g Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation The amount of I2 in the reaction mixture can be determined from the intensity of the violet color of I2; the more intense the color, the more I2 in the reaction vessel. When 3.80 mol HI was placed in a 5.00-L vessel at a certain temperature, the equilibrium mixture was found to contain 0.443 mol I2. What is the value of Kc for the decomposition of HI at this temperature

Equilibrium constant, Kc = 0.023

Explanation:

Equation for the decomposition of Hydrogen iodide is given below:

2HI > H₂ + I₂

Initially, the number of moles of the reactant and the products are given as follows:

n(HI) = 2 * 3.800 moles = 7.600 moles

nH₂) = 0.000 moles

n(I₂) = 0.000 moles

At equilibrium, the equation becomes: 2HI <> H₂ + I₂

Number of moles of the reactant and the products are given as follows:

n(HI) = 7.600 - (0.886 + 0.886) moles = 5.828 moles

nH₂) = 2 * 0.443 = 0.886 moles

n(I₂) = 2 * 0.443 = 0.886 moles

Equilibrium constant, Kc = [H₂] [I₂] / [HI]²

Equilibrium constant, Kc = (0.886) * (0.886) / (5.828)²

Equilibrium constant, Kc = 0.023

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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