Half-Life is the time it takes for half of a reactant to decay in a certain time period.
Understanding Half-Life
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Half-Life Concept 1
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Now half life is just the time it takes to lose half of a reactant to decay or decompose in a certain time period. Now here the way half life function really depends on the order. Remember, we have 0th order, 1st order, and second order. Now if we're talking about 0 order, we're gonna say for 0 order reactions, we use the following equation for half life. Here, half life equals the initial concentration of our reactant divided by 2 times k. Here a sub o is again our initial reactant concentration, k equals our rate constant, and remember for for, for 0th order processes we'd say that it's units of molarity 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 gonna say half life does depend on the initial concentration of our reactant, and it gets shorter as the concentration decreases. Because think about it, if this number starts off at, let's say, 100, 100 divided by 2 k gives us a certain value. As this number starts to drop, that means that half life starts to drop. The smaller your half life, the less time it takes for you to lose half of your initial amount. Now here, 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. K? So we'd 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 a y versus x. So here that's why half life is on the y and time is on the x axis respectively. Right? So keep in mind that this is the half life equation for zero order process where half life equals initial concentration of your reactant divided by 2 k.
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example
Half-Life Example 1
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The reverse Haber reaction has a rate constant of 1.45 times 10 to the negative 6 molarity 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 2 moles per liter. Alright. So here we know that this is a 0th order reaction because, remember, for a 0th order reaction, the units for your rate constants are molarity times time inverse. So the fact that it's molarity times seconds inverse tells us this is a 0 order reaction. So here that means that half life equals the initial concentration of your reactant divided by 2 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 2 times our rate constant here, molarity times seconds inverse, The molarities will cancel out and our time our half life will be in seconds. So this comes out to be 8.5 2 times 10 to the 3 seconds. So this would be our final answer.
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Problem
Problem
Decomposition of a certain substance Y at 45°C was found to be zero order. What is the half-life of substance Y if it took 15.5 minutes to decompose 67% of this substance? [Y]0 = 0.25 M.
A
0.058 min
B
0.029 min
C
11.6 min
D
23.5 min
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Half-Life Concept 2
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When it comes to 1st order and half life, first realize 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 life equals l n 2 over k. Now if you plug in ln2ln2 in your calculator, you're gonna get 0.693, and it'll still be over k. K. So just be aware of that. You may see on your formula sheet instead of ln 2 you may see 0.693, that's where it comes from. Now, here we're gonna say ln2 is our constant in this case, we're gonna say k is our rate constant, and here we'd say since it's dealing with first order it'd be units of time inverse. T of course is time, when it's t half, it's half life. Now what can we say about this half life equation and concentration? Well if we take a look, half life equals ln2 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 half phi is just gonna stay flat because k is gonna be a number that's not gonna change and ln2 is always the same number as well. So half life is gonna stay consistent throughout the whole reaction. Now, here remember if we do a plot it's of y versus x, y here would be half y 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 of rate law equations.
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example
Half-Life Example 2
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Here we're told that the rate constant for the following reaction was found to be 2.3 times 10 to negative 3 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 inverse, this is a giveaway that it is a first order rate law process. And since it's first order, that means half life equals ln2 over k. So ln2 divided by our rate constant k which is 2.3 times 10 to the minus 3 seconds inverse. 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 no just extra information given to us. At times, we're given a lot of So that's just extra information given to us. At times, we're given a lot of info in 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.
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Problem
Problem
Radioactive plutonium-239 (t1/2 = 2.41 × 105 yr) is used in nuclear reactors and atomic bombs. If there are 5.70 × 102 g of plutonium isotope in a small atomic bomb, how long will it take for the substance to decay to 3.00 × 102 g?
A
2.23×105 yr
B
2.60×105 yr
C
517 yr
D
600 yr
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Half-Life Concept 3
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For reactions with second order, we use the following equation. So for second order, rate law reactions, half life equals 1 over k times the initial concentration. So again a sub o is initial concentration, k equals our rate constant in. Since it's a second order process, the units for k would be molarity inverse times time inverse. T equals time, but if we're dealing with t half 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? Well, 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 your 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, exclusively because they're more consistent. Right? So keep this in mind in terms of a formula and how the formula relates to the initial concentration of reactants, and how the concentration can affect the length of the half life for any second order rate law process.
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example
Half-Life Example 3
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The half life of a certain reaction with second order was found to be 0.45 seconds. What was the initial Alright. So Alright. So they tell us it's second order, which means that half life equals 1 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 1. Divide both sides now by k in half life, and we've isolated our initial concentration. So initial concentration equals 1 over your rate constant k times your half life. Alright. So now knowing this, let's solve. So initial concentration equals 1 over we're told that our half life is 0.45 seconds. But how do we figure out k? 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 k here is this value. So we plug in 3.5 times 10 to the minus 3. Doing that gives us our initial concentration which is 634.92 molar. Since both of our numbers have 2 sig figs, this comes out to 630 molar. So this would be our initial concentration of our reactant.
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Problem
Problem
Use the data below to determine the half-life of decomposition of NOCl reaction which follows 2nd order kinetics.