An equilibrium mixture in a 10.0 L flask contains 7.0 mol HI(g) and 1.0 mol each
of I2(g) and H2(g). If 2.5 mol H2, 5.0 mol I2 and 2.0 mol HI are added to that
equilibrium mixture how many moles of each gas will be present when equilibrium
is reestablished?
First, convert mols to M.
(HI) = 7/10 = 0.7M
(I2) = 1/10 = 0.1M
(H2) = 1/10 = 0.1M
Next calculate the Kc.
2HI ==>H2 + I2
Kc = (H2)(I2)/(HI)^2
Kc = (0.1)(0.1)/(0.7)^2 = approx 0.02
Then do and ICE chart.
............2HI ==> H2 + I2
I...........0.7....0.1...0.1
add.........0.2....0.25..0.5
Determine which way the reaction will shift and you do that by looking at Qc.
Qc = (0.25)(0.5)/(0.2)^2 = approx 3
Since Qc is larger than Kc, you know the numerator is too large and the denominator is too small; therefore, the reaction will shift to the left in order to diminish the products and increase the reactants.
............+2x......-x....-x
E.........0.9+2x..0.35-x..0.6-x
Now substitute the E line into Kc expression and solve for x and evaluate 0.90+2x
Thank you, but I have one more question. In your solution, wouldn't the equilibrium constant equation be [HI]^2 / [H2][I2]
Wouldn't HI be the product and H2 and I2 be the reactants?
I have made mistakes like that but in this case I didn't. It could be either way. Which ever way you do it will work out.
If you assume you started with HI, then
2HI ==> H2 + I2
and the EQUILIBRIUM MIXTURE will be as given. Then you work out Kc for that reaction. You add the quantities listed and use this Kc to work out the details for the new equilibrium.
If you assume you started with H2 and I2 and it gives HI
H2 + I2 ==> 2HI
and the EQUILIBRIUM MIXTURE will still be as given BUT the Kc for that reaction will not be the same as the one above (actually Kc for this reverse reaction will be 1/Kc original). Then when you add in the quantities the reaction still will shift (but not in the same direction) and that + the new Kc makes it all come out in the wash.
You might want to sit down and do it both ways and check it out. That might be a good "extra credit" problem.
Would x equal .114?
I obtained, using the equilibrium as
2HI ==> H2 + I2 as
K = (H2)(I2)/(HI)^2 = (0.1)(0.1)/(0.7)^2 = 0.0204 which would be
1/0.0204 = 49 for the way you suggested.
To solve this question, we need to use the balanced equation for the reaction between hydrogen iodide (HI), hydrogen (H2), and iodine (I2). The equation is as follows:
2 HI (g) ⇌ H2 (g) + I2 (g)
The equilibrium constant expression for this reaction is:
Kc = [H2] * [I2] / [HI]^2
Given the initial amounts of each gas in the equilibrium mixture (7.0 mol HI, 1.0 mol I2, and 1.0 mol H2), we can calculate the initial value of the equilibrium constant, Kc_initial, using these values.
Now, let's calculate Kc_initial:
Kc_initial = (1.0 * 1.0) / (7.0^2) = 0.02041
To determine how many moles of each gas will be present when equilibrium is reestablished after adding 2.5 mol H2, 5.0 mol I2, and 2.0 mol HI, we need to calculate the changes in the number of moles for each reactant and product.
Let's denote the change in the number of moles for each substance as Δn:
Δn(H2) = 2.5 - 1.0 = 1.5 mol
Δn(I2) = 5.0 - 1.0 = 4.0 mol
Δn(HI) = 2.0 - 7.0 = -5.0 mol
Now, we can use the equation:
Kc_final = (([H2] + Δn(H2)) * ([I2] + Δn(I2))) / ([HI] + Δn(HI))^2
Substituting the values:
Kc_final = ((1.0 + 1.5) * (1.0 + 4.0)) / (7.0 + (-5.0))^2
Kc_final = (2.5 * 5.0) / (2.0)^2
Kc_final = 12.5 / 4.0
Kc_final = 3.125
Now, to find the number of moles of each gas when equilibrium is reestablished, we can set up an expression using the value of Kc_final:
3.125 = ([H2] + Δn(H2)) * ([I2] + Δn(I2)) / ([HI] + Δn(HI))^2
Solving this expression will give us the value of [H2], [I2], and [HI].
Please note that to solve this equation, you may need to use an iterative method such as the quadratic formula or a calculator that can handle equations with multiple variables.