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
