -----> It is correct too dor the same reason as the first one.

So answer is A.

You are contradicting yourself. If Carbon has 4 bonds and  its atoms are arranged tetrahedrally then it should have a LARGER BOND ENERGY than graphite.

Therefore, wouldnt this eliminate option 2 ?

Can you explain how option 1 is correct in more detail ?

[Deadly_king]

You are contradicting yourself. If Carbon has 4 bonds and  its atoms are arranged tetrahedrally then it should have a LARGER BOND ENERGY than graphite.

Therefore, wouldnt this eliminate option 2 ?

Can you explain how option 1 is correct in more detail ?

Nope........am not contradicting myself dude. Let me get it clear for you.

Bond energy is not the 4 C-C bonds present but only one of them. A carbon atom in carbon carries 4 bonds, so the force of attraction needs to be divided exactly in 4. While in graphite the force of attraction is divided in only 3.

So the bond energy in graphite will be larger.

Option 1.

Enthalpy change of Atomisation can also be considered as the energy required for breaking of bonds to form C_(g). So since bond energy for graphite is larger, so will be atomisation energy.

[elemis]

Okay, I see.

It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.

So what if its endothermic ? How did you realise graphite requires more energy ?

[Deadly_king]

Okay, I see.

So what if its endothermic ? How did you realise graphite requires more energy ?

Hmm.........I had the same explanation I gave you, in mind.

[elemis]

http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w04_qp_1.pdf

Question 13. I understand the answer is D and that you need a strong acid and base, but why ? What is the chemistry behind all of this ?

Why wouldnt a strong acid and weak base give a SHARP colour change ?

What chapter does this appear in the chem book by Brian Ratcliff ?