SECOND ORDER ⇔ SECOND ORDER THAT GOES TO EQUILIBRIUM

Chapter 2 Introduction to Kinetic Processes in Materials 41 Chapter 3 Second-Order and Multistep Reactions 61 Chapter 4 Temperature Dependence of the Reaction Rate Constant 95

3.4 SECOND ORDER ⇔ SECOND ORDER THAT GOES TO EQUILIBRIUM

3.4.1 G

^eneRal

Whether the reaction is a second order forward and second order back, or first order forward and second order back, or second order forward and first order back, the results are all similar and the procedures used to get the solutions are essentially the same for all three cases. The reason for this is that the rates of reaction, dx/dt (x here is concentration to minimize symbol complexity), are all proportional to a second-order polynomial in x. From the results in Section 3.3.2, it is tempting to write for a second-order reaction,

0 20 40 60 80 100

0.2 0.4 0.6 0.8 1.0

[B]₀

[A]_e [B]_e [A]₀

Time (s) [A]

[B]

k = 0.04 s⁻¹

Concentration (arbitrary units)

FIGURE 3.4 First-order reaction that goes to equilibrium—one that has a first-order back reaction.

d A

dt k A A _e

[ ]

_{= −}

( [ ]

⁻

[ ] )

^2. _(3.22)

Unfortunately, this is incorrect and solutions to second-order equations that go to equilibrium unfortunately are more complex.

The detailed procedure carried out in Appendix A.2 works for second order ⇔ second order, second order ⇔ first order, and first order ⇔ second order because all involve the solution of quadratic equations. In Appendix A.2.4, the general solution for second order ⇔ second order is found. The reaction is assumed to be the simple reaction:

AB or

d A

dt k A k B

[ ]

_{= −} ¹

[ ]

^{2+ 2}

[ ]

^2. (3.23)

To minimize the notation, let [A] = x, and [A]0 = x0 and [B]0 = 0 so that Equation 3.23 becomes dx

dt = −k x1 2+k x

(

−x

)

2 0

2. (3.24)

Now this is a relatively simple differential equation to solve because the variables are separable, all the “xs” on the left, and “ts” on the right, so that standard integration gives a solution without recourse to special differential equation solution techniques.

3.4.2 e

^quilibRium

At equilibrium, x x= e so Equation 3.24 becomes

−^{k x}1 e2+^{k x}

(

−^xe

)

⁼

2 0

2 0

or

k x k x x k

k

x x x

e e

e e 1 2

2 0

2

1 2

0

=

(

−

)

± = −

.

Note the ± sign in front of the term on the left that is necessary because the square root is being taken. This leads to two mathematically possible equilibrium values as will be seen. At this point, the initial concentration of A, x0, could be obtained and substituted into Equation 3.24 and eliminated as was done for the first-order reaction in Section 3.3. However, this does not help much to solve this problem. In any event, the equilibrium concentration can be obtained from

x k

k x x

x k k

k x

e e

e

± =

 ±

 

 =

1 2

0

2 1

2

0

which leads to

x x k

k k

e=

±



 



0 2

2 1 (3.25)

and the algebra is already getting messy and the differential equation, Equation 3.24, has yet to be solved. The general solution, involving much algebra, is left to the Appendix.

3.4.3 s

^PeciFic

e

^xamPle

To make things simpler, and hopefully more transparent, start with a specific case of k1 = 2, k2=1,and x0=1. So Equation 3.25 becomes

x x k

k k

e=

±



 

 = ±



 



0 2

2 1

1 0 1 0 1 0 2

. .

.

which gives xe=0 41421. and−2 41421. . Clearly, xe must be greater than zero so the first xe = 0.41421 must be the equilibrium value. (Note that if k2 > k1, then xe > x0, which is physically impossible if [B]0 = 0 as assumed.) The differential equation, Equation 3.24 with these numerical values, becomes,

dx

dt x x

dx

dt x x x

dx

dt x x

= − + −

( )

= − + − +

= − − +

2 1

2 1 2 2 1

2 2

2 2

2 .

(3.26)

This is easy to solve by finding the roots to the quadratic equation on the right and one of these roots will be the equilibrium concentration of A from which the equilibrium concentration of B can be calculated. So

x b b ac

= − ± ²a−4 2

or x

x and

= ± +

−

= −

2 4 4 2

2 41421. 0 41421. .

