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Why is PCL3 and Cl2 produced in equal amounts?
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Mad Man
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#1
charco
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#2
PCl5 ==> PCl3 + Cl2
for every PCl3 formed there must also be one Cl2 formed
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Mad Man
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#3
(Original post by charco)
Look at the balanced equation:
PCl5 ==> PCl3 + Cl2
for every PCl3 formed there must also be one Cl2 formed
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Look at the balanced equation:
PCl5 ==> PCl3 + Cl2
for every PCl3 formed there must also be one Cl2 formed
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charco
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#4
(Original post by Mad Man)
So in terms of moles they are the same in terms of products and reactants so should PCl5 be the same as the others?
So in terms of moles they are the same in terms of products and reactants so should PCl5 be the same as the others?
If x mol of PCl5 breakdown then x mol of the other two will be formed, but the mol of PCl5 must be (original mol - x)
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#5
(Original post by charco)
No, because PCl5 has to break down to give the others.
If x mol of PCl5 breakdown then x mol of the other two will be formed, but the mol of PCl5 must be (original mol - x)
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No, because PCl5 has to break down to give the others.
If x mol of PCl5 breakdown then x mol of the other two will be formed, but the mol of PCl5 must be (original mol - x)
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So the partial pressures must sum to give AB?
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Mad Man
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#6
(Original post by charco)
No, because PCl5 has to break down to give the others.
If x mol of PCl5 breakdown then x mol of the other two will be formed, but the mol of PCl5 must be (original mol - x)
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No, because PCl5 has to break down to give the others.
If x mol of PCl5 breakdown then x mol of the other two will be formed, but the mol of PCl5 must be (original mol - x)
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charco
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#7
(Original post by Mad Man)
So do you mean AB --> A + B?
So the partial pressures must sum to give AB?
So do you mean AB --> A + B?
So the partial pressures must sum to give AB?
In terms of moles.
If the initial mol of PCl5 is n
and the mol of PCl5 decomposed is x
Then the final mol of each component:
PCl5 = (n-x)
PCl3 = x
Cl2 = x
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Mad Man
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#8
(Original post by charco)
I never mentioned partial pressures ...
In terms of moles.
If the initial mol of PCl5 is n
and the mol of PCl5 decomposed is x
Then the final mol of each component:
PCl5 = (n-x)
PCl3 = x
Cl2 = x
-----------------------------------------------------------------------------------------------
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I never mentioned partial pressures ...
In terms of moles.
If the initial mol of PCl5 is n
and the mol of PCl5 decomposed is x
Then the final mol of each component:
PCl5 = (n-x)
PCl3 = x
Cl2 = x
-----------------------------------------------------------------------------------------------
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charco
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#9
(Original post by Mad Man)
Perfect. I get it now
Perfect. I get it now
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Mad Man
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#10
(Original post by charco)
I never mentioned partial pressures ...
In terms of moles.
If the initial mol of PCl5 is n
and the mol of PCl5 decomposed is x
Then the final mol of each component:
PCl5 = (n-x)
PCl3 = x
Cl2 = x
-----------------------------------------------------------------------------------------------
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I never mentioned partial pressures ...
In terms of moles.
If the initial mol of PCl5 is n
and the mol of PCl5 decomposed is x
Then the final mol of each component:
PCl5 = (n-x)
PCl3 = x
Cl2 = x
-----------------------------------------------------------------------------------------------
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(Original post by charco)
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Can I use it for the decomposition of Calcium Carbonate?
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charco
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#11
(Original post by Mad Man)
So if CaCO3--> CaO + O2
So does this only work for homogeneous systems?
Can I use it for the decomposition of Calcium Carbonate?
So if CaCO3--> CaO + O2
So does this only work for homogeneous systems?
Can I use it for the decomposition of Calcium Carbonate?
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Mad Man
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#12
(Original post by charco)
You cannot have the concentration of a solid, or a pure liquid.
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You cannot have the concentration of a solid, or a pure liquid.
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#13
(Original post by Mad Man)
I think get it so for (aq) or (g) equations.
I think get it so for (aq) or (g) equations.
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Mad Man
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#14
(Original post by charco)
The moles relationship works of course, but you cannot do equilibrium calculations without (aq) or (g)
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The moles relationship works of course, but you cannot do equilibrium calculations without (aq) or (g)
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#15
(Original post by Mad Man)
So all species have to have the same state of matter?
So all species have to have the same state of matter?
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#16
This is from the CGP textbook page 315. I'm confused about the state of matter stuff now.
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#18
(Original post by Mad Man)
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This is from the CGP textbook page 315. I'm confused about the state of matter stuff now.
This is from the CGP textbook page 315. I'm confused about the state of matter stuff now.
This is the equilibrium constant with regard to partial pressures. It can only apply to gases.
Kc applies to solutions or gases.
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#19
(Original post by charco)
Notice they say kp.
This is the equilibrium constant with regard to partial pressures. It can only apply to gases.
Kc applies to solutions or gases.
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Notice they say kp.
This is the equilibrium constant with regard to partial pressures. It can only apply to gases.
Kc applies to solutions or gases.
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