Asked by katie
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!
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!
Answers
Answered by
DrBob222
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.
(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.
Answered by
katie
why is it 2-2x? isn't it only just -2x?
Answered by
katie
oh i get it, it came from the initial... thank you !
Answered by
DrBob222
That's right. You start with 2.0 (intial), take away -2x (the change) and end up with 2-2x (equilibrium).
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