The equilibrium pressure of O 2 to maintain 1 : 1 mole ratio of SO 3 to SO 2 has to be calculated. Concept Introduction: Equilibrium constant ( K p ) : In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant ( K p ) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction Consider the reaction where A reacts to give B. aA ⇌ bB Rate of forward reaction = Rate of reverse reaction k f P A a =k r P B a On rearranging, P B b P A a = k f k r =K p Where, k f is the rate constant of the forward reaction. k r is the rate constant of the reverse reaction. K p is the equilibrium constant.

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Chapter 22, Problem 22.55P

(a)

Interpretation Introduction

Interpretation:

The equilibrium pressure of $O2$ to maintain $1:1$ mole ratio of $SO3 to SO2$ has to be calculated.

Concept Introduction:

Equilibrium constant $(Kp)$ :

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant $(Kp)$ is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkfPAa =krPBa$

On rearranging,

     $PBb PAa=kfkr=Kp$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kp$ is the equilibrium constant.

(b)

Interpretation Introduction

Interpretation:

The equilibrium pressure of $O2$ to maintain $95:5$ mole ratio of $SO3 to SO2$ has to be calculated.

Concept Introduction:

Equilibrium constant $(Kp)$ :

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant $(Kp)$ is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkfPAa =krPBa$

On rearranging,

     $PBb PAa=kfkr=Kp$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kp$ is the equilibrium constant.

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