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Friday, June 21, 2019

ENERGY IN BOHR'S ORBITS:


­­Total energy of (E) of an electron revolving in nth orbit is equal to sum of kinetic energy and Potential energy.
We know the electron revolve around nucleus due balancing of two forces columbic and centrifugal forces

This is the famous Bohr’s equation applicable to Hydrogen like atoms or ions as He+1, Li+2 , Be+3 
etc.
The factor (4 pi epsilon zero) is known as permittivity factor and its numerical value is 1.11268*10-10C2N-1M-2  ( In CGS Unit K= 1)
Pi= 22/7= 3.424, me=9.109 *10-31 kg, e = 1.602 *10-10 C and h= 6.626*10-34 j-s
Calculation of En in SI Unit:
Bohr’s energy in electron volt:
We know that, 1eV = 1.602 *10-19 J hence
Energy in term of Rydberg’s Constant:

Relation between Total energy (TE), Kinetic energy (KE) and Potential energy (PE):


Important conclusions:
(1) The minus sign for the energy of an electron in an orbit represents attraction between the +vely charged nucleus and negatively charged electron.
(2) Energy of an electron at infinite distance from the nucleus is zero.
(3) As an electron approaches the nucleus, the electrical attraction increases, energy of electron decreases and it becomes negative.
(4) Energy of an electron increases as the value of ‘n’ increases i.e.
(5) Value of ‘n’ remaining unchanged, the amount of energy associated with an electron remains unaltered.
(6) Energy of electron in first, second, third and fourth orbit are –13.6, –3.4, –1.5, and –0.85 eV/atom respectively.
(7) Although the energy of electron increases with increase in the value of ‘n’ (orbit), yet the difference of energy between successive orbits decreases. Thus E2 – E1 > E3 – E2 > E4 – E3 > E5 – E4 >, etc….

Wednesday, June 19, 2019

PARALLEL REACTION: ILLUSTRATION: GENERAL TYPE (4):

Consider a parallel reaction follow first order chemical kinetics
Let t=0 concentration of (A) is [A]0 and after time (t) concentration of (A) and (B)  are [A]t and [B]t respectively .
Concentration of [B] and [C] after time t:


Percentage Yield of Product:

ILLUSTRATIVE EXAMPLE (1): An organic compound A decomposes following two parallel first order mechanisms:

Calculate the concentration ratio of C to A, if an experiment is allowed to start with only A for one hour.
SOLUTION:

ILLUSTRATIVE EXAMPLE (2): An organic compound A decomposes by following two parallel first order mechanisms:


Select the correct statement(s)
(A) If three moles of A are completely decomposed then 2 moles of B and 1 mole of C will be
formed.
(B) If three moles of A are completely decomposed then 1 moles of B and 2 mole of C will be
formed.
(C) half life for the decomposition of A is 20 min
(D) half life for the decomposition of B is 0.33 min
SOLUTION: (BC)
Generalisation of parallel reaction:

PARALLEL REACTION: ILLUSTRATION TYPE (3):

Consider a parallel reaction follow first order chemical kinetics
Let t=0 concentration of (A) is [A]0 and after time (t) concentration of (A) and (B)  are [A]t and [B]t respectively .

Percentage Yield of Product:
PARALLEL REACTION: ILLUSTRATION: GENERAL TYPE (4): for Continue reading click here...

PARALLEL REACTION: ILLUSTRATION TYPE (2):

Let t=0 concentration of (A) is [A]0  and after time concentration of (A) and (B)  are [A]t and [B]t respectively .
[B]t and [C]t - Concentration of [B] and [C] after time t:
Concentration of (A) after time t is:
Similarly concentration of (C) after time t is:
Percentage Yield of Product:
Time of completion and half life of reactant:

PARALLEL REACTION: ILLUSTRATION TYPE (3): for Continue reading click here ...

PARALLEL PATH OR CONCURRENT ELEMENTARY REACTIONS

The chemical reaction in which a substance reacts or decomposed in more than one way are called parallel or side reaction.
Some examples of parallel reactions:

PARALLEL REACTION: ILLUSTRATION TYPE (1):
Let a chemical reaction in which reactant (A) give two products (B) and (C) and both the reactions are first order reactions. Initially at t=0, pure (A) was present with concentration [A]0 and after time t concentration of (A) is [A]t and concentration of (B) is [B]t and concentration of (C) becomes [C]t
[A]t -Concentration of [A] after time t:
[B]t -Concentration of [B] after time t:
[C]t -Concentration of [C] after time t:
Graphical representation of variation of conc of [A], [B], [C] with time: If K1>>> K2 then [B] is the main product and [C] is the side product
Ratio of concentration of (A) and (C) After time t is:
Percentage Yield of Product:
Time of completion and half life of reactant:
Question for Illustration (1):  Cu64 (Half life=22.8 hours) decay by beta emission (38%), beta+ emission (19%) and electron capture (43%). write the decay product and calculate Half lives for each of the decay process. [IIT 2002]
Solution:
(1):
 (2Above are parallel reactions occurring from Cu64   
T1 T2 and T3 are the corresponding partial half lives and also K= K1 + K2 + K3 (for parallel reaction) 
PARALLEL REACTION: ILLUSTRATION TYPE (2): for Continue reading click here ...