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Half - Life of a Reaction Video Lecture - NEET
FAQs on Half - Life of a Reaction Video Lecture - NEET
| 1. What is the definition of half-life in a reaction? |  |
Ans. The half-life of a reaction is the time it takes for the concentration of a reactant to decrease by half. It is a measure of the rate at which the reaction proceeds.
| 2. How is the half-life of a reaction determined? | |
Ans. The half-life of a reaction can be determined by conducting multiple experiments and measuring the concentration of the reactant at different time intervals. The data is then plotted on a graph, and the time it takes for the concentration to decrease by half is identified as the half-life.
| 3. What factors can affect the half-life of a reaction? | |
Ans. Several factors can influence the half-life of a reaction, including temperature, concentration of reactants, presence of catalysts, and the nature of the reactants themselves. Higher temperatures generally lead to shorter half-lives, while increased concentrations and the use of catalysts can decrease the half-life of a reaction.
| 4. How does the concept of half-life relate to the stability of a reaction? | |
Ans. The concept of half-life is closely related to the stability of a reaction. A reaction with a longer half-life is considered more stable, as it takes a longer time for the concentration of the reactant to decrease significantly. On the other hand, a reaction with a shorter half-life is considered less stable, as the concentration of the reactant decreases rapidly.
| 5. Can the half-life of a reaction be altered or manipulated? | |
Ans. Yes, the half-life of a reaction can be altered or manipulated by changing the conditions under which the reaction takes place. By adjusting factors such as temperature, concentration, and the presence of catalysts, it is possible to speed up or slow down the rate of a reaction, consequently affecting its half-life.
Text Transcript from Video
all right so here we are going to talk
about half-life so what is half-life
half-life is defined as the time it
takes the concentration of a reactant to
decrease to a half of its initial value
so for instance if I have an initial
concentration of I don't know 10 molar
and I have a half-life T 1/2 of 60
seconds then the concentration of a
after sick after 60 seconds is going to
be 1/2 of 10 molar or 5 molar so that is
just a basic summary of how half-life
works now it turns out that depending on
the order of the reaction each order of
the reaction whether it's zero for a
second or whatever each order is going
to have its own unique half-life
expression so let's start with zero
order half-life so
we have zero-order half-life and if we
look at what the integrated rate law is
for a zero order reaction we say that
the concentration of a at time T equals
minus KT plus the initial concentration
of a and since we're working with
half-life here our concentration of a at
time T is going to be one-half of the
initial concentration so we're going to
plate replace this term here this
concentration of a at time T with the
concentration of a initial over two
because that's just the definition of
half-life so basically if we rearrange
this equation and solve for T we'll get
an expression for the half-life of a
zero order reaction so now all we all we
have left to do is just like I said
algebraically solve for T so if I
subtract both sides of the equation by
the initial concentration we can rewrite
this equation as the following we say
that minus KT is equal to the
concentration in the initial
concentration over two minus just the
initial concentration
and we don't want - tea we want just tea
so we're going to multiply both sides of
the equation by minus 1 so that means
that everything on both sides of the
equation is going to get multiplied by
negative 1 so that means this is going
to turn into a positive and this is
going to turn into a negative and this
negative will turn into a positive so an
equivalent way of saying this is KT is
equal to the negative of the initial
concentration over 2 plus the initial
concentration so on the right-hand side
of the equation we're subtracting the
1/2 of the initial concentration from
the initial concentration itself and if
we have one of something and we subtract
one half of that same thing we'll just
be left with 1/2 of whatever it is so we
can rewrite the right-hand side of the
equation by saying that KT is equal to
simply the initial concentration over 2
so if we divide both sides of this by K
we'll get that T or T 1/2 the half-life
is equal to the initial concentration
over 2 K so now given any three of these
terms or excuse me given any two of
these terms we can solve for one unknown
so suppose I know the half-life and I
know the initial concentration well we
can solve for the rate constant if we do
that also if I know the initial
concentration and I know the rate
constant then we can solve for the
half-life so now let's move on to a
first-order half-life
so now we're going to look up the first
order integrated rate law the first
order integrated rate law says that the
natural logarithm of the initial
concentration excuse me the natural
logarithm of the concentration at time T
is equal to minus KT plus the natural
logarithm of the initial concentration
and once again we can replace this
concentration of a at time T term with
the initial concentration over two so
now we have a natural logarithm of the
of one half of the initial concentration
on the left hand side of the equation
so now our goal again is to solve for T
so to start with I'm going to subtract
both sides of the equation by the
natural logarithm of the initial
concentration and we'll get that minus
KT is equal to the natural logarithm of
the initial concentration over to minus
the natural logarithm of the initial
concentration and when it comes to
natural logarithms or rather any
logarithms logarithms have a general
property that says that if I have the Ln
of a minus the Ln of B that expression
