Kinetics of First and Second Order Reactions

First Order Kinetics

A reaction of the first order is represented as:
X → Y
where X is the reactant and Y the product. The rate of the reaction will be directly proportional to the concentration, i.e.,

$$- \frac{dC}{dt} = kC \tag{XXII.3}$$

in which C is the concentration of the reactant at any time t and k is a constant, called the velocity constant or specific reaction rate.

Rewriting,

$$- \frac{dC}{C} = kdt$$

On integration,

$$- \ln C = kt + z \quad (\text{constant})$$

If at the start of the reaction, the initial concentration of the reactant is C₀, then we have at t = 0,

$$C = C₀$$

Substituting in,

$$- \ln C₀ = z \Rightarrow -\ln C = kt - \ln C₀$$

i.e.,

$$\ln \frac{C₀}{C} = kt, \quad \text{or} \quad \frac{C}{C₀} = e^{-kt}$$

Hence,

$$\begin{array}{cccc}& C=C_0{e}^{-kt}& & \text{<br>}\end{array}$$

The concentration (C) therefore diminishes exponentially with time.

We may also write,

$$C₀ - C = C₀ \left[1 - e^{-kt} \right] \tag{XXII.4a}$$

The rate equation may also be conveniently expressed in an alternative form, by expressing the rate in terms of the product. When x moles per unit volume of product Y are formed from the reactant, the concentration of the reactant is a - x, when a is the initial concentration of the reactant. So,

$$\frac{dx}{dt} = kC = k(a - x)$$

i.e.,

$$\frac{dx}{a - x} = kdt$$

On integration,

$$- \ln (a - x) = kt + z' \quad (\text{constant})$$

When t = 0, x = 0, hence

$$- \ln a = z'$$

i.e.,

$$- \ln (a - x) = kt - \ln a$$
$$k = \frac{1}{t} \ln \frac{a}{a - x} \quad \text{or} \quad x = a(1 - e^{-kt}) \tag{XXII.5}$$

The fractional extent of the reaction at time t, is

$$\begin{array}{cccc}& \frac{x}{a}=1-e^{-kt}& & \text{<br>}\end{array}$$

Even when the initial concentration a is not known but the concentrations at two intervals t₁ and t₂ are known, the rate equation can be derived. If x1 and x2 are the concentrations of product at time t₁ and t₂, then the corresponding concentrations of reactants would be (a - x₁) and (a - x₂). So,

$$kt₁ = \ln \frac{a}{a - x₁} \quad \text{and} \quad kt₂ = \ln \frac{a}{a - x₂}$$ $$k(t₂ - t₁) = \ln \frac{a - x₁}{a - x₂}, \quad \text{or} \quad k = \frac{1}{t₂ - t₁} \ln \frac{a - x₁}{a - x₂} \tag{XXII.6}$$

Some interesting characteristics of the first order reactions are:

(i) In a first order reaction,

$$C = C₀ e^{-kt}$$

the reaction cannot be complete; for, C would become zero only at infinite time, t.

(ii) The quantity

$$\frac{a}{a - x} \quad \text{(or } \frac{C₀}{C})$$

is a ratio of concentrations, so its value will be the same whatever units are employed to express the concentrations, e.g., moles per litre, gms per c.c., partial pressures etc.

It follows that the velocity constant

$$k \left( = \frac{1}{t} \ln \frac{a}{a - x} \right)$$

will have the dimension of reciprocal time, sec⁻¹.

(iii) The equation  is

$$\ln (a - x) = -kt + \ln a$$

or

$$\log (a - x) = - \frac{k}{2.303} t + \log a$$

If log(a - x) is plotted graphically against time, it would give a straight line. So when a plot constructed from experimental values of log(a - x) and t is found to be linear, the reaction is of the first order. The slope will be -k/2.303 from which the velocity constant k is known.

(iv) The time required for half the reactant to change can be easily evaluated. Let t₁/₂ be the time when x = a/2
Then

$$t_{1/2} = \frac{1}{k} \ln \frac{a}{a - x} = \frac{1}{k} \ln \frac{a}{a/2} = \frac{2.303 \log 2}{k} = \frac{0.693}{k}$$

The period of half-decomposition is thus a constant for a given reaction and is independent of initial concentration. The time required is often called half/value period, or ‘half-life’ in the case of radioactive changes.

Not only half-value period, the time necessary to complete any definite fraction (φ) of the reaction is independent of the initial concentration in first order kinetics. For,

$$t_\phi = \frac{2.303}{k} \log \frac{a}{a - a \phi} = \frac{2.303}{k} \log \frac{1}{1 - \phi} = \text{constant}$$