Consider the following equilibrium:
2HI(g) *are in equilibrium with* H2(g) = I2(g) Keq = 81.0
A 2.00L container is initially filled with 4.00 mol HI. Calculate the [HI] at equilibrium.
*I don't know how to put the equilibrium sign on the computer*
Can someone please explain to me how to do this? Thank you in advance!
To do the equilibrium sign you simply use the < the > and == like this <==>. Most of use just use a straight arrow, either --> or ==> and let it go.
(HI) = mols/L = 4.0/2.00 = 2.00M
Set up an ICE chart as below:
........2HI ==> H2 + I2
I.......2.0M.....0....0
C......-2x.......x....x
E......2-2x......x....x
Substitute the E line into Keq expression and solve for x, then 2-2x.
why is it 2-2x? isn't it only just -2x?
oh i get it, it came from the initial... thank you !
That's right. You start with 2.0 (intial), take away -2x (the change) and end up with 2-2x (equilibrium).
To solve this question, we need to use the equilibrium expression and the given information to set up an equation that we can solve for the concentration of HI at equilibrium.
First, let's define the initial and equilibrium concentrations of HI as [HI]_initial and [HI]_equilibrium, respectively.
Given:
Initial concentration of HI, [HI]_initial = 4.00 mol in a 2.00 L container
Equilibrium constant:
Keq = 81.0
For the given reaction:
2HI(g) ⇌ H2(g) + I2(g)
The stoichiometry of the balanced equation tells us that for every 2 moles of HI, we get 1 mole of H2 and 1 mole of I2. Therefore, at equilibrium, the concentration of each of these species is related to the concentration of HI.
Since the molar ratio of HI to H2 is 2:1, at equilibrium, the concentration of H2 is half the concentration of HI:
[H2]_equilibrium = 0.5 * [HI]_equilibrium
Similarly, the molar ratio of HI to I2 is also 2:1:
[I2]_equilibrium = 0.5 * [HI]_equilibrium
Now, we can set up the equilibrium expression using the given equilibrium constant, Keq:
Keq = ([H2]_equilibrium * [I2]_equilibrium) / ([HI]_equilibrium)^2
Substituting the expressions for [H2]_equilibrium and [I2]_equilibrium:
Keq = (0.5 * [HI]_equilibrium) * (0.5 * [HI]_equilibrium) / ([HI]_equilibrium)^2
Simplifying the equation:
Keq = 0.25 * [HI]_equilibrium^2 / [HI]_equilibrium^2
Keq = 0.25
Now, we can solve for [HI]_equilibrium. Rearranging the equation:
[HI]_equilibrium^2 = Keq / 0.25
Taking the square root of both sides:
[HI]_equilibrium = √(Keq / 0.25)
Substituting the given value of Keq = 81.0:
[HI]_equilibrium = √(81.0 / 0.25)
[HI]_equilibrium = 18.0
Therefore, the concentration of HI at equilibrium, [HI]_equilibrium, is 18.0 mol/L.