# Half-life of a first-order reaction | Knetics | Chemistry | Khan Academy

This is one form of the kinetic equation for a first order reaction. We will continue with the math, probably talking about half-life. On the left we have a natural logarithm of the concentration of A at time t minus the natural logarithm of the initial concentration A. This is the same as the natural logarithm of the concentration of A on the initial concentration of A. It’s just a property of logarithms. And that equals -Kt, where K is the rate constant. We will now get rid of the natural logarithms. We will raise both sides to a degree.

We will anti-logarithm and we will free ourselves from the natural logarithm. Now on the left we have the concentration on the initial concentration. On the right we have a degree – Kt. We will multiply both sides by the initial concentration of A. We obtain that the concentration of A at time t is equal to the initial concentration of A in e ^ – Kt. It’s a little easier to think about the chart now. We can put the concentration of A on Oh, and the time on Oh. And this takes the form of an exponential function.

Here I have plotted a decay indicator function, just to show what it looks like. Let’s look at this point in the graph. This is at t = 0. When t = 0, what is the concentration? Let’s replace t = 0. We obtain that the concentration is equal to the initial concentration is zero. is zero is 1. So this is 1. The initial concentration is evident at this point, and time is zero. Obviously, this is our initial concentration. I’ll write it down here. That’s the point. And when time tends to infinity, the concentration of A branches to zero. So somewhere here obviously the concentration of A tends to zero. This is the logic behind the decay function graph. Now let’s see what the half-life is. This is the definition of half-life. This is the time for which the concentration of starting materials decreases by half from the original concentration. If this is the initial concentration, what would be the concentration when half of A reacted? We will have the initial concentration divided by two. We substitute this value for concentration. The half-life symbol is t 1/2. We replace this with time.

When time equals half-life, the concentration is equal to half of the initial concentration. Let’s replace and find the half-life. On the left we have the initial concentration divided by two. And this equals the initial concentration of A is more of a degree -K, half-life, replace with t 1/2. And now we just have to find t 1/2. We will find the half-life for a first-order response. Let’s make some space downstairs. We can immediately reduce the initial concentration of A.

So we have 1/2 is equal to is of degree – Kt1 / 2. Now we have to get rid of it here. Therefore, we will log both sides. A natural logarithm of 1/2 is equal to a natural logarithm –Kt 1/2. That’s how we got rid of it. Now we have a natural logarithm of 1/2 equals -Kt1 / 2. We find t 1/2, which is the half-life. So t 1/2 equals minus the natural logarithm of 1/2 divided by K. I take the calculator to find out how natural is a log of 1/2. Let me make room. The natural logarithm of 0.5 is –0.693. We have the downside to this, so we get a positive half-life. Half life is … let me rewrite it here … The half-life, t 1/2 is 0.693 divided by K, which is the velocity constant. This is the half-life of a first-order response. Let’s look at it. If K is a constant and 0.693 is also a constant, So half life is a constant. The first half-life of the reaction does not depend on the initial concentration of A.

We will always have the same half-life. Let’s take an example. Let’s go back to the chart and think about the half-life. Let’s have some initial concentration, let me change the colors, so we can track it. I will present the initial concentration with eight points. Let’s have eight particles to start the process with. This is obviously a theoretical reaction. We will wait until half of the reagent is gone. We lost half of the reagent, so we only have four particles. There are only four of us left. Where will this be on our schedule? This point is the initial concentration. This is the initial concentration, so we need half of it.

This will be right here on the graph. We go here and find this point. We come down to Oh. And I’ll put some time here. Let it be 10 seconds, 20 seconds, 30 seconds, 40 and so on. We see that it took 10 seconds, to reduce the concentration by half. So the first half-life is 10 seconds. Let’s put it here, t 1/2 is 10 seconds. I repeat, this is a fictional reaction, just to explain what half life is. Okay, now we have 4 particles. How long will it take for half to react? Two react, two remain. Let’s see the graph, it will be here. This point is half of that, we go down and down the x-axis. How long did it take to halve the concentration? Again, 10 seconds. So here it is again 10 seconds. This half-life is 10 seconds. We can do it again. Half of the starting material reacts again. And here we go on the chart, we go down. How long did only one particle remain from two particles? Again, it took us 10 seconds.

So the half-life is again 10 seconds. So the half-life does not depend on the initial concentration. It doesn’t matter if we start with eight particles or two. The half-life is always 10 seconds. This represents half life in first order reactions.