Integrated Rate Equations

The branch of chemistry which deals with the study of the speeds or the rates of chemical reactions, the factors affecting the rates of the reactions and the mechanism by which the reactions proceed is known as chemical kinetics. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism, a molecular-Level view of the path from reactants to products.

What is Rate law and Rate Constant?

The rate law is defined as the molar concentration of the reactants with each term raised to some powers which may or may not be the same as the stoichiometric coefficient of the reactant in a balanced chemical equation.

Rate = k[A]^α[B]^β

where α and β are the concentration of the A and B respectively.

Rate constant may be defined as the rate of reaction when the molar concentration of all reactants is unity i.e., 

[A]=[B]= 1 mol/litre.

Therefore,

Rate = K.

Significance of rate constant: As explained above it depend on the molar concentration of the reactant, so it means the greater the value of the rate constant, the more quickly is the reaction takes place.

Some Important Features of the Rate of reaction are,

• Rate constant measures the rate of the reaction.
• As the temperature increases, the value of the rate constant also increases.
• The unit of the rate constant depends on the order of the reaction (time, concentration).

Order of reaction

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]

In this case rate of reaction depends on one concentration term. Hence it is a first-order reaction.

Significance of order of reaction: The order of reaction explain how the concentration of the reactants is affected by the rate of reaction.

The order of the reaction is basically integers numbers like 0,1,2 but for some complex species, it can be infractions.

• Zero-order: The rate of reaction does not depend on the concentration of the reactants. In general,

Rate ∝ [A]⁰

• First-order: The rate of reaction is proportional to the concentration of the reactants. In general,

Rate ∝ [A]¹

• Second-order: The rate of reaction is proportional to the square of the concentration of the reactants. In general,

Rate ∝ [A]²

Integrated rate equations

For the general reaction,

a A + b B ⇢ c C + d D

Rate = d[R]/dt = k[A]^α [B]^β

This form of the equation is called the differential rate equation. This form is not convenient to determine the rate law and hence the order of the reaction. This is because the instantaneous rate has to be determined to form the slope of the tangent at time t in the plot of concentration versus time. 

To overcome the above difficulty, we integrate the differential equation for the reaction of any order. This gives us an equation related directly to the experimental data, i.e., time, concentrations at different times and rate constant.

Integrated rate equation for Zero-order reactions

Consider the general reaction: A ⇢ products

If it is a reaction of zero-order, Rate = – d[A]/dt = k[A]⁰ =k or d[A] = – k dt

Integrating both sides, we get: 

[A] = – kt + I              …(i)

• where I is constant of integration.

Substituting this value of I in Eqn. (i), we get 

[A] = – kt + [A]0          …(ii)

or 

kt = [A]0 – [A] 

or

k = 1/t {[A]0 – [A]}           …(iii)

Some important characteristics of reactions of zero-order-

• Any reaction of zero-order must obey equation, (ii) As it is an equation of a straight line (y = mx + c), the plot of [A] versus t will be a straight line with slope = – k and intercept on the concentration axis = [A]0,
• Half-life period: The half-life period (t_1/2) is the time in which half of the substance has reacted.

This implies that when [A] = [A]0/2 , t = t1/2.

Substituting these values in Eqn. (iii), we get

t_1/2 = 1/k  { [A]0 – [A]0/2 } = [A0]/2k  

i.e., 

t_1/2 =[A]0/2k                       …(iv) 

Thus, the half-life period of a zero-order reaction is directly proportional to initial concentration, i.e., t_1/2 ∝ [A]0.

• Units of rate constant (k): Form eqn. (iii), k = molar conc./time = mol L^-1 /time = mol L^-1  time^-1 .

Integrated rate equation for First-order reactions

A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. Thus, we may have

• For the reaction: A ⇢ products, rate the reaction ∝ [A]. 
• For the reaction: 2A ⇢  products, rate the reaction ∝ [A] only.
• For the reaction: A + B ⇢  products, rate the reaction ∝ [A] or [B] only.

