Equlibrium (1 Viewer)

[Cheezy-G]

Can some one please explain the exothermic, endothermic and temperature changes in Equilibrium. I get everything else in equlibrium but this part is something, I can't get my head around. It would be a great help.

Thanks

 

[funking_you]

Just download the document attached.

Cheers,
George

 

[Dreamerish*~]

hmmm well take the carbon dioxide equilibrium:

CO_2(g) + H₂O_(l) --> H₂CO_3(aq) + heat

if you add more heat, there will be too much heat in the right-hand-side therefore the equilibrium will shift to the left, producing more CO_2(g).

if you take away heat, there will be too little heat in the right-hand-side therefore the equilibrium will shift to the right to creat more heat, producing more H₂CO_3(aq)

 

[xiao1985]

hehehe this is gonna be fun... now guys, there's a challenge: wut happens when u freeze the system?!

i am looking for eq and explaination =p

 

[Dreamerish*~]

hehehe this is gonna be fun... now guys, there's a challenge: wut happens when u freeze the system?!

i am looking for eq and explaination =p

aiyerr... i've learnt and forgotten.

CO_2(g) + H₂O_(l) --> H₂CO_3(aq) + heat

if you freeze it... then the equilibrium will shift to the uh... right? and uh... yeah...

you can tell how confident i am now

 

[Will Hunting]

Chill out, man, it's coooooool!

Lowering the temperature of the eqm system will increase the solubility of CO2, by LCP causing an eqm shift to the right and the formation of more H2CO3. This occurs by the following:

CO2(g) + H2O(l) ---> H2CO3(aq)

Lowering the temperature of the eqm mixture below the melting point of water, however, will cause solid ice to freeze out of solution, leading to the evolution of CO(sub)2 gas from solution and its enclosure within teenie weenie confined spaces, then... BOOOOOOM!!! (or something a little less ostentatious)

H2CO3(aq) ---> H2O(s) + CO2(g)

 

[xiao1985]

Chill out, man, it's coooooool!

Lowering the temperature of the eqm system will increase the solubility of CO2, by LCP causing an eqm shift to the right and the formation of more H2CO3. This occurs by the following:

CO2(g) + H2O(l) ---> H2CO3(aq)

Lowering the temperature of the eqm mixture below the melting point of water, however, will cause solid ice to freeze out of solution, leading to the evolution of CO(sub)2 gas from solution and its enclosure within teenie weenie confined spaces, then... BOOOOOOM!!! (or something a little less ostentatious)

H2CO3(aq) ---> H2O(s) + CO2(g)

BLING BLING

i think we got a winner=p
soz dreamish... the problem is more profound than it first seemed =p

tho the other equilibrium shud be H2O (l) <---> H2O (s)

and when freezing, the equilibrium will shift to the right... which inturn influences the CO2 equilibrium as well...