Half-Life

now half life is just the time it takes to lose half of a reacting to decay or decomposed in a certain time period. Now here the way half by function really depends on the order. Remember we have zeroth order, first order and second order. Now if we're talking about zero order, we're gonna say for zero order reactions, we use the following equation for half life here half life equals the initial concentration of our reactant divided by two times K. Here a sub zero is again our initial reacting concentration K equals our rate constant. Remember for for um for zero order processes we'd say that its units of polarity times time, inverse here, T in this case is time. But when we have t half that really stands for half life. Now if we take a look at this we'd say what happens to half life? If our initial concentration gets blank as concentration decreases. Well here we're going to say that since half life is part of the equation, we're going to say half life does depend on the initial concentration of our reactant and it gets shorter as the concentration decreases. Because I think about it if this number starts off at let's say 100 100 divided by two K. Gives us a certain value. As this number starts to drop. That means that half life starts to drop the smaller half life, the less time it takes for you to lose half of your initial amount now here um the time gets shorter over time. So what would that look like graphically we'd say that as time progresses that our time gets shorter and shorter. Okay, so we see it with a negative slope decreasing over time. And here remember we have our half life as the Y axis and our time as the X. Axis. And remember when we're doing a plot it's always of Y versus X. So here that's why half life is on the line and time is on the X axis, respectively. Alright, so keep in mind that this is the half life equation for a zero order process, where half life equals initial concentration of your reactant divided by two K.

the reverse haber reaction has a rate constant of 1.45 times 10 to negative six polarity times seconds inverse at 25 degrees Celsius. Now calculate the half life for the reaction, where the initial concentration of ammonia equals 2.47 times 10 to the negative two moles per liter. Alright, so here we know that this is a zero order reaction. Because remember for a zero order reaction, the units for your rate constants are polarity times time inverse. So the fact that its polarity times seconds in verse tells us this is a zero order reaction. So here that means that half life equals the initial concentration of your reactant divided by two K. So this becomes a simple plug and chug type of question. We have the initial concentration of our reactant in terms of ammonia divided by two times our rate constant here, malaria, t times seconds inverse. The polarities will cancel out and our time our half life will be in seconds. So this comes out to be 8.52 times to the three seconds. So this would be our final answer.

now when it comes to first order in half life first realized that all radioactive processes follow a first order rate law. So for reactions with first order, we use the following equation and that is half black equals Ln two over K. Now, if you plug in Ln two Ln two in your calculator, you're gonna get 20.693 And it will still be over K. Okay, so just be aware of that. You may see on your formula sheet instead of Ln two, you may see .693. That's where it comes from. Now. Here we're gonna say Ln two is our constant. In this case we're going to say K is our rate constant. And here we'd say since it's dealing with first order it would be units of time. And verse T of course is time when it's T. Half its half life. Now, what can we say about this half life equation and concentration? Well, if we take a look, half life equals Ln two over K. We don't see initial concentration of our reactant anywhere. So that means half life does not depend on the initial concentration of our reactant. And as a result of this, it's going to be constant throughout the whole reaction. So that means how five she's gonna stay flat because K. Is going to be a number that's not gonna change. And Ellen too is always the same number as well. So half life is going to stay consistent throughout the whole reaction. Now, you remember if we do a plot of Y versus X. Why here would be half life? So that's why it's on the Y axis, and X. Here is T for time. That's why it's on over here on the X axis. So keep these little facts in the back of your mind when dealing with half life and first order rate law equations.

here, we're told that the rate constant for the following reaction was found to be 2.3 times 10 to negative three seconds inverse at 35 degrees Celsius. Here, if the initial concentration of nitrogen dioxide is 1.4 times 10 to the negative one, what is the half life of this reaction? Or on this reaction? Remember since our rate constant K. Has units of just time in verse. This is a giveaway that it is a first order rate law process. And since its first order, that means half life equals Ln two over K. So, Alan two divided by a rate constant K. Which is 2.3 times 10 to the -3 seconds in verse When we punch that in, that gives us 301 seconds. This would be our final answer. Now, notice what else did I give us? I gave us the initial concentration of our reactant. But remember that's nowhere to be seen within our formula for half life for a first order rate law process. So, that's just extra information given to us at times were given a lot of info and a question that doesn't necessarily mean we have to use all of it. This is one of those times. The only part that's important here is our rate constant K. Which we plug in and that helps us to find our half life at the very end

for reactions with second order, we use the following equation. So for second order rate law reactions, half life equals one over K times the initial concentration. So again, a sabo is initial concentration K equals our rate constant in since it's a second order process, the units for K would be polarity inverse times time, inverse T equals time. But if we're dealing with T have that equals half life now, what can we say about half life when it comes to the initial concentration and what happens to it as concentration decreases? We're looking at this formula for second order rate law processes. We say that the half life is dependent on initial concentration because it's found within the formula and it gets longer as the concentration decreases. So think about it. This number here is getting smaller and smaller. So one is getting divided by a smaller number, which means overall the number is larger. So you're half life is larger. So we'd expect that the half life for a second order process to increase over time, meaning that as time progresses it takes longer and longer for you to lose half of your substance. Okay, so it's a little bit weird. That's why when it comes to radioactive processes, we stick to first order processes um exclusively because they're more consistent. Right? So keep this in mind in terms of the formula. And how the formula relates to the initial concentration of reactant and how the concentration can affect the length of the half life for any second order rate law process

The half life of a certain reaction with second order was found to be .45 seconds. What was the initial concentration of the reactant? If the slope of the straight line for this reaction is 3.5 times 10 to the -3. Alright, so they tell us it's second order which means that half life equals one over K. Times initial concentration here we need to solve for initial concentration. So what we do is we multiply both sides by K. Times initial concentration which becomes K. Times initial concentration times half life equals one, divide both sides. Now by K. And half life. And we've isolated our initial concentration. So initial concentration equals one over your rate constant K. Times your half life. Alright, so now knowing this let's solve so initial concentration equals one over. We're told that our Half life is .45 seconds. But how do we figure out? Kay remember when we dealt with the integrated weight loss when it comes to K. It's equal to the slope of a straight line. So them telling us the slope of the straight line is them really telling us what K. Is. So okay here is this value. So we plug in 3.5 times 10 to the -3. Doing that gives us our initial concentration which is 6:34.92 molars. Since both of our numbers have two sig figs, this comes out to 630 moller. So this would be our initial concentration of our reactant