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# Half-life of a second-order reaction

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- We've already talked about
the definition for half-life. Remember half-life is symbolized by t 1/2. The half-life is the
time that it takes for the concentration of
a reactant to decrease to half of its initial concentration. We've also already talked about the integrated rate law or the
integrated rate equation for a second order reaction. So here's one form of it. One over the concentration of A minus one over the initial concentration of A is equal to the rate
constant k times the time. So if we're talking about the half-life so that would be when
time is equal to t 1/2 so we're going to plug
t 1/2 in for the time what's the concentration of A? Well using the definition for half-life, the concentration of A should be half of the initial concentration. So our initial
concentration would be this. So it's half of our initial concentration. So we're gonna plug that into here. Alright let's plug those
in and see what we have. We have one over the
initial concentration of A divided by two minus one over the initial concentration of A is equal to the rate constant k times the half-life. So one the left side this would just be two over the initial concentration of A minus one over the
initial concentration of A is equal to k t1/2. So what is two over the
initial concentration of A minus one over the initial
concentration of A? That would just be one over
the initial concentration of A and that's equal to the rate constant k times the half-life. So now we can solve for the half-life. Just divide both sides by k. So we get the half-life is equal to one over k times the initial concentration of A. And so here's our equation for the half-life for a second order reaction. Notice this is very different for the half-life for a first order reaction. For a first order reaction we saw that the half-life was constant but here the half-life isn't constant because the half-life depends on the initial concentration of A. Now let's look at a graph
of concentration versus time for a second order reaction so we can understand this concept
a little bit better. So when time is equal to zero, alright, this point right here
on the graph would be the initial concentration of A. So here is our initial concentration of A. And let's just do a made-up reaction here. Let's say we're starting
with eight particles that are eight molecules so here we have one, two, three, four, five, six, seven, eight. Alright, so if we wait
for the concentration to decrease to half its initial, what are we left with? We were left with four molecules. Alright so we're left with four molecules so I'll draw those in there and how long did it take
to go from eight to four? Alright we can find that on our graph. If this point represents
the initial concentration, half that would be right here. That would be the initial
concentration of A divided by two. So we find that point on our graph and we drop down to here on the x-axis and it took one second. Alright so our first
half-life is one second. So let me write that in here. So this represents our first
half-life which is one second. How long does it take
for the second half-life? So how long does it take to go from four molecules to two molecules? Let me go ahead and change
color and be consistent here. So we're going from four
molecules to two molecules. How long does it take to do that? Well if this is our
starting concentration now, what's half of that? So on our graph half of
that would be right here. And this would represent
the initial concentration divided by four now. So here's half of this and if we go over, we can find this point
on our graph, right? So that point on our
graph would be right there and we drop down and we can see we're at time is equal to three. So how long did it take
this time for our half-life? This time our half-life
would be two seconds. Right? Our half life is two seconds. It's twice the first half-life and we can understand that by plugging in to our half-life equation over here. Alright so if we say the
first half-life is one second, so we say the first
half-life is one second here, let's go ahead and plug in
for the second half-life. So the second, the second half-life would be equal to this would be one over the rate constant k but now our initial concentration, right, our initial concentration
is not this, right? It's this. It's the initial
concentration divided by two. Alright so go back over here, this would be the initial
concentration of A divided by two. So for the second half-life
this would be equal to two over k times the
initial concentration of A. And we can see that this, that is twice this up here. If this is our first
half-life, which is one second, this is twice that, which of
course would be two seconds. So this makes sense,
both looking at the graph and also thinking about the
equation for our half-life. So our second half-life is
twice as long as the first and each half-life is going to be twice as long as the one before it. So if we wait one more half-life,
so the third half-life, we're going from two
molecules to one molecule so now our starting
concentration would be right here so what's half of that? That would be right here so we go over and we find this point on our graph. So we find that point on
our graph right about there and we drop down and
we can see the time is equal to seven seconds. So what's the third half-life? This would be the third
half-life right here on our graph and obviously that would
be four seconds, alright? So four seconds which is
twice the preceding half-life. Alright so let's see if we can figure this out a little bit more. So what does this mean? This means in the early
stages of your reaction you have a higher concentration
of your reactant, alright? And the higher concentration means that those molecules can collide better and so therefore the reaction goes faster and so if we have an
increased concentration we have an increased reaction rate, right? Therefore the faster the
reactant is being consumed and therefore the shorter the half-life, the shorter the time it
takes for the concentration of the reactant to decrease to half its initial concentration.