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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

The equilibrium constant for the dissociation of iodine molecules to iodine atoms

I2(g) ⇄ 2 I(g)

is 3.76 × 10−3 at 1000 K. Suppose 0.105 mol of I2 is placed in a 12.3-L flask at 1000 K. What are the concentrations of I2 and I when the system comes to equilibrium?

Interpretation Introduction

Interpretation: The concentration of I2andI when the system comes to equilibrium has to be determined.

Concept introduction:

  • Equilibrium constant: At equilibrium the ratio of products to reactants has a constant value. And it is represented by the letter K. 

    For a general reaction, aA+bBcC+dD

    The equilibrium constant Kc = [C]c[D]d[A]a[B]b, where a, b, c and d are the stoichiometric coefficients of reactant and product in the reaction. Concentration value for solid substance is 1.

    If the value of Kc and the concentration of any of the reactant of a reaction is known then the concentration of product can be determined by multiplying Kc with the concentration of reactant.

  • Concentration =AmountofsubstanceVolume
Explanation

The balanced equation for the dissociation iodine molecules to iodine atoms is,

     I2(g)2I(g)

The concentration of I2,[I2] = 0.105mol12.3L=8.5×103mol/L

Initial concentration of reactant I2,[I2]=8.5×103x

Final concentration is (8.5×103x)2x

Substituting this concentrations of reactant and product in the equilibrium constant equation,

Thus,

Kc=[I]2[I2]=(2x)28.5×103x

Concentration of I,[I] = 2x

Concentration of I2,[I2]= 8.5×103x

The equilibrium constant for this reaction is 3.76×103 at 1000K

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