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## The Integrated Rate Equation

We have just learnt that the rate of change of concentration of the reactant is directly proportional to that of concentration of the reactant. For a general reaction, A → product. The rate law is Rate = \(\frac{-d[A]}{dt}\) = k[A]^{x}

Where k is the rate constant, and x is the order of the reaction. The above equation is a differential equation, \(\frac{-d[A]}{dt}\), so it gives rate at any instant. However, using the above expression, we cannot answer questions such as how long will it take for a specific concentration of A to be used up in the reaction? What will be the concentration of reactant after a time ‘t’?. To answer such questions, we need the integrated form of the above rate law which contains time as a variable.

**Integrated Rate Law for a First Order Reaction**

A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following first order reaction,

A → Product

Rate law can be expressed as

Rate = k[A]^{1}

Where, k is the fist order rate constant.

\(\frac{-d[A]}{dt}\) = k[A]^{1}

⇒ \(\frac{-d[A]}{[A]}\) = k dt …………… (1)

Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A_{0}] to [A] at the later time.

– ln[A] – (- In[A_{0}]) = k (t-0)

– ln[A] + In[A_{0}] = kt

ln(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt ……….. (2)

This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303. 2.303 log (\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt

k = \(\frac{2.303}{t}\)log(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) ………… (3)

Equation (2) can be written in the form y = mx + c as below

ln[A_{0}]-ln[A] = kt

ln[A] = ln[A_{0}]-kt

⇒ y = c + mx

If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated. Examples for the first order reaction

(i) Decomposition of Dinitrogen Pentoxide

N_{2}O_{5}(g) → 2NO_{2}(g) + \(\frac{1}{2}\)O_{2}(g)

(ii) Decomposition of Sulphurylchloride

SO_{2}Cl_{2}(l) → SO_{2}(g) + Cl_{2}(g)

(iii) Decomposition of the H_{2}O_{2} in aqueous solution; H_{2}O_{2} → H_{2}O(l) + \(\frac{1}{2}\)O_{2}(g)

(iv) Isomerisation of Cyclopropane to Propene.

**Pseudo First Order Reaction:**

Kinetic study of a higher order reaction is difficult to follow, for example, in a study of a second order reaction involving two different reactants; the simultaneous measurement of change in the concentration of both the reactants is very difficult.

To overcome such difficulties, A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,

Rate = k [CH_{3}COOCH_{3}] [H_{2}O]

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e.,concentration of water remains almost a constant.

Now, we can define k [H_{2}O] = k’; Therefore the above rate equation becomes

Rate = k'[CH_{3}COOCH_{3}]

Thus it follows first order kinetics.

**Integrated Rate law for a Zero Order Reaction:**

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.

A → product

The rate law can be written as,

Rate = k[A]°

\(\frac{-d[A]}{dt}\) = k(1) (∴[A]° = 1)

⇒ -d[A] = k dt

Integrate the above equation between the limits of [A_{°}] at zero time and [A] at some later time ‘t’,

Equation (2) is in the form of a straight line y = mx + c

i.e., [A] = – kt + [A_{°}]

⇒ y = c + mx

A plot of [A] Vs time gives a straight line with a slope of – k and y – intercept of [A_{°}].

Fig 7.4: A plot of [A] Vs time for a zero order reaction A → product with initial concentration of [A] = 0.5M and k = 1.5 × 10^{-2}mol^{-1}L^{-1}min^{-1}

**Examples for a Zero Order Reaction:**

**1. Photochemical reaction between H _{2} and I_{2}**

**2. Decomposition of N _{2}O on hot Platinum Surface**

N_{2}O(g) ⇄ N_{2}(g) + \(\frac{1}{2}\)O_{2}(g)

**3. Iodination of Acetone in Acid Medium is Zero Order With Respect to Iodine.**

Rate = k [CH_{3}COCH_{3}][H^{+}]