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Hydrogen iodide, HI, decomposes at moderate temperature according to the equation 2HI (g) H2 (g) +...

Hydrogen iodide, HI, decomposes at moderate temperature according to the equation 2HI (g) H2 (g) + I2 (g) When 4.00 mol HI was placed in a 5.00-L vessel at 458C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc for the decomposition of HI at this temperature?

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Answer #1

concentration of HI = 4.00 / 5.00 = 0.800 M

2 HI (g)   <------------> H2 (g)   + I2 (g)

0.800                            0              0

0.80 - 2x                       x               x

At equilibrium :

moles of I2 = 0.442 mol

concentration of I2 = 0.442 / 5 = 0.0884 M

[H2] = x = 0.0884 M

[HI] = 0.623 M

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

   = (0.0884)^2 / (0.6232)^2

Kc = 0.0201

value of Kc = 0.0201

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