Not surprisingly, the roots of the quadratic equation are the equilibrium concentrations. As a result, Equation 3.26 can be written in terms of its factors:

dx

dt =

(

^{2 41421.} +x

) (

^{0 41421. −}x

)

^. (3.27) Just as a check, multiplying these two factors gives

2 41421 0 41421 2 41421 0 41421 0 99999 2

2

2

. . . .

.

× − + −

− −

x x x

x x

which is pretty close to 1 2− x x− ²! The integration of Equation 3.27 is now straight forward. First, separate variables:

dx

x x dt

2 41421. 0 41421. .

(

+

) (

⁻

)

^{= (3.28)}

Now separate the denominator into two terms that are easily integrated, the method of partial fractions,

1

2 41421. + 0 41421. 2 41421. 0 41421.

(

^x

) (

^−x

)

⁼

(

^{A +x}

)

⁺

(

^{B −x}

)

multiplying both sides by the denominator of the left-hand side A

(

0 41421. −x

)

⁺B

(

2 41421. ⁺x

)

⁼1

and equate the left- and right-hand factors of x and the constant gives −Ax Bx+ =0or A B= because there is no x-term on the right-hand side, and 0 41421. A+2 41421. A=2 82842. A=1 or A=1 2 82842. =B. Equation 3.28 can now be written as

dx x

dx

x dt

2 41421 0 41421 2 82842

. . .

(

+

)

⁺

(

⁻

)

⁼

and integrated to

ln .

(

2 41421+^x

)

⁻ln .

(

0 41421^−x

)

⁼2 82842. ^{t C+}

where C is just a constant of integration. Taking the exponent of both sides of the equation gives 2 41421

0 41421

2 82842

. .

+ .

( )

(

−^xx

)

^{= ′C e t}

where C′ is just another constant. When t = 0, x = x0, so

′ =

(

+

)

(

−

)

⁼

(

+

)

(

−

)

^{= −}

C x

x 2 41421 0 41421

2 41421 1

0 41421 1 5 8284

0 0

. .

.

. .

and the solution becomes

2 41421

0 41421. 5 8284 ^{2 82842}

. + . ^.

( )

(

−^xx

)

^{= − ×e t=D (3.29)}

where at any given time t the right-hand side of this equation is a constant D. It is now necessary to manipulate Equation 3.29 to get “x” explicitly. So, Equation 3.29 can be written as

2 41421 0 41421 2 41421 0 41421

1 0 41

. .

. .

( ) .

(

+

)

^{= ×}

(

⁻

)

+ = × −

+ =

x D x

x D Dx

x D 4421 2 41421 0 41421 2 41421

1

0 41421 2 41421

1 1

× −

= × −

+

= −

+ D

x D

D

x D

D .

. .

. .

and replacing Equation 3.29 for D gives the sought for solution

x e

e

t

= + × t

− ×

−

−

0 41421 0 41421 1 0 17157

2 82842 2 82842

. .

.

.

. (3.30)

when t = 0, x = 0.99999 which is pretty close to 1.0 and when t = ∞, x = xe = 0.41421. The concentra- tion of B, [B], is just

[ ]

B ⁼x⁰⁻x and B^{[ ]}e^{=0 58579. .} Both [A] and [B] are plotted in Figure 3.5.

3.4.4 t

^he

G

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s

^OlutiOn

Following exactly the same procedure as in Section 3.4.3 for the specific case, the general solution to Equation 3.24 is solved in Appendix A.2.4 and is

x

x k k k e

k k k k e

x k k t

x k k

=

(

+

) {

⁺

}

−

( (

−

) (

⁺

) )

−( )

−

0 2 2 1

2

1 2 1 2

2

1 1

0 2 1

0 2 1

/

/ (( )^{t. (3.31)}

3.4.5 n

^umeRical

s

^OlutiOn

Given the amount of algebra and elementary calculus involved in obtaining the general solution to a relatively simple differential equation of importance to kinetics, what about higher order reac- tions and noninteger orders? In general, it is generally neither reasonable nor possible to solve such problems analytically—at least not in the time it might take—so numerical techniques are used.