is equal to the Ln of a divided by B so
we're going to apply that to our
equation here if I have the Ln of 1/2
the initial concentration minus the Ln
of the initial concentration well we can
rewrite this and just say that the
right-hand side of the equation is just
going to be equal to the natural
logarithm of the initial concentration
over to that whole term over the initial
concentration
and erase that real quick
so if we take the natural logarithm of
the initial concentration over 2 and
divide that whole thing by the initial
concentration well it looks to me like
the concentration of the initial
concentration terms will cancel and
we'll get Ln of 1/2 so really what we're
left with is minus KT is equal to the
natural logarithm of 1/2
so now what we're going to do is we're
going to multiply both sides of the
equation by negative 1 because again we
don't want minus T we want just T and
we'll get that KT is equal to the
negative of the natural logarithm of one
half and the negative the negative of
natural logarithm of one half is the
same as just the regular positive nap
natural logarithm of two that is another
property of logarithms so now we're in a
position where we have KT is equal to
the natural logarithm of 2 divided by K
on both sides and we'll end up getting
that the half-life T one half is equal
to the natural logarithm of 2 over K the
natural logarithm of two is out of 0.693
or something like that so this is also
equal to 0.693 something something
something over K and notice that for a
first order reaction the expression for
the half-life is independent of the
concentration the initial concentration
so that means we don't have to know what
the initial concentration is to
determine the half-life that turns out
to be very useful when it comes to
things like radiometric dating all right
so now let's look at second order
reaction half-life
second so again what we're going to do
is we're going to refer to the
integrated rate law and in the
integrated rate law for a second order
reaction we say that one over the
concentration of a at time T equals KT
plus 1 over the initial concentration
and of course the concentration of a at
time T since we're dealing with
half-life is going to be one-half of the
initial concentration so I can say that
1 over the initial concentration over to
1 over this whole term is equal to the
right-hand side of the equation and this
looks kind of ugly so if we're dividing
by a denominator of 2 that's the same
thing as multiplying by just 2 so I'm
going to say that I'm going to rewrite
this left-hand side of the equation and
just say that 2 over the initial
concentration of a is equal to the rest
of this stuff KT plus 1 over the initial
concentration this is just a way of
rewriting 1 over the initial
concentration over 2 so now what we're
going to do again what we'd like to do
is solve for T so we're going to
subtract both sides of the equation by 1
over the initial concentration and we'll
end up getting that KT is equal to 2
over the initial concentration minus 1
over the initial concentration
and since these denominators are both
the same we can subtract the numerators
so if I have two of something and I
subtract one of that same thing I'll
just end up with 1 so 2 over initial
concentration minus 1 over initial
concentration is simply just going to be
1 over the initial concentration and
lastly we are going to divide both sides
by K and we'll get that the half-life T
1/2 is equal to 1 over K times the
initial concentration so this is the
half of the expression for half-life for
a second-order reaction so what do we do
in this video well we started out with
the integrated rate law for each order
and then we substituted the
concentration of a at time T term by
saying that if we're dealing with
half-life the concentration of a at time
T is just going to be 1/2 of the initial
concentration and then we just
algebraically solve for T so as long as
you know the integrated rate laws you
can figure out the half-life you don't
just have to memorize the expressions
about half-life so what is half-life
half-life is defined as the time it
takes the concentration of a reactant to
decrease to a half of its initial value
so for instance if I have an initial
concentration of I don't know 10 molar
and I have a half-life T 1/2 of 60
seconds then the concentration of a
after sick after 60 seconds is going to
be 1/2 of 10 molar or 5 molar so that is
just a basic summary of how half-life
works now it turns out that depending on
the order of the reaction each order of
the reaction whether it's zero for a
second or whatever each order is going
to have its own unique half-life
expression so let's start with zero
order half-life so
we have zero-order half-life and if we
look at what the integrated rate law is
for a zero order reaction we say that
the concentration of a at time T equals
minus KT plus the initial concentration
of a and since we're working with
half-life here our concentration of a at
time T is going to be one-half of the
initial concentration so we're going to
plate replace this term here this
concentration of a at time T with the
concentration of a initial over two
because that's just the definition of
half-life so basically if we rearrange
this equation and solve for T we'll get
an expression for the half-life of a
zero order reaction so now all we all we
have left to do is just like I said
algebraically solve for T so if I
subtract both sides of the equation by
the initial concentration we can rewrite
this equation as the following we say
that minus KT is equal to the
concentration in the initial
concentration over two minus just the
initial concentration
and we don't want - tea we want just tea
so we're going to multiply both sides of
the equation by minus 1 so that means
that everything on both sides of the
equation is going to get multiplied by
negative 1 so that means this is going
to turn into a positive and this is
going to turn into a negative and this
negative will turn into a positive so an
equivalent way of saying this is KT is
equal to the negative of the initial
