综合率方程
研究化学反应的速度或速率、影响反应速率的因素以及反应进行的机制的化学分支称为化学动力学。除了有关反应发生速度的信息外,动力学还揭示了反应机制,即从反应物到产物的路径的分子水平视图。
什么是速率定律和速率常数?
速率定律被定义为反应物的摩尔浓度,每一项都提高到一些幂,这可能与平衡化学方程式中反应物的化学计量系数相同或不同。
Rate = k[A]α[B]β
where α and β are the concentration of the A and B respectively.
速率常数可以定义为当所有反应物的摩尔浓度为一时的反应速率,即
[A]=[B]= 1 摩尔/升。
所以,
率 = K。
速率常数的意义:如上所述,它取决于反应物的摩尔浓度,因此速率常数的值越大,反应发生的越快。
反应速率的一些重要特征是,
- 速率常数测量反应的速率。
- 随着温度升高,速率常数的值也增加。
- 速率常数的单位取决于反应的顺序(时间、浓度)。
反应顺序
The sum of power to which the molar concentrations in the rate law equation are raised to express the observed rate of the reaction is known as the order of the reaction.
For example: Considering a chemical equation,
2N2O5 ⇢ 4NO2 + O2
Rate of reaction, R = [N2O5]
在这种情况下,反应速率取决于一个浓度项。因此,它是一级反应。
反应顺序的意义:反应顺序解释了反应物的浓度如何受反应速率的影响。
反应的顺序基本上是整数,如 0、1、2,但对于一些复杂的物种,它可能是违规的。
- 零级:反应速率不取决于反应物的浓度。一般来说,
比率∝ [A] 0
- 一阶:反应速率与反应物浓度成正比。一般来说,
比率 ∝ [A] 1
- 二阶:反应速率与反应物浓度的平方成正比。一般来说,
比率 ∝ [A] 2
综合速率方程
For the general reaction,
a A + b B ⇢ c C + d D
Rate = d[R]/dt = k[A]α [B]β
这种形式的方程称为微分速率方程。这种形式不方便确定速率定律,因此不方便确定反应的顺序。这是因为必须确定瞬时速率以在浓度与时间的关系图中形成时间 t 处的切线斜率。
为了克服上述困难,我们对任意阶反应的微分方程进行积分。这为我们提供了一个与实验数据直接相关的方程,即时间、不同时间的浓度和速率常数。
Z ero 级反应的综合速率方程
考虑一般反应: A⇢产品
如果是零级反应, Rate = – d[A]/dt = k[A] 0 =k or d[A] = – k dt
将双方积分,我们得到:
[A] = – kt + I …(i)
- 其中 I 是积分常数。
将 I 的这个值代入 Eqn。 (i),我们得到
[A] = – kt + [A]0 …(ii)
要么
kt = [A]0 – [A]
要么
k = 1/t {[A]0 – [A]} …(iii)
零级反应的一些重要特征
- 任何零级反应都必须服从方程, (ii) 因为它是一个直线方程 (y = mx + c),所以 [A] 与 t 的曲线将是一条斜率 = – k 和截距的直线在浓度轴上 = [A]0,
- 半衰期:半衰期(t 1/2 )是一半物质发生反应的时间。
这意味着当 [A] = [A]0/2 时,t = t1/2。
在方程式中替换这些值。 (iii),我们得到
t 1/2 = 1/k { [A]0 – [A]0/2 } = [A0]/2k
IE,
t 1/2 =[A]0/2k …(iv)
因此,零级反应的半衰期与初始浓度成正比,即 t 1/2 ∝ [A]0。
- 速率常数 (k) 的单位:形式 eqn。 (iii),k = 摩尔浓度/时间 = mol L -1 /time = mol L -1 time -1 。
一级反应的综合速率方程
如果反应速率仅取决于一个浓度项,则称该反应为一级反应。因此,我们可能有
- 对于反应:A ⇢ 产物,评价反应 ∝ [A]。
- 对于反应:2A ⇢ 产物,仅对反应 ∝ [A] 进行评分。
- 对于反应:A + B ⇢ 产物,仅评价反应 ∝ [A] 或 [B]。
Let us consider the simplest case, viz., A ⇢ products.
Suppose we start with moles per litre of the reactant A. After time t, suppose x moles per litre of it have decomposed. Therefore, the concentration of A after time t = (a – x) moles per litre. Then according to the law of mass action,
Rate of reaction ∝ (a – x),
i.e.,
dx/dt ∝ (a – x)
or
dx/dt = k (a – x) …(i)
where k is called the rate constant or the specific reaction rate for the reaction of the first order.
The expression for the rate constant k may be derived as follows:
Equation (i) may be rewritten in the form
dx/a-x =k dt …(ii)
Integrating equation (ii), we get
-In (a – x) = kt + I …(iii)
where I is a constant of integration.
