Rate Equation and Rate Essay

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Chapter 3 The Basis of Chemical Dynamics

2K(s) + 2H2O(l)

H 2 (g) +

2K+(aq) + 2OH (aq) + H2(g)

Δ rGm, 298K = −404.82 kJ·mol−1

1

Δ rGm, 298K = −228.59 kJ·mol−1
O 2 (g) =
H 2 O(g)
2

1．The concentrations of the reactants：
Steel wool burns with difficulty in air, which contains 20 percents O2 , but burst into a brilliant white flame in pure oxygen.
2．The temperature at which the reaction occur：
The rates of chemical reactions increase as temperature is increased. It’s for this reason that we refrigerate perishable food such as milk.
3．The presence of a catalyst：
The rates of many reactions can be increased by adding a substance known as a catalyst.
The physiology of most living species depends crucially on enzymes, protein molecules that act as catalysts, which increase the rates of selected biochemical reactions.
4．The surface area of solid or liquid reactants or catalysts：
Reactions that involve solids often proceed faster as the surface area of the solid is increased. For example, a medicine in the form of a tablet will dissolve in the stomach and enter the bloodstream more slowly than the same medicine in the form of a fine powder.

§3-1

The Rates of Chemical Reactions

1．Definition：通常以单位时间内反应物浓度的减少或生成物浓度的增加来表示。根据时

48

CCl
4

2．Units：mol · dm−3 · s−1、mol · dm−3 · min−1 或者 mol · dm−3 · hr −1
3．Average rate（平均速率） v = −Δ[反应物] / Δt
4．Instantaneous rate（瞬时速率）： lim{− ∆[反应物] / ∆t} =t
−d[反应物] / d
∆t → 0

1 d[A] a dt

1 d[B]

=−

b dt

1 d[G]

=

g dt

=

1 d[H] h dt

∴−

1 d[A] a dt

=−

1 d[B] b dt

=

1 d[G] g dt

=

1 d[H] h dt

Sample Exercise：The decomposition of N2O5 proceeds according to the equation：
2N2O5(g)

4NO2(g) + O2(g)

If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2×10−7 mol · dm−3 · s−1，what is the rate of appearance of (a) NO2； (b) O2?
Solution：∵ −

1 d[N 2 O5 ]

2 d[NO 2 ]

=

1 d[NO 2 ]

=

d[O 2 ]

dt
4 dt dt 2d[N 2 O5 ]

=

= 4.2 × 10 −7 = × 10 −7 mol · dm−3 · s−1

8.4 dt dt d[O 2 ]
1 d[N 2 O5 ] 1
=

= 4.2 × 10 −7 = × 10 −7 mol · dm−3 · s−1
×
2.1 dt 2 dt 2
When we speak of the rate of a reaction without specifying a particular reactant or product, we

will mean it in this sense.

In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy in order to react. According to the collision model, this energy comes from the kinetic energies of the colliding molecules.
N

1．分子运动速率分布（Maxwell-Boltzmann

*

N△E

distribution）

d

*

a

(1) 图 3.1 中横坐标为动能(kinetic energy)，纵坐标

b

c

E1 E2

E平

Ec

Ek

kinetic eneryg

Fig. 3.1 Distribution of kinetic energies in gas molecules
49

N

⋅E =
1 ，即 S t = 1。图 3.1 中阴影部分面积为 Sabcd = N */N，即在温度 T
N ⋅E

energy）。E 平表示 T 温度时的平均能量。

(2) 升高温度（如图 3.2），大动能的