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Topic: Law of mass action  (Read 10868 times)

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777888

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Law of mass action
« on: November 24, 2004, 11:13:13 AM »
Hi, I don't know how to do this question...
1) C4H10(l) <-> C4H10(g)
Why the equilibrium law expresiion yields a constant value at a particular temperature and will the concentration of butane gas in the system relate to quantity of liquid butane present?

2)For, the Le Chatelier's principle, after applying a stress, the system adjusts in a way to resist the change. After adjusting, will that be a new equilibrium state?

Thanks!
« Last Edit: November 24, 2004, 09:45:07 PM by 777888 »

Offline Donaldson Tan

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Re:Law of mass action
« Reply #1 on: November 24, 2004, 12:16:01 PM »
2. YES
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

777888

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Re:Law of mass action
« Reply #2 on: November 24, 2004, 12:42:05 PM »
Oh, I see! :)
Can someone help me with question 1?
Thank you!

777888

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Re:Law of mass action
« Reply #3 on: November 24, 2004, 09:46:40 PM »
3)What is the equilibrium law expression of the combustion of hydrogen?

[In the expression, do we omit H2O?]

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Re:Law of mass action
« Reply #4 on: November 25, 2004, 05:50:29 PM »
isnt (1) obvious? if you have a system whereby liquid butane is in equilibrium wiith gas butane, then the molar concentration of gas butance is related to the molar concentration of liquid butane. [YES]

3. we don't omit water. water is omitted for Ka and Kb for acid and bases because the concentration of water changes negligibly during the dissociation process. Hence, [H2O] approximates to a constant, so we ignored it for Ka and Kb. [H2O] isnt necessary a constant for combustion of hydrogen.

"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

777888

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Re:Law of mass action
« Reply #5 on: November 25, 2004, 10:11:30 PM »
isnt (1) obvious? if you have a system whereby liquid butane is in equilibrium wiith gas butane, then the molar concentration of gas butance is related to the molar concentration of liquid butane. [YES]

3. we don't omit water. water is omitted for Ka and Kb for acid and bases because the concentration of water changes negligibly during the dissociation process. Hence, [H2O] approximates to a constant, so we ignored it for Ka and Kb. [H2O] isnt necessary a constant for combustion of hydrogen.


My text book says we omit liquid and solid(condensed states) almost all the time because these states have no concentrations! These values refer to density of solids and liquids and are incorporated into the value of K(Equilibrium constant)...

For this reason, I am not sure about question 1.

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Re:Law of mass action
« Reply #6 on: November 26, 2004, 10:29:46 AM »
forgive my quick slip. the concentration of liquid butane, aka its density, is dependent on the pressure and temperature of the liquid butane. The vapour pressure of liquid butane is actually the partial pressure of the gas butane in equilibrium with liquid butane at the specific pressure and temperature.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

777888

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Re:Law of mass action
« Reply #7 on: November 26, 2004, 12:47:19 PM »
So for question1, the concentration of butane gas in the system is NOT related to the quantity of liquid butane present, am I right?

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Re:Law of mass action
« Reply #8 on: November 26, 2004, 01:25:56 PM »
not directly related
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

777888

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Re:Law of mass action
« Reply #9 on: November 26, 2004, 05:24:22 PM »
Oh I see! Thanks!

While omitting solids and liquids in the equilibrium law expression, why would K still be a constant??
Say CaCO3(s) <-> CaO(s) + CO2(g)
The Equilibrium Law Expression is:
K=[CO2]

Then how can K be a constant when [CO2] is changing?

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Re:Law of mass action
« Reply #10 on: November 26, 2004, 10:45:04 PM »
because the chemical reaction is at equilibrium. even if you increase the amount of solid components, their concentration aka density, remains the same. so there wont be a decrease or increase in the partial pressure of carbon dioxide,
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

777888

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Re:Law of mass action
« Reply #11 on: November 27, 2004, 12:35:24 AM »
Would K change if we increase the concentration of CO2 directly?

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Re:Law of mass action
« Reply #12 on: November 27, 2004, 12:24:54 PM »
CaCO3(s) <-> CaO(s) + CO2(g)

If you increase [CO2], the system tends to decrease the amount of CO2 present. This is in accordace to Le Chatelier's Principle. K is a temperature dependent constant.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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