As simple example of one of these numerical procedures is the so-called Euler’s method (Celia and Gray 1992; Blanchard et al. 2012) that involves treating the derivative as a ratio of differences:

dx dt= ∆ ∆x t. This is an example of a finite difference approximation that can be used to solve both ordinary and partial differential equations under certain conditions (Celia and Gray 1992; Kee et al.

2003). It is a particularly useful technique because the solutions can frequently be obtained with simple spreadsheet calculations.

For example, consider the specific differential equation given in Equation 3.26 dx

dt = − −x² 2x+1 (3.32)

which required a fair amount of algebra and calculus to get the final result, Equation 3.30 with the initial condition that x = 1.0 at t = 0. Putting Equation 3.32 into difference form,

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Time [A]

[A]₀ = 1.0

[B]_e = 0.58579

[A]_e = 0.41421 k₁ = 2.0

k₂ = 1.0

[B]

Concentration

FIGURE 3.5 Second-order reaction with a second-order back reaction AB with the specific values for the constants in the equation: [A]₀ = 1.0, [B]₀= 0, k₁= 2.0, and k₂= 1.0. The calculated points and the equilibrium concentrations are shown.

∆

∆ ∆

x ∆

t

x x

t x x

t t t

t t

=

(

⁺ −

)

_{= − −}2 2 ₊1 and rearranging gives

xt+^∆t=xt+^∆t

(

− −xt^{2 2}xt+^{1 .}

)

^(3.33)

That is, for a uniform time-step, Δt, the value of xt+∆t at the next time-step, is just the value at the pre- vious time-step, xt, plus Δt times the value of the differential equation at the previous time-step. For the plots of [A] and [B] in Figure 3.5 from Equation 3.30, a relatively large time-step of Δt = 0.05 s was used to calculate these curves. This is somewhat larger than desirable for the evaluation by Euler’s method but it will be used to illustrate the method. So, because x0 = 1.0, x1 at Δt is calculated by

x x t x x

x x

1 0 02

0 1

1

2 1

1 0 0 05 1 2 1 0 9

= +

(

− − +

)

= + − − +

=

∆

. . ( )

. and x2  at t = 2Δt,

x x t x x

x x

2 1 12

1

2 2

2

2 1

0 9 0 05 0 9 2 0 9 1 0 9 0

= +

(

− − +

)

= + − − +

= −

∆

. . ( . ) ( . ) . .

( )

00805 0 8195 x2= .

and so on. The values of x(t) at these two times calculated with the exact solution, Equation 3.30, are x1 = x(0.05) = 0.9092 and x2 = x(0.1) = 0.8342, less than 2% difference between the numerical and exact solution. This difference could be made considerably smaller with a smaller Δt. Figure 3.6 shows how the finite difference approximation can be easily and quickly calculated with a spreadsheet. The first column is just n increasing going down the column as far as desired; the second column is time, nΔt, where Δt is fixed (but does not have to be); and column three, x is calculated with Equation 3.33.

A powerful feature of a spreadsheet solution is only the first row for n = 0 needs to be inserted and then simply copied vertically down to a desired end point. Also, Δt can be introduced as a separate constant and varied to get as close as possible to the analytical solution. Figure 3.7 gives plots of [A]

Time x

0 1Δt 2Δt

(n+1)Δt nΔt x_n

x_n+1

= −x² − 2x + 1 Differential equation dx

dt

x_n+1 = x_n + Δt(−x_n − 2x_n + 1) Differential equation

Spreadsheet columns n

1 2 0

n n+1

x₂ x₁ x₀

...

...

...

...

... ...

2

x₁ = x0 + Δt(−x02 − 2x0 + 1)

x_n+1 = x_n + Δt(−x_n² − 2x_n + 1)

FIGURE 3.6 Spreadsheet layout for the numerical calculation of the concentrations as a function of time, for the specific reaction equation plotted in Figure 3.5, with Euler’s finite difference method of approximation. The values of x_n from the difference equation are plotted in the x column.

for both the exact and a spreadsheet numerical solutions to Equation 3.32 with Δt = 0.5. The two solu- tions are pretty close and get closer at longer times because, in this case, the derivative is getting smaller. Clearly, the numerical solution is the easier and faster way to solve this relatively simple problem and numerical solutions become mandatory as the kinetics become more complex, as indeed they do, even for simple reactions.

3.5 PARALLEL AND SERIES REACTIONS