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sdas [7]
3 years ago
10

g Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation The amount of I2 in the reaction mixture ca

n 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
Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer:

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

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