Course Content
CHAPTER 1: THE SOLID STATE
THE SOLID STATE
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CHAPTER 2: SOLUTION
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CHAPTER 3: ELECTROCHEMISTRY
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CHAPTER 4: CHEMICAL KINETICS
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CHAPTER 5 : SURFACE CHEMISTRY
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CHAPTER 6: GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS
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CHAPTER 7: THE p-BLOCK ELEMENTS
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CHAPTER 8: THE D & F BLOCK ELEMENTS
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CHAPTER 9: COORDINATION COMPOUNDS
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CHAPTER 10: HALOALKANES AND HALOARENES
Topic Name 10 Haloalkanes and Haloarenes 10.1 Classification 10.2 Nomenclature 10.3 Nature of C–X Bond 10.4 Methods of Preparation of Haloalkanes 10.5 Preparation of Haloarenes 10.6 Physical Properties 10.7 Chemical Reactions
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CHAPTER 11: ALCOHOLS, PHENOLS AND ETHERS
Topic Name 11 Alcohols, Phenols and Ethers 11.1 Classification 11.2 Nomenclature 11.3 Structures of Functional Groups 11.4 Alcohols and Phenols 11.5 Some Commercially Important Alcohols 11.6 Ethers
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Class 12th Chemistry Online Class For 100% Result

### Zeroth order reaction

If the rate of reaction is independent of concentration of the reactant participating in the reaction then the reaction is called Zeroth order reaction.

A –> B

At time t = 0 concentration of A (reactant) is a and B (product) is 0. At time t = t the concentration of A (reactant) is (a-x) and that of B (product) is x.

-d[A]/dt = k0[A]0 => dx/dt = k0(a-x)0

dx/dt = k0

∫ 0x  dx =k0 ∫0 xdt

X = k0t

PROBLEM.   The decomposition of NH3on platinum surface is zero order reaction. What are the rates of production of N2and H2if k = 2.5 x 10-4mol-1L s-1?

SOLUTION.    2NH –> N2 + 3H2

Rate of zero order reaction

-1/2 (d[NH3]/dt) = d[N2]/dt = 1/3 (d[H2]/dt)

-1/2 (d[NH3]/dt) = d[N2]/dt = 1/3 (d[H2]/dt)= k = 2.5 x 10-4 mol L-1 s-1

Rate of production of N2

d[N2]/dt = 2.5 x 10-4 mol L-1 s-1

Rate of production of H2

d[H2]/dt = 3 x 2.5 x 10-4 mol L-1 s-1

### First order reaction

If the rate of reaction depends on the concentration of single reactant participating in chemical reaction raised to the first power then it is called a first order reaction.

A –> B

At time t = 0 concentration of A (reactant) is a and B (product) is 0. At time t = t the concentration of A (reactant) is (a-x) and that of B (product) is x.

-dx/dt ∝ (a-x) = dx/dt = k1(a-x)

∫ 0x  dx/(a-x) = k1∫ 0t dt

dx/dt = k0(a-x)0

dx/dt = k0

∫0 x dx = k0∫ 0t dt

ln (a/a-x) = k1t => t = 1/ k1 ln (a/a-x) = 2.303/ k1 log (a/a-x)

k1 = 2.303 log (a/a-x)

PROBLEM.   A first order reaction has a rate constant 1.15 10-3s-1. How long will 5 g of this reactant take to reduce to 3 g?

SOLUTION.   From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10 – 3s – 1

We know that for a 1st order reaction,

t = (2.303/k)log[R0]/[R]

(2.303/1.15X10-3)log/

(2.303/1.15X10-3) X 0.2219 = 444.38 s = 444 s

### Second order reaction

A reaction with order equal to two is called a second order reaction.

r = k[A]2

or r = k [A][B]

## Pseudo order reaction

The reaction that appears to be an nth order reaction but belongs to some different order is called Pseudo order reaction.

For example, a pseudo first order reaction is a chemical reaction between two reactants participating in a chemical reaction and therefore should be a second order reaction. But it resembles to be a first order reaction due to the presence of reactants in negligible quantity.

Let R` + R“ –> P

Rate = k[A]1[B]1

Order of reaction = 2.

Let us consider another reaction,

CH3Br + OH→ CH3OH+Br

Rate law for this reaction is

Rate = k [OH][CH3Br]

Rate = k [OH][CH3Br] = k(constant)[CH3Br]=k′[CH3Br]

As only the concentration of CH3Br would change during the reaction, the rate would solely depend upon the changes in the CH3Br reaction.

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