Quantcast

need help w/ g.chem question!!!

DATBooster | The Ultimate DAT Resource
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

avian777

Full Member
10+ Year Member
Joined
Jul 11, 2008
Messages
39
Reaction score
0
anyone can help me with this?? thanks
heat + C(s) + CO2(g) < ---> 2CO (g)
If the pressure on the system is increase and the temperature kept constant, what will be the result?
the answer is: the amount of CO will decrease and that of C and CO2 will increase.

I thought that the amount of CO is indeed decreased and the amount of CO2 is increased but it has no effect on C(s) since this is not considered in the equilibrium equation
Please help me to explain this!
 

le14

Full Member
10+ Year Member
7+ Year Member
Joined
Mar 28, 2008
Messages
79
Reaction score
0
When you increase the pressure, the result is the increase to the side with less mols of gas so to the left so C (s) and CO2(g) will increase whereas 2CO(g) will decrease. Le Chanteliers Principle favors all the reactants in this instance.
 

Sugafoot79

Full Member
10+ Year Member
Joined
Feb 4, 2009
Messages
72
Reaction score
0
anyone can help me with this?? thanks
heat + C(s) + CO2(g) < ---> 2CO (g)
If the pressure on the system is increase and the temperature kept constant, what will be the result?
the answer is: the amount of CO will decrease and that of C and CO2 will increase.

I thought that the amount of CO is indeed decreased and the amount of CO2 is increased but it has no effect on C(s) since this is not considered in the equilibrium equation
Please help me to explain this!

You are correct in a sense. Since we are not considering C(s), increasing pressure will tend to the side with the less number of moles (Le Chatelier's Principle). Thus, the amount of CO2(g) (1mole) will increase, while the amount of CO(g) will decrease (2 moles). Disregard C(s), it is not included in the equilibrium

Hope this helps!:D
 
Top