concentration over 2 plus the initial
concentration so on the right-hand side
of the equation we're subtracting the
1/2 of the initial concentration from
the initial concentration itself and if
we have one of something and we subtract
one half of that same thing we'll just
be left with 1/2 of whatever it is so we
can rewrite the right-hand side of the
equation by saying that KT is equal to
simply the initial concentration over 2
so if we divide both sides of this by K
we'll get that T or T 1/2 the half-life
is equal to the initial concentration
over 2 K so now given any three of these
terms or excuse me given any two of
these terms we can solve for one unknown
so suppose I know the half-life and I
know the initial concentration well we
can solve for the rate constant if we do
that also if I know the initial
concentration and I know the rate
constant then we can solve for the
half-life so now let's move on to a
first-order half-life
so now we're going to look up the first
order integrated rate law the first
order integrated rate law says that the
natural logarithm of the initial
concentration excuse me the natural
logarithm of the concentration at time T
is equal to minus KT plus the natural
logarithm of the initial concentration
and once again we can replace this
concentration of a at time T term with
the initial concentration over two so
now we have a natural logarithm of the
of one half of the initial concentration
on the left hand side of the equation
so now our goal again is to solve for T
so to start with I'm going to subtract
both sides of the equation by the
natural logarithm of the initial
concentration and we'll get that minus
KT is equal to the natural logarithm of
the initial concentration over to minus
the natural logarithm of the initial
concentration and when it comes to
natural logarithms or rather any
logarithms logarithms have a general
property that says that if I have the Ln
of a minus the Ln of B that expression
is equal to the Ln of a divided by B so
we're going to apply that to our
equation here if I have the Ln of 1/2
the initial concentration minus the Ln
of the initial concentration well we can
rewrite this and just say that the
right-hand side of the equation is just
going to be equal to the natural
logarithm of the initial concentration
over to that whole term over the initial
concentration
and erase that real quick
so if we take the natural logarithm of
the initial concentration over 2 and
divide that whole thing by the initial
concentration well it looks to me like
the concentration of the initial
concentration terms will cancel and
we'll get Ln of 1/2 so really what we're
left with is minus KT is equal to the
natural logarithm of 1/2
so now what we're going to do is we're
going to multiply both sides of the
equation by negative 1 because again we
don't want minus T we want just T and
we'll get that KT is equal to the
negative of the natural logarithm of one
half and the negative the negative of
natural logarithm of one half is the
same as just the regular positive nap
natural logarithm of two that is another
property of logarithms so now we're in a
position where we have KT is equal to
the natural logarithm of 2 divided by K
on both sides and we'll end up getting
that the half-life T one half is equal
to the natural logarithm of 2 over K the
natural logarithm of two is out of 0.693
or something like that so this is also
equal to 0.693 something something
something over K and notice that for a
first order reaction the expression for
the half-life is independent of the
concentration the initial concentration
so that means we don't have to know what
the initial concentration is to
determine the half-life that turns out
to be very useful when it comes to
things like radiometric dating all right
so now let's look at second order
reaction half-life
second so again what we're going to do
is we're going to refer to the
integrated rate law and in the
integrated rate law for a second order
reaction we say that one over the
concentration of a at time T equals KT
plus 1 over the initial concentration
and of course the concentration of a at
time T since we're dealing with
half-life is going to be one-half of the
initial concentration so I can say that
1 over the initial concentration over to
1 over this whole term is equal to the
right-hand side of the equation and this
looks kind of ugly so if we're dividing
by a denominator of 2 that's the same
thing as multiplying by just 2 so I'm
going to say that I'm going to rewrite
this left-hand side of the equation and
just say that 2 over the initial
concentration of a is equal to the rest
of this stuff KT plus 1 over the initial
concentration this is just a way of
rewriting 1 over the initial
concentration over 2 so now what we're
going to do again what we'd like to do
is solve for T so we're going to
subtract both sides of the equation by 1
over the initial concentration and we'll
end up getting that KT is equal to 2
over the initial concentration minus 1
over the initial concentration
and since these denominators are both
the same we can subtract the numerators
so if I have two of something and I
subtract one of that same thing I'll
just end up with 1 so 2 over initial
concentration minus 1 over initial
concentration is simply just going to be
1 over the initial concentration and
lastly we are going to divide both sides
by K and we'll get that the half-life T
1/2 is equal to 1 over K times the
initial concentration so this is the
half of the expression for half-life for
a second-order reaction so what do we do
in this video well we started out with
the integrated rate law for each order
and then we substituted the
concentration of a at time T term by
saying that if we're dealing with
half-life the concentration of a at time
T is just going to be 1/2 of the initial
concentration and then we just
algebraically solve for T so as long as
you know the integrated rate laws you
can figure out the half-life you don't
just have to memorize the expressions
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