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, please 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.
To summarize, in order to find the equilibrium concentrations of each reactant and product in the given reaction, you start by setting up an ICE (Initial, Change, Equilibrium) chart.
The equation for the reaction is: CO2(g) + H2(g) <=> H2O(g) + CO(g)
Given:
Initial moles: CO2 = 0.5, H2 = 0.5
Volume: 1 liter
Equilibrium constant (K): 2
Now we can set up the ICE chart:
Initial:
CO2 = 0.5
H2 = 0.5
H2O = 0
CO = 0
Change:
CO2 = -x
H2 = -x
H2O = +x
CO = +x
Equilibrium:
CO2 = 0.5 - x
H2 = 0.5 - x
H2O = x
CO = x
Since the equilibrium constant K is given as 2, we can set up the equation:
K = (CO)(H2O) / (CO2)(H2)
Substituting the equilibrium concentrations from the ICE chart:
2 = (x)(x) / (0.5 - x)(0.5 - x)
Now we can solve for x:
2(0.5 - x)(0.5 - x) = x^2
2(0.25 - x + x^2) = x^2
0.5 - 2x + 2x^2 = x^2
2x^2 - x^2 - 2x + 0.5 = 0
x^2 - 2x + 0.5 = 0
Solving this quadratic equation will give us the value of x, which represents the equilibrium concentration of CO2 and H2O. In this case, the solutions are approximately x = 0.29 and x = 1.71.
Since we are dealing with moles, the concentrations of CO2 and H2 will be 0.5 - 0.29 = 0.21, and the concentrations of H2O and CO will be 0.29.
Therefore, the equilibrium concentrations of each reactant and product are approximately:
CO2: 0.21
H2: 0.21
H2O: 0.29
CO: 0.29