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21
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Linh_ on May 26, 2024, 02:35:02 AM »
I just don't understand why can you separate the equations for CH4.

And also why do you add the products to the concentration equation?

Thanks for your help
22
High School Chemistry Forum / Re: Valence electrons and metallic bond strength
« Last post by Hunter2 on May 26, 2024, 01:22:04 AM »
Quote
and protons have the same charge as electrons

This is wrong the charges are in opposit. Protons are positiv and electrons are negativ.
In a metal the oxidation number is Zero.

What do you mean with
Quote
Why do cations with greater oxidation number form stronger metallic bonds with itself?
.
In compounds of metal and nonmetals we have ionic bonds, where also a oxidationnumber is existing. But normally no metalic bonds.
23
High School Chemistry Forum / Valence electrons and metallic bond strength
« Last post by sd79812 on May 25, 2024, 08:47:41 PM »
Why do cations with greater oxidation number form stronger metallic bonds with itself?

Yes, you're increasing number of delocalized electrons when you have metallic bonds in a higher oxidation number material, but AFAIK the delocalized electrons also have to make up for more protons repulsing, and protons have the same charge as electrons, so shouldn't the added gluing effect of more valence electrons just cancel out because you'd have to counter the protons repulsing so valence electrons have absolutely no effect on metallic bonding and metals all have the same bonding force with themselves respectively?
24
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Hunter2 on May 25, 2024, 04:08:33 PM »
It's because two reactions take place which create CH4.

The System is feed with C2H6 and H2

C2H6 will be cracked in two CH3° radicals.

This react with hydrogen according
CH3° + H2 => CH4 + H°
1. Methane source and a hydrogen radical
And H° react again with ethane according
C2H6 + H° => CH4 + CH3°
2. Methane source and again a methyl radical

Then the game starts again until all Ethane and hydrogen is consumed or the Termination reaction takes place.


25
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Linh_ on May 25, 2024, 03:24:55 PM »
Why is it possible to separate the expression for CH4 to 2 different expressions?
Shouldn't it be one that depends on both reactions?
26
Undergraduate General Chemistry Forum / Re: Reverse Diffusion Demonstration
« Last post by S_Ch_S on May 25, 2024, 03:24:18 PM »
Many thanks for your response! I will look into what you are describing. My tank was held vertically by two clamps, maybe the glass broke from the tension of the clamps and the fact that it happened after I turned the voltage off is just a coincidence?
27
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Hunter2 on May 25, 2024, 12:46:50 PM »
No , you should not add, you should substitute.

CH4 = k2*CH3°*H2/H°

CH4 = k3*C2H6*H°/CH3°

H° = CH4*CH3°/(k3*C2H6)

Fill in this  for H° in first equation.

Solve for CH4 .


28
Undergraduate General Chemistry Forum / Re: Reverse Diffusion Demonstration
« Last post by Corribus on May 25, 2024, 09:20:51 AM »
The failure after the battery being turned off could just be a matter of kinetics. Under Borek's theory of corrosion, corrosive substances generated during the cell operation could take some time to cause enough damage to reach the point of failure.
29
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Linh_ on May 25, 2024, 08:56:40 AM »
First is CH4 = k2*CH3°*H2/H° and CH4 = k3*C2H6*H°/CH3° ?
CH4 = k2*CH3°*H2 + k3*C2H6*H°, from the two propagation reactions.
Also the suggested mechanism is part of the question and not something I chose.

Can you explain more please? substituting here doesn't lead to anything and I don't know what to do
30
Analytical Chemistry Forum / Re: Kinetic chain reaction mechanism
« Last post by Hunter2 on May 25, 2024, 07:56:21 AM »
First is CH4 = k2*CH3°*H2/H° and CH4 = k3*C2H6*H°/CH3° ?

Only this two equations have to be combined.
Substitute H° in first equation with second equation and solve for CH4.

Initiation and Termination is in my opinion not necessary. It's not part of the chain reaction.
Termination have more possibilities like 2 H° => H2 or CH3° + H° => CH4



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