Vietnam suplementary

Reactive Systems
Supplementary on
Rates of Reactions,
Differential and Integrated
Rate Laws


First-order reactions
For first-order reactions:

AB

The time rate of change of the concentration of A, (given by [A]) is:

d[𝐴]
d[𝐵]
= −𝑘 𝐴 =
d𝑡
d𝑡

Differential rate law,
describing rates

∞
Rearrangement of this equation yields:

0

𝐴
ln
𝐴

𝑡
0

= −𝑘𝑡

d[𝐴]
=−
𝐴

𝐴 𝑡 = 𝐴 0𝑒 −𝑘𝑡

Integral rate law, describing concentrations

∞

𝑘𝑑𝑡
0


First-order reactions
Plots the [A(t)]/[A(0)] in the upper panel and the natural log of [A(t)]/[A(0)] in the lower
panel as a function of time for a first-order reaction. Note that the slope of the line in the
lower panel is -k and that the concentration falls to 1/e of its initial value after a time =1/k, often called the lifetime of the reactant. A related quantity is the time it takes for
the concentration to fall to half of its value, obtained from:

𝐴 (𝑡 = 𝜏1/2 ) 1
= = exp⁡(−𝜏1/2 )
[A]0
2

𝜏1/2

ln⁡(2)
=
𝑘

The quantity 1/2 is known as the
half-life of the reactant.


Pseudo first-order reactions
For pseudo first-order reactions:

A + B  products

The time rate of change of the concentration of A, (given by [A]) is:

d[𝐴]
= −𝑘 𝐴 [𝐵]
d𝑡

If [B] remains constant (by either
setting [B]0 >> [A]0 or recycling B):

∞
Rearrangement yields:

𝐴
ln
𝐴

0
𝑡
0

= −𝑘 𝐵 0𝑡

𝐴 𝑡 = 𝐴 0𝑒 −𝑘 𝐵 0𝑡

d[𝐴]
=−
𝐴

∞
0

𝑘 𝐵 0𝑑𝑡


Opposing reactions, equilibrium
For pseudo first-order reactions:

A⇋B

The time rate of change of the concentration of A, (given by [A]) is:

d[𝐴]
d[𝐵]
=−
− 𝑘1 𝐴 + 𝑘−1 𝐵

d𝑡
d𝑡
Defining the equilibria as:
and

[𝐴]𝑒 = [𝐴]0 −𝑥𝑒

[𝐵]𝑒 = [𝐵]0 +𝑥𝑒

𝐴 𝑡 = 𝐴 0 + 𝑥𝑒 𝑒 −(𝑘1+𝑘−1)⁡𝑡
This defines the characteristic time for
a system to reach equilibrium. It is
dominated by the largest rate constant.


Parallel reactions
For parallel reactions:

A  C (k1)

and

B  D (k2)


Consecutive reactions and steadystate approximation
For consecutive reactions:

A  B (k1)

and

B  C (k2)


Consecutive reactions and steadystate approximation
For parallel reactions:

A  C (k1)

For k1 = 10k2

and

B  D (k2)

For k2 = 10k1