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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

You place 0.0300 mol of pure SO3 in an 8.00-L flask at 1150 K. At equilibrium, 0.0058 mol of O2 has been formed. Calculate Kc for the reaction at 1150 K.

2 SO3(g) ⇄ 2 SO2(g) + O2(g)

Interpretation Introduction

Interpretation:

The equilibrium constant Kc for the reaction under the given conditions 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
  • ICE (reaction initial concentration equilibrium) table is mainly used to calculate the value of K for a reaction. This table contains the concentration of reactant and product in various stage of reaction.
Explanation

The balanced chemical equation for the given reaction is,

    2SO3(g)2SO2(g)+O2(g)

The concentration of SO3=0.0300mol8.00L=0.00375mol/L

The concentration of O2(x)=0.0058mol8.00L=0.000725mol/L

By using these concentration ICE table for this reaction can be constructed as follows,

Reaction                           2SO3(g)2SO2(g)+O2(g)
Initial concentration(mol/L)0

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