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Shown here is a plot of $[\mathrm{A}]_{t}$ versus $t$ for the reaction$A \longrightarrow$ product. (a) Determine the order and the rate constant of the reaction. (b) Estimate the initial rate and the rate at 30 s.

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(a) first-order, $k=0.035 \mathrm{s}^{-1}$(b) initial rate $=0.035 \mathrm{M} / \mathrm{s}$, rate at $30 \mathrm{s}=0.012 \mathrm{M} / \mathrm{s}$

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

Rice University

University of Maryland - University College

University of Kentucky

Lectures

22:42

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Hello. So did they were going to be looking at this graph right here, off the concentration of a reactive over time. So we see AT T equals zero. We have, ah, one concentration of one. Now let's take a look at when? Oh, how much time passes? When the concentration is a half, we see that it's a 20 seconds. Now let's see how much time passes when it's around 2.5 and we'll see that it's around 40. And now let's see, half of 20.25 is put 1 to 5, and that's around 60. So it looks like the half life off this graph is 20 seconds. And if you know that the A constant half light means that this is a first order reaction. So this looks like this is a graph of a first order reaction and the half life is 20 seconds. If we know the half life, we can figure out Okay, so the half life is equal to the natural log of two over K, the right constant. So the rate constant is just a natural log of two over have life, which in this case is 20 seconds. So we will see if we do the math that it looks like our rate constant is around zero 0.0 three five and for seconds. So now let's try to estimate the rates. So what is the rate initially, how would we estimate this? Well, we would sort of draw an imaginary line, and we would try to approximate. All right, so it looks like if we draw this American imaginary line that we would go from 12.7 in 10 seconds. So 0.3 over 10. That's how much we changed. And we see Well, we changed negative zero point 3/10. Still, it looks like we have a 0.3 decrease in a. So it looks like this is our rate right here initially. Now let's take a look at our rate at around 30 seconds. So 30 seconds. Let's draw an imaginary line, you and we'll approximate that. We're going from 0.75 to 0. And from 15 0 to 50. So wait, We're decreasing by 0.75 and 50 seconds is passing. So looks like we're decreasing moderate off 0.15 That means that our rate is approximately 0.15 mil Arat e per second. So is disappearing at 0.15 polarities for second. So this is just an approximation based off of basically tangent lines to those points, there is a

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