In the beginning, when t=0, x=0
Putting these values in equation (iii), we get
– In (a – 0) = k x 0 + I
or
– In a = I …(iv)
Substituting this value of I in equation (iii), we get
– In (a – x) = kt + (- In a)
or
kt = a – In (a -x)
or
kt = In a/a-x …(v)
or
k = 1/ t In a/a-x
or
k=2.303/t log a/a-x …(vi)
If the initial concentration is [A]0 and the concentration after time t is [A], then putting a = [A]0 and (a – x ) = [A],
Equation (vi) becomes:
k = 2.303/t log [A]0/[A] …(vii)
Putting a = [A]0 and (a – x) = [A] in eqn. (v),
We get,
kt = In [A]0/[A] …. (viii)
which can be written in the exponential form as:
[A]0/[A] = ekt
or
[A]/[A]0 =e-kt
or
[A] = [A]0 e-kt …(ix)
一级反应的一些重要特征——
- 任何一级反应都必须服从方程 (vi)、(vii) 和 (ix)。
- 半衰期: 任何部分反应完成所用的时间与初始浓度无关。
t = 2.303/k log a/a – x …(xiii)
- 当反应完成一半时,x = a/2。表示通过 t 1/2完成一半反应所需的时间,等式 (xiii) 变为
t 1/2 = 2.303/k log a/aa/2
= 2.303/k log 2
= 0.693/k
t 1/2 = 0.693/k
一阶积分速率方程 气相反应
Consider the general first-order gas-phase reaction:
A (g) ⇢ B (g) + C (g)
Suppose the initial pressure of A = P0 atm. After time t, suppose the pressure of A decreases by p atm.
Now, as 1mole of A decomposes to give 1 mole of B and 1 mole of C, the pressure of B and C will increase by p each. Hence, we have
A (g) ⇢ B (g) + C (g)
Initial pressure: P0 atm 0 0
pressure after time, t: (P0 – p) p atm p atm
Total pressure of the reaction mixture after time t,
Pt = (P0 – p) + p + p = P0 + p atm
p = Pt – P0
So, pressure of A after time t (PA) = P0 – p = P0 – (Pt – P0) = 2 P0 – Pt
But initial pressure of A (P0) ∝ initial conc. of A, i.e., [A]0
and pressure of A after time t(PA) ∝ conc. of A at time t, i.e., [A]
Substituting these values in the first order rate equation,
k = 2.303/t log [A]0/[A],
we get
k = 2.303/t log P0/2P0 – Pt
综合费率法
这是研究化学反应动力学的最常用方法。例如,考虑反应:
n A ⇢ 产品
如果我们从 A 的摩尔/升开始,在时间 t,x 摩尔/升发生反应,使得时间 t 之后的浓度为 (a – x) 摩尔/升,那么
如果反应是一级反应, dx/dt = k(a – x) ,如果反应是二级反应, dx/dt = k (a – x) 2以此类推。
这些微分方程可以积分得到速率常数的表达式。下面给出了零级、一级和二级反应:
- 对于零阶, k = 1/t {[A 0 ] – [A]}
- 对于一阶, k = 2.303/t log [A 0 ]/[A]
- 对于二阶, k = 1/t {1/[A] – 1/[A 0 ]}
积分方法的优点是这些积分形式的方程包含不同时间的反应物浓度,因此可以求解,仅从一次实验运行的数据中找到 k 的值,而无需从不同的初始值开始浓度。此外,它们可用于确定反应的任何部分完成的时间。
示例问题
问题 1:在 373 K 时,N 2 O 5热分解的半衰期为 4.6 秒,并且与 N 2 O 5的初始压力无关。计算该温度下的比速率常数。
解决方案:
Since the half-life period is independent of the initial pressure, this shows that the reaction is of the first order.
For a reaction of the first order, we know that t1/2 = 0.693/k
or
k = 0.693/t1/2 = 0.693/4.6s
= 0.1507 s-1
问题 2:发现一级反应具有速率常数 k = 5.5 x 10 -14 s -1 。求反应的半衰期。
解决方案:
For a first order reaction, t1/2 = 0.693/k
t1/2 = 0.693/5.5×10-14s-1
= 1.26 x 1013 s-1.
问题 3:证明在一级反应的情况下,发生 99.9% 反应所需的时间大约是一半反应所需时间的十倍。
解决方案:
For reaction of first order,
t1/2 = 2.303/k log a/a – a/2
= 2.303/k log 2
= 2.303/k (0.3010)
t99.9% = 2,303/k log
= a/a-0.999a
t99.9% = 2.303/k log 10-3
= 2.303/k x 3
Therefore, t99.9%/t1/2 = 3/0.3010 ≅ 10
问题4:一级反应中N 2 O 5的初始浓度N 2 O 5 (g) ⇢ 2 NO 2 (g)+ 1/2 O 2 (g) 为1.24 x 10 -2 mol L -1在 318 K。60 分钟后 N 2 O 5的浓度为 0.20 x 10 -2 mol L -1 。计算 318 K 时的反应速率常数。
解决方案:
k = 2.303/t log [A]0/[A] = 2.303/t log [N2O5]0/[N2O5]t
2.303/60min log 1.24 x 10-2 mol L-1/0.2 x 10-2 mol L-1
= 2.303/60 log 6.2 min-1 = 2.303/60 x 0.7924min-1
= 0.0304 min-1.
问题 5:发现一级反应的速率常数 k = 7.39 x 10 -5 sec -1 。求反应的半衰期(log 2 = 0.3010)。
解决方案:
For a first order reaction, k = 2.303/t log a/a-x
For t = t1/2, x = a/2
t1/2 = 2.303/k log a/ a-a/2
= 2.303/k log 2
= 2.303/7.39×10-5s-1 x 0.3010
= 9.38 x 103 s-1.
问题6:为什么高阶反应的数量较少?
解决方案:
A reaction takes place because the molecules collide. The chance for a large number of molecules or ions to collide simultaneously is less. Hence, the reaction of higher order is less.
问题7:为什么乙酸乙酯与NaOH的水解是二级反应,而与HCl的水解是一级反应?
解决方案:
The rate of hydrolysis of ethyl acetate by NaOH depends upon the concentration of both while that by HCl depends only upon the concentration of ethyl acetate.