Question

Suppose a 250.mL flask is filled with 0.60mol of I2 and 1.4mol of HI . The following reaction becomes possible: H2(g) + I2 (g) -> 2HIg The equilibrium constant K for this reaction is 0.644 at the temperature of the flask. Calculate the equilibrium molarity of HI . Round your answer to one decimal place.

Answer 1

[I2] = 0.6/0.25 = 2.4

[HI] = 1.4/0.25 = 5.6

[H2] = 0

in equilibrium

[I2] = 2.4 - x

[HI] = 5.6 + x

[H2] = 0 - x

from here, we can see that x will be negative, since H2 cant be necative

K = [HI]^2 / [H2][I2]

0.644 =(5.6 + x)^2 / (2.4 - x)(-x)

0.644*(2.4 - x)(-x) = 5.6^2 +5.6*2x + x^2

-1.5456x + 0.644x^2 = 31.36 + 11.2x + x^2

(1-0.644)x^2 + (-1.5456-11.2)x -31.36 = 0

0.356x^2 -12.745x  -31.36 = 0

x = -2.31

then

[I2] = 2.4 - -2.31 = 3.29

[HI] = 5.6 + -2.31 = 4.71

[H2] = 0 - -2.3 1= 2.21