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Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant (Kc) has to be calculate given elementary of reversible reaction.

Concept Information:

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions.   Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x.

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

Reversible Reaction and equilibrium conversion: The equilibrium, the concentration of the species is related by the relationship dictated by thermodynamic equation.

  Kc=CB2CA

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant value should be determined given liquid-phase reaction.

Concept Information:

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions.   Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x.

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

(c)

Interpretation Introduction

Interpretation:

Equilibrium constant for the given reaction has to be written as a function of conversion, numerically evaluating all possible symbols.

Concept Information:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x.

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

(d)

Interpretation Introduction

Interpretation:

Equilibrium constant for the given reaction has to be written as a function of conversion, numerically evaluating all possible symbols.

Concept Information:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x.

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

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4.59 Using the unilateral z-transform, solve the following difference equations with the given initial conditions. (a) y[n]-3y[n-1] = x[n], with x[n] = 4u[n], y[− 1] = 1 (b) y[n]-5y[n-1]+6y[n-2]= x[n], with x[n] = u[n], y[-1] = 3, y[-2]= 2 Ans. (a) y[n] = -2+9(3)", n ≥ -1 (b) y[n]=+8(2)" - (3)", n ≥ -2
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