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If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49?
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The given equation for the dissociation of HI is:
HI ⇌ H+ + I-
Given that 2/9 of 1 mol of HI dissociates, we can calculate the concentration of H+ and I- ions at equilibrium.
Calculating the Equilibrium Concentrations of H+ and I-
Let x be the concentration of H+ ions and I- ions at equilibrium.
HI dissociates into H+ and I- ions in the ratio of 1:1. Therefore, the concentration of H+ and I- ions at equilibrium will be x/2.
At equilibrium, the concentration of undissociated HI will be (1 - 2x/2) = (1 - x).
The equilibrium constant expression for the dissociation of HI is:
Kc = [H+][I-]/[HI]
Substituting the equilibrium concentrations of H+ and I- ions, we get:
Kc = (x/2)(x/2)/(1 - x)
Calculating the Value of Kc
Given that 2/9 of 1 mol of HI dissociates, the initial concentration of HI is 1 mol and the concentration of H+ and I- ions at equilibrium is x/2.
Therefore, the concentration of undissociated HI at equilibrium is (1 - x) = (7/9) mol.
Substituting these values in the expression for Kc, we get:
Kc = (x/2)(x/2)/(7/9)
Kc = (81/196)(x^2)
At equilibrium, the concentration of H+ and I- ions will be equal to each other, i.e., x/2 = (7/18) mol.
Substituting this value in the expression for Kc, we get:
Kc = (81/196)[(7/18)^2] ≈ 0.409
Choosing the Correct Option
The equilibrium constant of disintegration of acid is given by Kc = [H+][I-]/[HI].
At equilibrium, the concentration of H+ ions and I- ions will be equal to each other, i.e., x/2 = (7/18) mol.
Therefore, [H+][I-] = (7/18)^2.
From the expression for Kc, we know that Kc = (81/196)(x^2).
Substituting the values of [H+][I-] and [HI] at equilibrium, we get:
Kc = [(7/18)^2]/(7/9) = 1/49
Therefore, the correct option is (d) 1/49.
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If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49?
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If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49?.
If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be a) 64 b) 1/64 c) 49 d) 1/49?.
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