If 1.00 mol each of carbon dioxide and hydrogen is initially injected into a 10.0-L reaction chamber at 986 degrees Celsius, what would be the concentrations of each entity at equilibrium?
CO2(g) + H2(g) „²„³ CO(g) + H2O(g) k = 1.60 for 986 degrees Celsius
Heres my work:
C = n/v
= 0.1 mol/L
CO2(g) + H2(g) „²„³ CO(g) + H2O(g)
I 0.1 0.1 0 0
C -x -x +x +x
E 0.1-x 0.1-x x x
Keq = [CO][H2O]/[CO2][H2]
1.60 = x(x)/(0.1 -x)(0.1 ¡V x)
This is the part I am stuck with please help thanks a lotļ
ok. You have
x*x/(0.1-x)(0.1-x) = 1.60. That is
X^2/(0.1-X)^2 =1.60
Take the square root of both sides.
X/(0.1-X) = sqrt 1.60
X/0.1-X)=1.265
X = 1.265(0.1-X)
X = 0.1265 - 1.265X. Rearrange to
X+1.265X = 0.1265
X = 0.1265/2.265 = 0.0558 which rounds to 0.056 M for CO and H2O.
0.1 - X = 0.1 - 0.0558 = 0.0442 which rounds to 0.044 for CO2 and H2.
I don't know how picky your prof is about significant figures so check these out. Check my work. Check my arithmetic. check it all.
Check the s.f. and key in the right number for s.f. The problem is solved correctly.
To solve this problem, you need to use the equilibrium expression and plug in the given values.
The equilibrium expression for this reaction is:
Keq = [CO][H2O] / [CO2][H2]
First, substitute the given value for the equilibrium constant (Keq = 1.60 for 986 degrees Celsius).
1.60 = x(x) / (0.1 - x)(0.1 – x)
Next, simplify the equation by multiplying both sides by (0.1 - x)(0.1 - x) to clear the denominator:
1.60 * (0.1 - x)(0.1 - x) = x(x)
Now, expand the brackets on the left side:
0.16 - 1.60x + 1.60x^2 = x^2
Simplify the equation:
0.16 = 3.20x^2
Divide both sides by 3.20 to isolate x^2:
0.16 / 3.20 = x^2
0.05 = x^2
Take the square root of both sides:
√0.05 = x
x ≈ 0.224
Now that you have found the value of x, you can substitute it back into the equation to find the concentrations of each entity at equilibrium. Using the values given in the initial setup:
[CO2] = 0.1 - x ≈ 0.1 - 0.224 ≈ -0.124 (which is not physically possible, so we discard this negative value)
[H2] = 0.1 - x ≈ 0.1 - 0.224 ≈ -0.124 (also not physically possible)
[CO] = x ≈ 0.224
[H2O] = x ≈ 0.224
Therefore, at equilibrium, the concentrations of CO and H2O would be approximately 0.224 mol/L, while the concentrations of CO2 and H2 would remain negligible (-0.124 mol/L, which is not physically possible).