Chemistry: Post your doubts here!

[romeopoomore]

What if we get this question CO(g) + 1/2O2(g) CO2(g) ∆H = -284 kJmol-1 , isn’t it oxygen gas is 0 , CO results to -284

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Yes the oxygen is 0 . The backward reaction gives +284 i.e the enthalpy of formation of Carbon monoxide from carbon dioxide is +284 but care is to be taken that this isn't the standard enthalpy of formation of neither CO nor CO2 ... I hope you understand that there's a difference as this are formed from non standard Conditions

 

[Madina2410]

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Yes the oxygen is 0 . The backward reaction gives +284 i.e the enthalpy of formation of Carbon monoxide from carbon dioxide is +284 but care is to be taken that this isn't the standard enthalpy of formation of neither CO nor CO2 ... I hope you understand that there's a difference as this are formed from non standard Conditions

I don’t understand why we have to do Hess law, what’s its importance any way if we can just do the subtraction method

 

[romeopoomore]

I don’t understand why we have to do Hess law, what’s its importance any way if we can just do the subtraction method

You don't have to apply the Hess Cycle if you don't see the necessity

 

[Madina2410]

Can someone explain this question to me please?

 

[romeopoomore]

View attachment 65243Can someone explain this question to me please?

 

[Madina2410]

Thank you , appreciate It

 

[Madina2410]

Help please

 

[romeopoomore]

View attachment 65246 Help please

View attachment 65246 Help please

I would go for option A because if you write the formulas they result in the same reaction i.e burning carbon results in Only Carbon dioxide under standard Conditions.Moreover it relates to standard Conditions and hence we can trust that it gives the standard enthalpy of formation of CO2
B is wrong for me because bond energies are based on averages and are not entirely standard e.g compare the C and O double bond in acetone,ethanal and carbon dioxide ... They will vary because of the environments the carbon is in .
C is wrong because the combustion of carbon may not necessarily be at standard conditions and hence the enthalpy change of reaction will not necessarily give the standard enthalpy of formation of carbon dioxide
D is wrong because the enthalpy change of combustion from Diamond is non standard

I believe A is the correct answer for those reasons what do you think ?

 

[Madina2410]

Do

I would go for option A because if you write the formulas they result in the same reaction i.e burning carbon results in Only Carbon dioxide under standard Conditions.Moreover it relates to standard Conditions and hence we can trust that it gives the standard enthalpy of formation of CO2
B is wrong for me because bond energies are based on averages and are not entirely standard e.g compare the C and O double bond in acetone,ethanal and carbon dioxide ... They will vary because of the environments the carbon is in .
C is wrong because the combustion of carbon may not necessarily be at standard conditions and hence the enthalpy change of reaction will not necessarily give the standard enthalpy of formation of carbon dioxide
D is wrong because the enthalpy change of combustion from Diamond is non standard

I believe A is the correct answer for those reasons what do you think ?
[/QUO
I was stu

I would go for option A because if you write the formulas they result in the same reaction i.e burning carbon results in Only Carbon dioxide under standard Conditions.Moreover it relates to standard Conditions and hence we can trust that it gives the standard enthalpy of formation of CO2
B is wrong for me because bond energies are based on averages and are not entirely standard e.g compare the C and O double bond in acetone,ethanal and carbon dioxide ... They will vary because of the environments the carbon is in .
C is wrong because the combustion of carbon may not necessarily be at standard conditions and hence the enthalpy change of reaction will not necessarily give the standard enthalpy of formation of carbon dioxide
D is wrong because the enthalpy change of combustion from Diamond is non standard

I believe A is the correct answer for those reasons what do you think ?

i was stuck in between A and B , well I guess I’ll rely on your answer because I’m a novice to AS

 

[romeopoomore]

Do

I am also doing AS 😁

 

[Madina2410]

The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.

what examples are there that makes hydrogen more electronegative hence giving it a negative -1 ?

 

[romeopoomore]

The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.

what examples are there that makes hydrogen more electronegative hence giving it a negative -1 ?

In Sodium Hydride

 

[romeopoomore]

In Sodium Hydride

Oxidation states (oxidation numbers)

Explains what oxidation states (oxidation numbers) are, and how to calculate them and make use of them.

www.chemguide.co.uk

 

[Madina2410]

Hello again, there’s a question that’s rather bothering . Can someone just explain why is it the “b” option

 

[Madina2410]

Hello again, there’s a question that’s rather bothering . Can someone just explain why is it the “b” optionView attachment 65252

And how does the box diagram look like for elements beyond potassium , how do we fill it diagrammatically

 

[Madina2410]

Thank you

Oxidation states (oxidation numbers)

Explains what oxidation states (oxidation numbers) are, and how to calculate them and make use of them.

www.chemguide.co.uk

 

[romeopoomore]

Hello again, there’s a question that’s rather bothering . Can someone just explain why is it the “b” optionView attachment 65252

We use what is called the Hund's rule to fill up the atomic orbitals. Electrons fill up lower energy levels before they can fill higher ones. In this case we have 1st,2nd and 3rd energy levels . So electrons will fill up 1 then 2 then 3 in that order.
One thing to note too is that electrons will fill empty atomic orbitals before they can start pairing.
The atom in question has 8 electrons and thus 2 will go to the 1s orbital and the remaining 6 will go to the 2s and 2p respectively. When the 6 fill up the 2nd energy level orbitals , 2 will fill the 2s leaving 4 electrons . 3 of those will fill up the 2p orbitals (1 per orbital) and the remaining 1 will go to pair with one of the electrons in the 2p orbitals since the 3s is of a higher energy level.

The reason for all of this is due to two reasons I know of . The first one regarding the order of the filling of energy levels is due to atoms wanting to have the least potential energy possible which makes them very stable . If the atom was to have an electon which would be in a higher energy level this would make it unstable as the electron would be further away from the attraction of the nucleus making it veryyyyyyyy susceptible to being lost thus making the atom unstable . (Which is why Francium is highly reactive as it has an electron in a higher energy level)
The reason now for the filling of atomic orbitals is in order to minimise repulsions between electrons occupying the same energy level. Electrons will tend to be further away since they have the same charge. If electrons are closer to one another they tend to push each other away which makes the atom unstable . I hope I have answered you.

 

[romeopoomore]

And how does the box diagram look like for elements beyond potassium , how do we fill it diagrammatically

There will be introduction of the d orbitals which will be tedious to explain here. But the same principles of orbital filling apply . I will post a link to a YouTube video

 

[romeopoomore]

There will be introduction of the d orbitals which will be tedious to explain here. But the same principles of orbital filling apply . I will post a link to a YouTube video

 

[Madina2410]

Thank you so much for clearing my doubts😄

There will be introduction of the d orbitals which will be tedious to explain here. But the same principles of orbital filling apply . I will post a link to a YouTube video