Half-Life - Online Tutor, Practice Problems & Exam Prep

Half-life, the time required for half of a substance to decay or decompose, varies with reaction order. In zero-order reactions, half-life is directly proportional to the initial reactant concentration and inversely proportional to the rate constant (k), shortening as concentration decreases. For first-order reactions, common in radioactive decay, half-life is constant, independent of initial concentration, and calculated using:

$T_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Second-order reactions have a half-life that increases as the initial concentration decreases, as it is dependent on both the rate constant and the initial concentration. Understanding these relationships is crucial for predicting reaction behavior over time and is represented graphically with half-life on the y-axis and time on the x-axis.

Half-Life Concept 1

Now half life is just the time it takes to lose half of a reacting to decay or decompose in a certain time period. Now here the way half life function really depends on the order. Remember we have 0^th order, first order and 2^nd order. Now if we're talking about 0 order, we're going to 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 2 * K. Here a₀ is again our initial reacting concentration, K equals our rate constant. And remember for 0^th 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 hat 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, a 100 / 2 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 the time gets shorter overtime. So what would that look like? Graphically, we'd say that as time progresses that our time gets shorter and shorter. 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 AY versus X. So here that's why half life is on the Y and time is on the X axis respectively.

Alright, 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 2K.

Half-Life Example 1

The reverse Haber reaction has a rate constant of 1.45×10-6 molarity times seconds inverse at 25°C. Now calculate the half life for the reaction where the initial concentration of ammonia equals 2.47×10-2 moles per liter.

Right. So here we know that this is a zeroth order reaction because remember for a 0^th 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 2K. So this becomes a simple plug and chug type of question. We have the initial concentration of R reactant in terms of ammonia divided by two times our rate constant, here molarity times seconds. Inverse the molarities will cancel out and our half life will be in seconds.

So this comes out to be 8.52×103 seconds. So this would be our final answer.

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. 12 min
• D. 23.5 min

Half-Life Concept 2

Now 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 ln2K. Now if you plug in ln2 in your calculator, you're going to get 0.693 and it'll still be over K. OK, so just be aware of that. You may see on your formula sheet, instead of ln2, you may see 0.693. That's where it comes from.

Now here we're going to say ln2 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 D be units of time. Inverse T of course is time. When it's T half it's half light. Now what can we say about this half life equation and concentration? Well, if we take a look, half life equals ln2K 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 five is just going to stay flat because K is going to be a number that's not going to change and ln2 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, it's of Y versus X Y here would be half fly. 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 1st order rate law equations.

Half-Life Example 2

Here we're told that the rate constant for the following reaction was found to be 2.3 * 10 to -3 seconds inverse at 35°C. Here. If the initial concentration of nitrogen dioxide is 1.4 * 10 to the -1, 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 its first order, that means half life equals ln2/K, so ln2/K, which is 2.3×10−3 s−1. 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. We're given a lot of info on 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.

Radioactive plutonium-239 (t_1/2 = 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

Which of the following statements is False?

• A. The average rate of a reaction decreases during a reaction.
• B. The rate of zero order reactions are not dependent on concentrations.
• C. The rate of a first order reaction is dependent on concentrations.
• D. The half-life of a first order reaction is dependent on the initial concentration of reactant.

Half-Life Concept 3

For reactions with second order we use the following equation. So for 2nd order rate law reactions half life equals 1K×C₀. So again C₀ is the 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 Thalf that equals half life.

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 2nd 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. OK, so it's a little bit weird. That's why when it comes to radioactive processes, we stick to 1st order processes 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 reactants and how the concentration can affect the length of the half life for any second order rate law process.

Half-Life Example 3

The half life of a certain reaction with second order was found to be, oh, 0.45 seconds. What was the initial concentration of a reactant if the slope of the straight line for this reaction is 3.5 * 10 to the -3? All right, so they tell us it's second order, which means that half life equals 1K×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 1rate constant K times half life.

All right. So now, knowing this, let's solve. So initial concentration equals 1rate constant K times half life. 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 * 10 to the -3. Doing that gives us our initial concentration, which is 634.92 molar. Since both of our numbers have two sig figs, this comes out to 630 molar, so this would be our initial concentration of our reactant.

Use the data below to determine the half-life of decomposition of NOCl reaction which follows 2nd order kinetics.

• A. 1.00×10⁴ s
• B. 1.80×10⁵ s
• C. 2.02×10⁴ s
• D. 1.022×10³ s

Here’s what students ask on this topic:

Which expression relates h, the number of half-lives, and y, the number of years?

The expression that relates h (the number of half-lives) and y (the number of years) is derived from the definition of a half-life, which is the time required for half of a sample of a radioactive substance to decay. The expression can be written as:

$h=\frac{y}{t_{1/2}}$

Here, t1/2 represents the half-life of the radioactive substance in years. This equation shows that the number of half-lives (h) is equal to the total time elapsed (y) divided by the half-life of the substance. If you know the half-life of a substance and the amount of time that has passed, you can use this expression to calculate how many half-lives have occurred during that period. This is particularly useful in fields like geology, archaeology, and physics, where understanding the decay of radioactive isotopes is important for dating materials and studying processes over time.

How can I find the number of half-lives?

To find the number of half-lives that have elapsed, you need to know two key pieces of information: the total time that has passed and the half-life duration of the substance in question. The half-life is the time it takes for half of the original amount of a radioactive substance to decay.

Here's a step-by-step method to calculate the number of half-lives:

1. Determine the half-life of the substance. This information is usually provided in the problem or can be found in scientific literature.
2. Measure or find out the total time that has passed since the substance began to decay.
3. Divide the total time by the half-life of the substance to find the number of half-lives.

For example, if the half-life of a substance is 5 years and 20 years have passed, you would calculate the number of half-lives as follows:

$\text{Number of half-lives}=\frac{\text{Total time passed}}{\text{Half-life duration}}$ $\text{Number of half-lives}=\frac{20\text{years}}{5\text{years}}$ $\text{Number of half-lives}=4$

So, four half-lives have

How do you calculate the number of half-lives?

To calculate the number of half-lives that have passed for a sample of a radioactive substance, you'll use the formula:

$\text{Number of half-lives}=\frac{\text{Total time elapsed}}{\text{Half-life of the substance}}$

Here's how you do it step by step:

1. Determine the total time that has elapsed since the substance began to decay.
2. Find out the half-life of the substance, which is the time it takes for half of the substance to decay. This information is typically provided in textbooks or reference materials.
3. Divide the total time elapsed by the half-life of the substance.

For example, if you have a substance with a half-life of 4 years and you want to know how many half-lives have passed in 16 years, you would calculate it as follows:

$\text{Number of half-lives}=\frac{\text{16 years}}{\text{4 years/half-life}}=\text{4 half-lives}$

This means that in 16 years, the substance has gone through 4 half-lives.