Asked by Presh
A 5.00L reaction vessel is filled with 1.00mol of H2, 1.00mol of I2, and 2.50mol of HI. If equilibrium constant for the reaction is 129 at 500k, what are the equilibrium concentrations of all species?
H2(g) + I2(g) <-> 2HI(g) Kc = 129
Answers
Answered by
DrBob222
(H2) = 1.00/5.00 = 0.2M
(I2) = 1.00/5.00 = 0.2M
(HI) = 2.50/5.00 = 0.5M
.........H2 + I2 ==> 2HI
I........0.2...0.2...0.5
C........-x....-x....+2x
E.......0.2-x..0.2-x..0.5+2x
Kc = (0.5+2x)^2/(0.2-x)(0.2-x)
Note: In the C line, how do we know it is -x. -x, and +2x instead of +x +x, and -2x;i.e., how do we which way the reaction will occur. You run a Qrxn = (HI)^2/(H2)(I2)= (0.5)^2/(0.2)^2 = about 6. Compare with K = 129 and that means the numerator is too small and denominator is too large so the reaction must go to the right.
(I2) = 1.00/5.00 = 0.2M
(HI) = 2.50/5.00 = 0.5M
.........H2 + I2 ==> 2HI
I........0.2...0.2...0.5
C........-x....-x....+2x
E.......0.2-x..0.2-x..0.5+2x
Kc = (0.5+2x)^2/(0.2-x)(0.2-x)
Note: In the C line, how do we know it is -x. -x, and +2x instead of +x +x, and -2x;i.e., how do we which way the reaction will occur. You run a Qrxn = (HI)^2/(H2)(I2)= (0.5)^2/(0.2)^2 = about 6. Compare with K = 129 and that means the numerator is too small and denominator is too large so the reaction must go to the right.
Answered by
Presh
What does x equal out too?
Answered by
Shelby
h2, i2 = 11.4
hi = 22.8
hi = 22.8