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In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B → C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B: d [ C ] d t = k [ A ] [ B ] (See Example 3.7.4.) Thus, if the initial concentrations are [A] = a moles/L and [B] = b moles/L and we write x = [C], then we have d x d t = k ( a − x ) ( b − x ) (a) Assuming that a ≠ b , find x as a function of t . Use the fact that the initial concentration of C is 0. (b) Find x ( t ) assuming that a = b. How does this expression for x ( t ) simplify if it is known that [ C ] = 1 2 a after 20 seconds?

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Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550
BuyFind

Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550

Solutions

Chapter
Section
Chapter 9.3, Problem 40E
Textbook Problem

In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B → C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B:

d [ C ] d t = k [ A ] [ B ]

(See Example 3.7.4.) Thus, if the initial concentrations are [A] = a moles/L and [B] = b moles/L and we write x = [C], then we have

d x d t = k ( a x ) ( b x )

(a) Assuming that ab, find x as a function of t. Use the fact that the initial concentration of C is 0.

(b) Find x(t) assuming that a = b. How does this expression for x(t) simplify if it is known that [ C ] = 1 2 a after 20 seconds?

Expert Solution

(a)

To determine

To find: The function x(t) by solving the initial value problem: dxdt=k(ax)(bx) with x(0)=0 and assuming ab.

Explanation of Solution

Given:

In a chemical reaction the molecules A and B are reactants. The product of reactant is C.

[C]. at time 0 is zero. That is, x(0)=0.

The concentration of A and B is given below,

d[C]dt=k[A][B] (1)

The initial concentration is given below,

[A] = a moles/L and [B] = b moles/L then x = [C]

Formula used:

Partial fraction of 1(ax)(bx)=A(ax)+B(bx).

Calculation:

Consider the differential equation,

dxdt=k(ax)(bx) (2)

Simplify the above equation as follows.

dxdt=k(ax)(bx)dx(ax)(bx)=kdt

Here, ab.

Take integral on both sides,

dx(ax)(bx)=kdt (3)

By using partial fraction of 1(ax)(bx),

1(ax)(bx)=A(ax)+B(bx)1(ax)(bx)=A(bx)+B(ax)(ax)(bx)1=A(bx)+B(ax)

Substitute x=a in above solution,

1=A(ba)A=1(ba)

Substitute x=b in above solution,

1=B(ab)B=1(ab)

Substitute A and B value in the obtained solution,

1(ax)(bx)=1(ba)(ax)+1(ab)(bx)=1(ba)(ax)+1(ab)(bx)=1(ba)(ax)1(ba)(bx)=1(ba)(1ax1bx)

Substitute 1(ax)(bx) value in equatin (3),

1(ba)(1ax1bx)dx=kdt1(ba)

Expert Solution

(b)

To determine

To find: The function x(t) by solving the initial value problem: dxdt=k(ax)(bx) with x(0)=0 and x(20)=a2 and assuming a=b.

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Chapter 9 Solutions

Calculus: Early Transcendentals
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