Deriving concentrations from data n Part A, you were given the equilibrium press

Question

Deriving concentrations from data n Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K. In Part B however, you will be given initial concentrations and only one equi concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula for K Part B The following reaction was performed in a sealed vessel at 797 oC H2 (g) +I (g 2HI(g) Initially, only H2 and I were present at concentrations of H2 3.65M and I 2.80M The equilibrium concentration of I2 is 0.0800 M. What is the equilibrium constant, Kc, for the reaction at this 2 temperature? Express your answer numerically Kc 0.6144 Submit Hints My Answers Give Up Review Part Incorrect, Try Again, 4 attempts remaining Provide Feedback Continue

Explanation / Answer

Let's make an ICE table for the reaction.

Molarity . . . H2(g) + I2(g) <=> 2HI(g)
initial . . . . . 3.65 . . . .2.8 . . . . . . .0 . .
change . . . . .-x . . . . . .-x . . . . . . .2x . .
at equ. . . . 3.65-x . . .2.8-x . . . . . 2x . .

At equilibrium, [HI] = 2x = 0.0800, so x = 0.0400.

[H2] = 3.65 - x = 3.65 - 0.04 = 3.61 M
[I2] = 2.80 - x = 2.80 - 0.04 = 2.76 M

Kc = [HI]^2 / ([H2][I2]) = (0.0800)^2 / ((3.61)(2.76)) = 6.42 x 10^-4 Ans.

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