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) Calculating a reactant from Kc:
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
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#2
(Original post by Shafxx)
) Calculating a reactant from Kc:
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
) Calculating a reactant from Kc:
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
0
reply
Report
#3
(Original post by Shafxx)
) Calculating a reactant from Kc:
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
) Calculating a reactant from Kc:
• This time you are told Kc at the start and initial moles and equilibrium moles.
• Remember to convert to equilibrium concentrations.
6.00 moles of PCl3 is mixed with an unknown amount of Cl2. 2.00 moles of PCl5 is made.
How many moles of Cl2 was added? Kc = 20. The experiment was carried out in a 2dm3
flask.
Reaction PCl3(g) + Cl2(g) PCl5(g)
Stoichiometry 1 1 1
Initial n moles 6.00 x 0.0
Reacted
n moles at equilibrium 2.00
[Eq]
• Write the equilibrium expression, put in the values, calculate Kc and work out the units:
I've worked out the equilibrium expression but what would the answer to the question?
The equilibrium concentration for PCl5 is 2/2 = 1
Write the expression: Kc = [PCl5] / [PCl3][Cl2]
Rearrange and fill in what you know to find [Cl2]: [Cl2] = 2[Cl2] = 1 / 20
[Cl2] = 0.025
Work out moles of Cl2 by multiplying by the volume of solution: 0.025 x 2 = 0.05
Work out initial Cl2 moles by adding the change in moles to the moles at equilibrium: 2 + 0.05 = 2.25mol
Tell me if any of this didn’t make sense, hope it did though!
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#4
(Original post by M2003J)
The equilibrium concentration for PCl3 is 4/2 = 2
The equilibrium concentration for PCl5 is 2/2 = 1
Write the expression: Kc = [PCl5] / [PCl3][Cl2]
Rearrange and fill in what you know to find [Cl2]: [Cl2] = 2[Cl2] = 1 / 20
[Cl2] = 0.025
Work out moles of Cl2 by multiplying by the volume of solution: 0.025 x 2 = 0.05
Work out initial Cl2 moles by adding the change in moles to the moles at equilibrium: 2 + 0.05 = 2.25mol
Tell me if any of this didn’t make sense, hope it did though!
The equilibrium concentration for PCl3 is 4/2 = 2
The equilibrium concentration for PCl5 is 2/2 = 1
Write the expression: Kc = [PCl5] / [PCl3][Cl2]
Rearrange and fill in what you know to find [Cl2]: [Cl2] = 2[Cl2] = 1 / 20
[Cl2] = 0.025
Work out moles of Cl2 by multiplying by the volume of solution: 0.025 x 2 = 0.05
Work out initial Cl2 moles by adding the change in moles to the moles at equilibrium: 2 + 0.05 = 2.25mol
Tell me if any of this didn’t make sense, hope it did though!
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#5
(Original post by mh2410)
Can you help me pls with this question![Name: 81935646-CC6F-4014-9A25-1D4BBC3387F3.jpeg
Views: 8
Size: 23.6 KB]()
because I’m getting 5 different products
Can you help me pls with this question
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#6
(Original post by M2003J)
The products you get are CH3-CH=CH-CH2-CH3 and CH2=CH-CH2-CH2-CH3. The first product displays cis/trans isomerism so has 2 isomers. Therefore, the answer is 3.
The products you get are CH3-CH=CH-CH2-CH3 and CH2=CH-CH2-CH2-CH3. The first product displays cis/trans isomerism so has 2 isomers. Therefore, the answer is 3.
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#7
(Original post by mh2410)
What about the structural isomers of CH2=CH-CH2-CH2-CH3
What about the structural isomers of CH2=CH-CH2-CH2-CH3
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#8
(Original post by macca_garlick)
Apart from but-1-ene and but-2-ene no other structural isomers are possible because the nucleophile (in this case :OH) has to attack the hydrogen atoms on the carbon adjacent to the carbon bearing the bromine. Therefore it is 3 as you can only get but-1-ene, E-but-2-ene and Z-but-2-ene
Apart from but-1-ene and but-2-ene no other structural isomers are possible because the nucleophile (in this case :OH) has to attack the hydrogen atoms on the carbon adjacent to the carbon bearing the bromine. Therefore it is 3 as you can only get but-1-ene, E-but-2-ene and Z-but-2-ene
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#9
(Original post by mh2410)
Appreciate it , thank you. In a nutshell , during these kind of reaction, not other isomers occur accept the concerned carbons ?
Appreciate it , thank you. In a nutshell , during these kind of reaction, not other isomers occur accept the concerned carbons ?
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