Suppose we seal I2(g) and H2(g) in a vessel at T = 400 K with partial pressures

Question

Suppose we seal I2(g) and H2(g) in a vessel at T = 400 K with partial pressures PI2 = 0.067 atm and PH2= 1.275 atm. At this temperature H2 and I2 do not react readily to form HI(g), although in principle if we waited long enough they would produce HI(g) at its equilibrium partial pressure. Instead of waiting for this to happen, we choose to heat the gases in the sealed flask to 600 K, where they readily react to form HI(g). For the equilibrium

the equilibrium constant has been experimentally determined to be Keq = 38.6 at 600 K. What will PH2, PI2 and PHI be when equilibrium is established at T = 600 K?

Explanation / Answer

The reaction will be

                 H2         + I2    -->          2HI

Pressures

Initial        1.275                 0.067                    0

Change -x                          -x                         2x

Equilb       1.275-x           0.067-x                2x

Keq = 38.6 = [HI]^2 / [H2] [I2] = (2x)^2 / (1.275-x) (0.067-x)

38.6 = 4x^2 / (0.085 - 1.275 x - 0.067x +x^2

3.281 - 49.215x - 2.586x + 38.6x^2 = 4x^2

3.28 -51.50x + 34.6x^2 = 0

On calculating the equation

x = 1.42 or x = 0.067

The possible value is x = 0.067

So Partial pressure of H2 = pH2 = 1.275-0.067 = 1.208

pI2 = 0.067-0.067 = 0

pHI = 2x = 2X0.067 = 0.134

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