Equilibrium Physics Help: Calculating Kp for a Hydrogen Iodide Reaction

AI Thread Summary
The discussion focuses on calculating the equilibrium constant Kp for the dissociation of hydrogen iodide (HI) into hydrogen (H2) and iodine (I2). The reaction is represented as 2HI(g) <--> H2(g) + I2(g), with b moles of HI initially and x moles dissociated at equilibrium. The participant derives the pressures for reactants and products based on the initial and equilibrium conditions, ultimately expressing Kp as (b^2) / (b-x)^2. The participant seeks validation of their solution and requests assistance from others in the forum. The discussion emphasizes the importance of accurately interpreting the dissociation of hydrogen iodide in the context of the equilibrium constant calculation.
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Homework Statement



In an experiment , b mole of hydrogen iodide were put into a sealed vessel under pressure p . At equilibrium, x mole of the hydrogen iodide had dissociated , the reaction being represented by the following equation ,
Express in terms of b and x of Kp .

Homework Equations



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


The Attempt at a Solution


Here is my solution :
Reactant:
Before experiment : b mole of hydrogen iodide under pressure , p
At equilibrium : x mole of hydrogen iodide
P(pressure) for hydrogen iodide : [(p/b).(x)]^2

Products:
Before experiment : unknown
At equilibrium : I conclude that mole ratio of total products is the ( b-x )
mol . Therefore ,
P(pressure) for gas hyrogen and gas iodine : [(b-x).(p/b)]^2


Answer:
Kp = ([(b-x).(p/b)]^2 ) / ([(p/b).(x)]^2)
= (b^2) / (b-x)^2


If I am wrong , please guide me .
 
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Nobody guide me here ?
 
Somebody help me please .
 
Pay careful attention to the question: "x mole of the hydrogen iodide had dissociated ".
 
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