Chemistry Paper 2

[Soulgamer]

yup, correct

No it was as, Ar is inert because it has a complete octet of electrons/ stable electronic configuration.

 

[aaditya menon]

No it was as, Ar is inert because it has a complete octet of electrons/ stable electronic configuration.

Dude, this is from a past paper. There are 3 points

complete octet, high activation energy and inert.

It was just 1 mark right? I wrote all the 3 points, bleh..

 

[gary221]

For the empirical formula part:
C was 37.5 %, H was 4.17% and O was 58.3% so,
C:H:O = 37.5/12 : 4.17/1 : 58.3/16
= 3.125 : 4.17 : 3.644 then divide all values by lowest value 3.125 so,
= 1 : 1.334 : 1.166 then multiply sll values by 6 to obtain simplest whole no. ratio
= 6 : 8 : 7
Hence, the answer was definitely C6H8O7!

Why did u multiply it by 6???

 

[leosco1995]

Why did u multiply it by 6???

So that everything is in a whole number (empirical formula is when the atoms are in their lowest proportionals which are still integers).

 

[leosco1995]

Are you guyz sure it was -129? Coz the enthalpy change of the reaction worked out to +129 for me and my friends. :-/
Variant 22

I am quite sure it was -129, see this.

x works out to be -129. This is how I solved it.

 

[fathimaash]

That works out to +129 dude. *Phew*, you scared me for a moment there.

Here's how you calculate it. ΔHreaction = ΔH products - ΔH reactants.
Δ products = -726
ΔHreactants = (-286*2)+(-283) = -855
Do the subtraction now, -726-(-855) = -726+855 = +129.

no its -129....
X-726=-855
so X= -129

 

[Pals_1010]

That works out to +129 dude. *Phew*, you scared me for a moment there.

Here's how you calculate it. ΔHreaction = ΔH products - ΔH reactants.
Δ products = -726
ΔHreactants = (-286*2)+(-283) = -855
Do the subtraction now, -726-(-855) = -726+855 = +129.

Applying Hess' Law,

ΔH = (-283) + (2x-286) + -(-726)
= -855 + 726
= - 129 kJmol-1

 

[leosco1995]

That works out to +129 dude. *Phew*, you scared me for a moment there.

Here's how you calculate it. ΔHreaction = ΔH products - ΔH reactants.
Δ products = -726
ΔHreactants = (-286*2)+(-283) = -855
Do the subtraction now, -726-(-855) = -726+855 = +129.

It IS -129.. look:

Both reactions will go towards the empty box and those will have equal values. So using vectors,

x - 726 = (2 * -286) + (-283)
x - 726 = -855
x = -129

E: I just realized that you did ΔH = ΔH products - ΔH reactants. Hess' Law isn't applied like that.

 

[Soulgamer]

Dude, this is from a past paper. There are 3 points

Incomplete octet, high activation energy and inert.

It was just 1 mark right? I wrote all the 3 points, bleh..

Trust me I've done a whole lot of pp and in our course we have never said that Ar makes covalent compounds as you suggest that the activation energy is high.
It was for only 2 mark i guess and you had to right it is 1)unreactive/inert
2) As it has a complete octet/stable electronic configuration. ( its not incomplete)

 

[Silent Hunter]

That works out to +129 dude. *Phew*, you scared me for a moment there.

Here's how you calculate it. ΔHreaction = ΔH products - ΔH reactants.
Δ products = -726
ΔHreactants = (-286*2)+(-283) = -855
Do the subtraction now, -726-(-855) = -726+855 = +129.

are you sure this green part is correct?

 

[Pals_1010]

are you sure this green part is correct?

No, it isn't

This formula is applicable only when we talk about ΔH of formation. Then it becomes,


ΔH of reaction = ΔH of formation of products - ΔH of formation of reactants

But, the date given in the question was ΔH of combustion. So. you have to construct a cycle and apply Hess' Law

 

[Pals_1010]

^ Sorry I don't know why the font changed

 

[Silent Hunter]

exactly..!! thats my point.... well the diagram uploaded also helps in this regard i.e. by drawing hess cycle.. and your method is correct too

 

[Soulgamer]

I used the formula:

Delta H reaction= Delta H combustion of reactants - Delta H combustion of products

The answer was -129KJ.mol

 

[aaditya menon]

Well, idk then. Guess i'm wrong. :-/

 

[Pals_1010]

I used the formula:

Delta H reaction= Delta H combustion of reactants - Delta H combustion of products

The answer was -129KJ.mol

If you used this formula, you can't get -129. It gives +129

 

[Soulgamer]

trust me it came negative as well as for those 200+ students at beacon house who used it.

 

[Pals_1010]

trust me it came negative as well as for those 200+ students at beacon house who used it.

How??

 

[AhsanAadil]

I am expecting 33 in p12, 47 in p22 and 21 in p33. What grade should I expect?

 

[Soulgamer]

How??

Quote me the values and I'll solve it here!