Keq = [0.25 - x] / [0.25 - x][2x]
= [0.25 - ] / [0.25 - ][2()]
=
What numbers need to be substituted
3 answers
If this refers to a problem I helped with earlier, you need to repost the entire problem. However, when I said to substitute into Keq, I meant to write the Keq expression and substitute the above number into it.
I2 + Cl2 ==> 2ICl (Is that the equation?)
initially:
(I2) = 0.5/2.0 = ??
(Cl2) = 0.5/2.0 = ??
(ICl)= 0
change:
(ICl) = +2x
(I2) = -x
(Cl2) = -x
equilibrium:
(ICl) = 0 + 2x = 2x
(I2) = (0.5/2.0) - x = 0.25 - x
(Cl2) = (0.5/2.0) - x = 0.25 - x
Set up Keq, substitute, and solve for x. Post your work if you get stuck. Check my work.
initially:
(I2) = 0.5/2.0 = ??
(Cl2) = 0.5/2.0 = ??
(ICl)= 0
change:
(ICl) = +2x
(I2) = -x
(Cl2) = -x
equilibrium:
(ICl) = 0 + 2x = 2x
(I2) = (0.5/2.0) - x = 0.25 - x
(Cl2) = (0.5/2.0) - x = 0.25 - x
Set up Keq, substitute, and solve for x. Post your work if you get stuck. Check my work.
No.
Keq = (ICl)^2/(I2)(Cl2)
(2x)^2/(0.25-2)(0.25-x)
Solve for x.
Keq = (ICl)^2/(I2)(Cl2)
(2x)^2/(0.25-2)(0.25-x)
Solve for x.