Friday
July 25, 2014

Homework Help: Chemistry``

Posted by Linda on Tuesday, May 8, 2007 at 6:31pm.

I answered this question but I did not get the right answer.

CO2 + H2 <=> H2O + CO
CO2 = 0.5 mole and H2 he 0.5 mole both sollutions were forced into a 1 Litre container.
K = 2
what is the equlibrium concentration of each reactant and product.

as I calculated I did the following (ICE chart)
Keq = (X)^2/(0.5-X)^2 and finally
x= 0.25
what did I do wrong how do I get the right answer



I can only guess at what went wrong because the problem is confusing. What do you mean by "both solutions??? were forced into a 1 liter container." Do you mean both gases (CO2 is a gas as well as H2) were placed into a 1 L container? Next, is the H2O a liquid or a gas at equilibrium? You need to know that? As to what went wrong.
If I solve the equation you have I find x = 0.29 and not 0.25. Next, x will be the (CO) but not the concentration of CO2 and H2. Those are 0.5 - x so you solved only part of the problem. Finally, if the H2O is a liquid, then it doesn't enter into the Keq expression. If this doesn't help, plese clarify the question.

ok let me re-write it

when 0.5 mole of CO2 and 0.5 mole of H2 were forced into a 1 litre reaction container, and the equilibrium was established
CO2(g) + H2(g) <=> H2O(g) + CO(g)
under the conditions of the experiment K=2

find the equilibrium concentration of each reactant and product.

ok let me re-write it

when 0.5 mole of CO2 and 0.5 mole of H2 were forced into a 1 litre reaction container, and the equilibrium was established
CO2(g) + H2(g) <=> H2O(g) + CO(g)
under the conditions of the experiment K=2

find the equilibrium concentration of each reactant and product.

OK. That is clearer. Since H2O is a gas it enters into the calculation, also. Your original equation is correct; i.e.,
x^2/(0.5-x)2 = 2.
Did you just make an arithmetic error?
sqrt of both sides gives
x/(0.5-x) = 1.414
x = 0.707 - 1.414x
2.414x = 0.707
x = 0.29, which, if 0.5 mol was given in the problem (and not 0.50), then we would round the 0.29 to 0.3
(CO2) = 0.5-0.3=0.2
(H2)=0.5-0.3=0.2
(H2O)=0.3
(CO)=0.3
Rounding like this won't produce a K of 2 if you plug the numbers back in it will if you use 0.29 and 0.21.

Thanks for clarifying it.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

CHEM - When 0.5 mole of CO2 and 0.5 mole of H2 were forced into a litre reaction...
AP Chemistry - The system CO2(g) + H2(g)<->H2O(g) + CO(g) is at ...
ap chemistry - The system CO2(g) + H2(g) *) H2O(g) + CO(g) is at equilibrium at ...
chemistry - CH4 + 2O2 ---> CO2 + 2H2O Delta H H2O= -285.8 kJ/mole Delta H CO2...
Chemistry - I am having such a hard time approaching this problem. It would be ...
Chemistry - I am having such a hard time approaching this problem. It would be ...
Chemistry, CO2 aq+H2O->H2CO3! - How many grams of CO2 are needed to ...
Chemistry, CO2 aq+H2O->H2CO3! - How many grams of CO2 are needed to ...
Chemistry, CO2 aq+H2O->H2CO3! - How many grams of CO2 are needed to ...
Chemistry, CO2 aq+H2O->H2CO3! - How many grams of CO2 are needed to ...

Search
Members