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Rate constant k from half-life example | Knetics | Chemistry | Khan Academy

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In this video we will look at how the speed constant K is for first order reactions. We will use this example, where we have a data table so we know the concentration of the starting material hydrogen peroxide, H2O2, and we have 1, 2, 3, 4 separate points in time, and the concentrations in each. The reaction we are considering is disintegration of hydrogen peroxide to water and oxygen. The reaction was carried out at 40 degrees Celsius. The first question is: what is K? Based on the data given, this concentration over time, we want to find the velocity constant K. There are two common ways we can answer the question. The first way is, and probably for most people, this is the first one to remember is to do it graphically. You can plot a schedule because we know that in first order reactions, we plot a natural logarithm of concentration over time. When you build the graph, you can take the integral shape of the kinetic equation for this reaction is a natural logarithm of concentration at any given time is equal to natural logarithm of the original concentration minus Kt.

We can graphically represent this data. We find a natural logarithm of these values, in terms of time, and the slope is –K. This is a good way to accomplish this particular task. But you have to offer, especially knowing that this particular reaction is first order, we can solve it in another way. What is specific to first-order reactions, is that their half-life, or t 1/2, is a constant. Not only is it a constant, but it is related to the rate constant K. The half-life does not change. And we have the formula: t 1/2 is equal to a natural log of 2 on K.

The natural logarithm of 2 can be easily calculated, so this equals 0.693 on the rate constant K. In a separate video, we explain how this formula is derived. So knowing by definition the half-life, it is the time since the initial concentration we go to half of it, so we just have to replace 1/2 at the initial concentration and find t. But if you are in the exam, you may already know this formula, or she was given to you, so we will just use it to find K, without graphically presenting our data. This is also convenient if you do not have a graphing exam calculator. We look at our data and see what we can find the half-life of the table. Between the first and second point … i’ll number them so it’s clear what i’m talking about. We go from line one to line two and the concentration is halved, and the elapsed time is 2.16 by 10 ^ 4 seconds. Just to be precise, let’s look at some of the other lines, and make sure that the half-life doesn’t really change.

If we look at lines two and three, concentration again halves, again halves between third and fourth. Looking at all the lines the concentration is halved, and the time is always 2.16 by 10 ^ 4 seconds. So the half-life is 2.16 by 10 ^ 4 seconds. I have to say that this is a slightly simplistic example. This data looks very correct. When you try first-order response, and you measure that data, your data is very unlikely to appear so perfect and that is totally normal. This is completely normal, so don’t be upset if your experimental data is not as perfect as this. This is actually a good sign. Anyway, we now know the half-life from our data table, and we can convert the formula to find K. To find K, it is equal to 0.693 divided by half-life. Because we know the half-life, we can substitute, and we get that K is 0.693 divided by 2.16 by 10 ^ 4 seconds. I enter the numbers into the calculator and I get that K equals 3.21 by 10 ^ (- 5), and units of measure are second to minus first degree.

The important thing to remember here is that for first order reactions where the time for half-life is a constant, you can use this convenient formula, to calculate K if you know the half-life, and vice versa. if you know the half-life you can always find K, and it’s always very easy to find him for a reaction first order based on graph or tabular data, finding the constant half-life.

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