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Topic: equillibrium again  (Read 4150 times)

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Offline Kaleyrvt

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equillibrium again
« on: January 14, 2007, 11:12:41 PM »
Consider the equilibrium represented by the following equation:
Heat  + H2O(l)    H2O(g)     in which T=373 K, P= 101 kPa
Which of the following will favour the reverse reaction?

a)   The total volume of the system is increased
b)   Boiling water is added to the system
c)   Water vapour is removed from the system
d)   The system is cooled


I think that the answer is c) becasue you originally had heat plus water(in liquid form) and the forward reaction is the gaseous state. Can someone please tell me if I am on the right track?

Offline enahs

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Re: equillibrium again
« Reply #1 on: January 14, 2007, 11:25:14 PM »
You are on the wrong track.

Liquid -> Gas

If you remove product (in this case, gas) you will encourage more of the reaction to take place to restore equilibrium.

Think about Le Chatelier's Principle:
"If a chemical system at equilibrium experiences a change in concentration, temperature, or total pressure; the equilibrium will shift in order to minimize that change."


Offline Kaleyrvt

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Re: equillibrium again
« Reply #2 on: January 15, 2007, 12:22:19 AM »
then the total volume is increased?

Man there is just way too much reading on this topic---My brain is on overload.

Hopefully this is right : ???

Offline breanainneire

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Re: equillibrium again
« Reply #3 on: January 15, 2007, 12:57:17 AM »
Ah, le Chatelier's principle... I just learned that a few months ago, and there were ample problems similar to yours where you do something to a system and you have to state if the reaction shifts to products or reactancts. If you haven't learned about le Chatelier's principle, you won't be able to solve a problem like this.

Offline thegame11

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Re: equillibrium again
« Reply #4 on: January 15, 2007, 07:15:25 PM »
Consider the equilibrium represented by the following equation:
Heat  + H2O(l)    H2O(g)     in which T=373 K, P= 101 kPa
Which of the following will favour the reverse reaction?

a)   The total volume of the system is increased
b)   Boiling water is added to the system
c)   Water vapour is removed from the system
d)   The system is cooled

I think that the answer is c) becasue you originally had heat plus water(in liquid form) and the forward reaction is the gaseous state. Can someone please tell me if I am on the right track?


a,b,c, are wrong.  Therefore, you can assume d is correct (but you really need to know whether its endo or exothermic rxn to be really sure)

a) if you increase the volume of the sys, the pressure goes down.  Therefore, to restore equilibrium you will shift the equilibrium in the forward direction to "make" more gas molecules.

b) boiling water = liquid water is added to system. Therefore, equilibrium will shift to the forward direction.

c) removing water vapour decrease [H2O(g)].  Therefore, equilibrium will shift to the forward direction.


Offline Kaleyrvt

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Re: equillibrium again
« Reply #5 on: January 15, 2007, 08:16:06 PM »
Thank you so much--For some reason I am having a HUGE issue with gettin the equillibrium theories into my brain---I have an entire unit, 5 HUGE lessons on it, so I apologize if I am back with more questions. It has been awhile since I have done chem ! Thanks again for your detailed explanations and help- it is rreally helping me understand.  :D

Offline xiankai

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Re: equillibrium again
« Reply #6 on: January 16, 2007, 08:03:55 AM »
Quote
Therefore, you can assume d is correct (but you really need to know whether its endo or exothermic rxn to be really sure)

in this, since heat is obviously a 'reactant', you can be sure ;)
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