Integrated Rate Equation for First order Reactions Video Lecture | Chemistry for JEE Main & Advanced

ahoy mateys let's find the integrated
rate law for a first-order reaction a
first-order reaction has the rates
proportional to both a rate constant K
and the concentration of a linearly
rather the exponent is 1 that was that's
what makes it a first-order reaction to
get the integrated rate law we are going
to need to rewrite rate as the rate of
disappearance of a over time this here
is the definition of rate and again it's
proportional to the rate constant times
the concentration of a separate your
variables is the next step what that
means is I'm going to leave da on the
left and move the concentration of a
tiller to the left as well notice it was
in top now it's in the denominator
because I had to divide both sides by a
and on the other side I'm going to move
my negative I'm going to leave my K
there now I'm going to move my DT to the
right as well notice it was in the
denominator now it's in the numerator
because we had to multiply both sides by
DT in order to get it over there now
that I'm here I'm actually literally
going to integrate both sides get it
integrated rate law let me show you more
accurately what I'm going to end up
doing I'm going to integrate 1 over a DA
on the left and I'm going to integrate 1
DT on the left it's a rule of integrals
that you're allowed to move constants
outside of it I'm left with one here the
same way if you factor an X out of X
you're left with one anyways what's the
integral of 1 over a I asked you you
tell me the answer is lon
of a
I asked you what the integral of one is
and you tell me it is T hey look you're
right because you're smart it goes from
zero to T in both cases if you're
familiar with integration you're
familiar with this notation that I'm
using what that really means is that the
lawn of a at t minus the lawn of a at
zero is equal to negative K T at time t
minus T at time zero obviously T is zero
at time zero beautiful now what I'm
going to do is rearrange this for you
just so you can emphasize what exactly
is going on lon a of T is negative KT
multiplying through plus the lawn of a
knot
what this means is that if i graph lawn
of a on my y-axis and T on my x-axis I'm
going to end up with a straight line
with a slope of negative K and a
y-intercept of lon a knot let me
emphasize what I mean I put time on my x
and i graph lawn of concentration on my
Y I'm going to get a straight line the
slope is going to be negative K and the
y-intercept will be at the lawn of a
knot how awesome is that the whole
reason we do integrated rate laws is to
turn something regular and boring like
this into something awesome like this I
want to show you one more thing that
only pops out in the first order
integrated rate law and that is I could
rearrange this another way to get a
of T equals this markers crap a knot e
to the minus KT that's using logarithm
rules over here undoing lawn with E ^
and then multiplying by eat both sides
by a knot all I want to emphasize here
is that a varies exponentially or rather
negative exponentially or whatever you
want to call it with T a decays
exponentially over time that's the
wording I was looking for beautiful you
guys should try doing this on your own
make sure you understand each step
especially if your teacher actually
cares about you doing it and until then
best of luck to you