• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • NCERT Solutions
    • NCERT Books
  • Class 10
  • Class 9
  • Class 8
  • Class 7
  • Class 6
  • Class 11
  • Class 12
  • MCQ Questions
    • CBSE MCQ

Learn Insta

RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

  • Extra Questions
  • CBSE Notes
  • RD Sharma Solutions
    • RD Sharma Class 12 Solutions
    • RD Sharma Class 11 Solutions
    • RD Sharma Class 10 Solutions
    • RD Sharma Class 9 Solutions
    • RD Sharma Class 8 Solutions
  • RS Aggarwal Solutions
    • RS Aggarwal Solutions Class 10
    • RS Aggarwal Solutions Class 9
    • RS Aggarwal Solutions Class 8
    • RS Aggarwal Solutions Class 7
    • RS Aggarwal Solutions Class 6
  • ML Aggarwal Solutions
    • ML Aggarwal Class 10 Solutions
    • ML Aggarwal Class 9 Solutions
    • ML Aggarwal Class 8 Solutions
    • ML Aggarwal Class 7 Solutions
    • ML Aggarwal Class 6 Solutions
  • English Grammar
    • Words with Letters
    • English Summaries
    • Unseen Passages

The Integrated Rate Equation

July 1, 2021 by Prasanna

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

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 [A0] to [A] at the later time.

The Integrated Rate Equation img 1

– ln[A] – (- In[A0]) = k (t-0)
– ln[A] + In[A0] = 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[A0]-ln[A] = kt
ln[A] = ln[A0]-kt
⇒ y = c + mx

The Integrated Rate Equation img 2

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

N2O5(g) → 2NO2(g) + \(\frac{1}{2}\)O2(g)

(ii) Decomposition of Sulphurylchloride

SO2Cl2(l) → SO2(g) + Cl2(g)

(iii) Decomposition of the H2O2 in aqueous solution; H2O2 → H2O(l) + \(\frac{1}{2}\)O2(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,

The Integrated Rate Equation img 3

Rate = k [CH3COOCH3] [H2O]

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 [H2O] = k’; Therefore the above rate equation becomes
Rate = k'[CH3COOCH3]
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’,

The Integrated Rate Equation img 4

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 × 10-2mol-1L-1min-1

The Integrated Rate Equation img 5

Examples for a Zero Order Reaction:

1. Photochemical reaction between H2 and I2

The Integrated Rate Equation img 6

2. Decomposition of N2O on hot Platinum Surface

N2O(g) ⇄ N2(g) + \(\frac{1}{2}\)O2(g)

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

The Integrated Rate Equation img 7
Rate = k [CH3COCH3][H+]

Filed Under: Chemistry

Primary Sidebar

  • Maths NCERT Solutions
  • Science NCERT Solutions
  • Social Science NCERT Solutions
  • English NCERT Solutions
  • Hindi NCERT Solutions
  • Physics NCERT Solutions
  • Chemistry NCERT Solutions
  • Biology NCERT Solutions
RS Aggarwal Solutions RD Sharma Solutions
RS Aggarwal Class 10 RD Sharma Class 10
RS Aggarwal Class 9 RD Sharma Class 9
RS Aggarwal Class 8 RD Sharma Class 8
RS Aggarwal Class 7 RD Sharma Class 11
RS Aggarwal Class 6 RD Sharma Class 12

Recent Posts

  • Mechanical Properties of Solids Class 11 Important Extra Questions Physics Chapter 9
  • Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
  • MCQ Questions for Class 9 Maths Chapter 2 Polynomials with Answers
  • MCQ Questions for Class 7 Maths Chapter 2 Fractions and Decimals with Answers
  • English Grammar for Class 6, 7, 8, 9, 10, 11 and 12
  • Unscramble Words With Letters | Make Words with These Letters
  • Fractions and Decimals Class 7 Notes Maths Chapter 2
  • MCQ Questions for Class 6 Maths Chapter 8 Decimals with Answers
  • MCQ Questions for Class 6 Maths Chapter 1 Knowing Our Numbers with Answers
  • RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
DMCA.com Protection Status

Footer

CBSE Library
NCERT Library
NCERT Solutions for Class 12
NCERT Solutions for Class 11
NCERT Solutions for Class 10
NCERT Solutions for Class 9
NCERT Solutions for Class 8
NCERT Solutions for Class 7
NCERT Solutions for Class 6
ML Aggarwal Class 10 ICSE Solutions
Concise Mathematics Class 10 ICSE Solutions
CBSE Sample Papers
cbse
ncert
English Summaries
English Grammar
Biology Topics
Microbiology Topics
Chemistry Topics
Like us on Facebook Follow us on Twitter
Watch Youtube Videos Follow us on Google Plus
Follow us on Pinterest Follow us on Tumblr
Percentage Calculator

Copyright © 2022 Learn Insta