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] = e^kt

or 

[A]/[A]₀ =e^-kt 

or 

[A] = [A]₀ e^-kt                   …(ix)

Some important characteristics of first-order reactions-

• Any reaction of the first order must obey equations (vi), (vii) and (ix).
• Half-life period: The time taken for any fraction of the reaction to complete is independent of the initial concentration.

t = 2.303/k log a/a – x         …(xiii)

• When half of reaction is completed, x = a/2. Representing the time taken for half of the reaction to be completed by t_1/2, equation (xiii) becomes 

t_1/2 = 2.303/k log a/a-a/2 

= 2.303/k log 2 

= 0.693/k 

t_1/2 = 0.693/k

Integrated rate equation for the first-order gas-phase reaction

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, 

P_t = (P₀ – p) + p + p = P₀ + p atm

p = P_t – P₀

So, pressure of A after time t (P_A) = P₀ – p = P₀ –  (P_t – P₀) = 2 P₀ – P_t

But initial pressure of A (P₀) ∝ initial conc. of A, i.e., [A]₀

and pressure of A after time t(P_A) ∝ 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 – P_t

Integrated rate law method

This is the most common method for studying the kinetics of a chemical reaction. For example, consider the reaction: 

n A ⇢ products 

If we start with a moles/litre of A and in time t, x moles/litre have reacted so that the concentration after time t is (a – x) moles/litre, then 

if the reaction is of first order, dx/dt = k(a – x) and if the reaction is of second order, dx/dt = k (a – x)² and so on.

These differential equations can be integrated to get expressions for the rate constants. These are given below for zero, first and second order reactions:

1. For zero order, k = 1/t {[A₀] – [A]} 
2. For 1st order, k = 2.303/t log  [A₀]/[A]
3. For 2nd order, k = 1/t {1/[A] – 1/[A₀]}

The advantage of the integrated method is that these integrated forms of equations contain the concentration of a reactant at different times and hence can be solved to find the value of k from the data of the run of one experiment only and need not start with different initial concentrations. Moreover, they can be used to find the time for any fraction of the reaction to complete.  

Sample Problems 

Problem 1: At 373 K, the half-life period for the thermal decomposition of N₂O₅ is 4.6 sec and is independent of the initial pressure of N₂O₅. Calculate the specific rate constant at this temperature.

Solution: 

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 t_1/2 = 0.693/k 

or 

k = 0.693/t_1/2 = 0.693/4.6_s

   = 0.1507 s^-1

Problem 2: A first-order reaction is found to have a rate constant, k = 5.5 x 10^-14s^-1. Find the half-life of the reaction.

Solution:

For a first order reaction, t_1/2 = 0.693/k 

t_1/2 = 0.693/5.5×10^-14s^-1

      = 1.26 x 10¹³ s^-1.

Problem 3: Show that in the case of a first-order reaction, the time required for 99.9% of the reaction to take place is about ten times that required for half the reaction.

Solution:

For reaction of first order, 

t_1/2 = 2.303/k log a/a – a/2 

      = 2.303/k log 2

     = 2.303/k (0.3010)

t_99.9% = 2,303/k log 

         = a/a-0.999a

t_99.9% = 2.303/k log 10^-3

         = 2.303/k x 3  

Therefore, t_99.9%/t_1/2 = 3/0.3010 ≅ 10

Problem 4: The initial concentration of N₂O₅ in the first-order reaction, N₂O₅(g) ⇢ 2 NO₂(g)+ 1/2 O₂(g), was 1.24 x 10^-2 mol L^-1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 x 10^-2 mol L^-1. Calculate the rate constant of the reaction at 318 K.

Solution: 

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.

Problem 5: A first-order reaction is found to have a rate constant k = 7.39 x 10^-5 sec^-1. Find the half life of reaction (log 2 = 0.3010).

Solution: 

For a first order reaction, k = 2.303/t log a/a-x

For t = t_1/2, x = a/2

t_1/2 = 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.

Problem 6: Why are reactions of higher-order less in number?

Solution:

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.

Problem 7: Why is the hydrolysis of ethyl acetate with NaOH is the reaction of second-order while with HCl, it is first order?

Solution:

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.