Kinetics - Derivations AWS

Fall 2017

Kinetics - Derivations Deriving the Intergrated Rate Laws Zero-Order Case

First, we begin with the zero-order rate equation.

Rate = k[A]n This is next written as the differential form in which n = 0, meaning the time component is now included

Rate = −

𝑑[𝐴]0 𝑑𝑡

=k

We then rearrange to get d[A] and dt on opposite sides

d[A] = −kdt We next perform our integration on both sides of the equation. [𝐴]

𝑡

∫[𝐴] 𝑑[𝐴] = −𝑘 ∫0 𝑑𝑡 0

Lastly, we solve for [A].

[A] – [A]0 = −k(t-0) [A] = [A]0 − kt

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Fall 2017 First-Order Case

First, we begin with the first-order rate equation.

Rate = k[A]n This is next written as the differential form in which n = 1, meaning the time component is now included

Rate = −

𝑑[𝐴] 𝑑𝑡

= k[A]

We then rearrange to get d[A] and dt on opposite sides 𝑑[𝐴] [𝐴]

= −kdt

We next perform our integration on both sides of the equation. [𝐴] 𝑑[𝐴]

∫[𝐴]

0 [𝐴]

𝑡

= −𝑘 ∫0 𝑑𝑡

ln[A] – ln[A]0 = -k(t-0) ln[A] – ln[A]0 = -kt

Lastly, we solve for ln [A].

ln[A] = ln[A]0 – kt

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Fall 2017

Second-Order Case First, we begin with the first-order rate equation.

Rate = k[A]n This is next written as the differential form in which n = 2, meaning the time component is now included 𝑑[𝐴]

Rate = −

𝑑𝑡

= k[A]2

We then rearrange to get d[A] and dt on opposite sides 𝑑[𝐴] [𝐴]2

= −kdt

We next perform our integration on both sides of the equation. [𝐴] 𝑑[𝐴]

∫[𝐴]

0 [𝐴]2

1 [𝐴]

– 1

[𝐴]

Lastly, we solve for

1

[𝐴]

𝑡

= −𝑘 ∫0 𝑑𝑡

1 [𝐴]0

–

= -k(t-0)

1 [𝐴]0

= -kt

. 1 [𝐴]

=

1 [𝐴]0

-kt

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Fall 2017

Half-Lives Zero-Order Case Half life means that half of the material is used up. 1

[A] = 2[A]0 We start with the integrated rate law (as it includes the time component). [A]=[A]o−kt We now substitute the first equation into the second. 1

[A]0=[A]o−k𝑡1⁄

2

2

Now solve for t1/2. [𝐴]0 2𝑘

= 𝑡1⁄

2

First-Order Case We start with the integrated rate law (as it includes the time component).

ln[A]1/2 = ln[A]0 – kt We know that using ex removes ln, so let’s do that [A]1/2 – [A]0 = 𝑒 −𝑘𝑡 [𝐴]1⁄

2

[𝐴]0

=

1 2

= 𝑒 −𝑘𝑡

Now take the ln once more ln 0.5 = −𝑘𝑡1⁄

2

ln 0.5 −𝑘

= 𝑡1⁄

2

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Fall 2017

Second-Order Case We start with the rate equation for a second-order rate

rate = k[A]2 = − We know that [A]1/2 =

𝑑[𝐴] 𝑑𝑡

𝐴0 2

[𝐴]

𝑡0 1 𝑑[𝐴] = −𝑘 ∫ 𝑑𝑡 2 [𝐴]0 [𝐴] 𝑡

∫

−1

−1

[𝐴]

[𝐴]0

1

1

[𝐴]

2[𝐴]

( )–( -( ) + ( -(

1 2[𝐴]

) = -kt

) = -kt1/2

) = -kt1/2

1 2𝑘[𝐴]

= t1/2

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