SOLUTION: The Arrhenius equation is
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Showing posts with label CHEMICAL KINETICS. Show all posts
Showing posts with label CHEMICAL KINETICS. Show all posts
Tuesday, June 9, 2020
In Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 10^13 sec–1 and 98.6 kJ mol–1 respectively. At what temperature, the reaction will have specific rate constant 1.1 × 10^–3 sec–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):
T1 T2 and T3 are the corresponding partial half lives and also K= K1 + K2 + K3 (for parallel reaction)
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):
[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 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.
(2) Above are parallel reactions occurring from Cu64
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):
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 ...
Wednesday, June 12, 2019
CONSECUTIVE OR SEQUENTIAL REACTION:
A chemical
reaction in which the product form is further decomposed into another product such
kind reaction of reaction is known as sequential reaction.
If rate
constant of reaction is K1 and K2 then the rate of reaction is for the reaction
For the
determination of concentration of A after time t, integrating equation (1)
For the
determination of concentration of B after time t,
Net rate of
formation of [B]
Multiply eK2t
to both side and integrating
For the
determination of concentration of [C] after time t,
Since by law
of mass balance [A]0=[A]t +[B]t+[C]t
Graphical
representation of concentration of A B and C after time t that is [A]t
, [B]t and [C]t respectively.
Case (1): K1 >>>K2
Concentration of [B]t:
when
K1>>>K2
in this case we can observed that reaction first (A to B) occurs first
and gives nearly to completion before reaction (B to C) take place. Thus nearly
all the (A) is converted to the intermediate (B) before any appreciable conversion
of (B to C), thus
Concentration of [C]t:
Graphical representation:
In general concentration of [A] decreases exponentially, and the concentration of [B] Initially increases up to a maximum and then decrease therefore and concentration of [C] increases steadily until it reaches its final value [A]0, when all A has changed into [C]
Case (2): K2
>>>K1
Concentration of [B]t:
Since K2>>>K1
then second term in parenthesis rapidly approaches zero while the first term is
still near unity consequently concentration of [B] approaches to K1/K2
[A] and decay more slowly according to
Since k1/K2
is very small, the maximum concentration of [B] is much less than [A]0
Concentration of [C]t:
Graphical representation:
Calculation of Maximum concentration of [B] and Maximum time:
Examples of consecutive reactions:
Illustrative Examples:
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