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A reaction is first order with respect to the reactant Y. The reaction has half-life of 5 minutes The initial concentration of Y is 1.20 mol dmWhat will be the conc...

Question

A reaction is first order with respect to the reactant Y. The reaction has half-life of 5 minutes The initial concentration of Y is 1.20 mol dmWhat will be the concentration of Y after 15 minutes?0.15 mol dm- 30.30 mol dm-30.40 mol dm 3 1.50 mol dm- 3

A reaction is first order with respect to the reactant Y. The reaction has half-life of 5 minutes The initial concentration of Y is 1.20 mol dm What will be the concentration of Y after 15 minutes? 0.15 mol dm- 3 0.30 mol dm-3 0.40 mol dm 3 1.50 mol dm- 3



Answers

It takes 15.4 minutes for the concentration of a reactant to drop to $5.0 \%$ of its initial value in a second-order reaction. What is the rate constant for the reaction in the units of $\mathrm{L} \mathrm{mol}^{-1} \min ^{-1}$ ?

So here we are giving the example of a specific reaction and this reaction is first ordered and were given information that this first order reaction has a half life of 15 minutes. So we know we can find the rate constant, room, the half life since this half life and the rate constant are related by the following equation. So we can find the rate constant which is equivalent to 0.046 inverse minutes. And we know that this agrees with the units for a first quarter uh great equation. Okay, so now we're asked to find uh the amount of our remaining concentration. So I'm going to rewrite our rate law a bit. So this would be the final fraction of the final concentration over the initial concentration. So this would be expressing the fraction of remaining are remaining stuff. In this case, fracture, remaining re agents is equivalent to negative Katie, mainly. This has just arrived from the fact that when you subtract, when you have to log subtracted, this is equivalent to natural log in the interior A divided by B. Just the log rules. So in this case we can exponentially eight this and we can find that the fraction of this uh the final concentration with respect to the initial concentration is eating the negative Katie. And we're going to let this process run for two hours. However, we have to convert hours, two minutes since we chose our very constant in minutes. So two hours has 120 minutes. And now we have all the information we need to substitute into this equation here. Yeah. And as a result of substituting everything into this equation, we can find that the fraction remaining is about 0.0040 and its unit list since the units for concentration cancel out. And this is our final answer.

Okay here's the integrated um equation for first order. We're told that at 30.5 seconds the concentration is .0451. Mhm. Yeah. And then in um what is it? 45 seconds is .0321. Yeah. Yeah appreciate it. Mhm. So if you compare these two times the time if you say this is like the initial time and this is the time after um the reaction is progress. It's A total time difference of um 14.5 seconds. Yeah and then we can just plug in our numbers. So the we have the natural log of the concentration at time T which is point okay 0321 is equal to the natural log of the initial concentration. Mhm. Okay -K. Which we're solving for because we don't know times the time that has elapsed which is 14.5 seconds. So K is going to be an inverse seconds. Um if we rearrange this equation we have the natural log of 0.3 to one minus the natural log of .0451 it's going to be equal to negative Katie. So we're gonna divide by negative t. Which is negative 14.5. And that's going to give us Okay? Yeah. All right so we have 2.321. We're going to take the natural log of that shit Is equal to negative 3.43 you're a minus sign and then point 0451. Take the natural log of that. It's negative three points 099. Why do I negative 14.55. Mhm. All right. And then we can simplify that. Yeah anything like that And it comes out 2.02-8 inverse seconds. Now we want the half life which is just the Ln of two divided by that K. Value. You know And that will give us half life in seconds. So let's eat the natural life too. Divided by .02-8 comes out to 30.4 seconds. All right. Um Okay So we calculated the half life and then we want how many seconds after the whole reaction starts does it take for the reacting constitution to decrease 2.01. So we'll go back to this equation. Let's use this as the initial concentration but then we'll just have to add 3.5 seconds. Um which already progressed to get to this point to get the total time. So we have natural log of .01 is our final concentration. Yeah Equals the natural log of. Our initial 1. .0451. Yeah I'm going to I mean thank you minus K. Which we solved for In the 1st part. Okay and then times the time. So then the time is just gonna be we'll rearrange to solve like we did before. Okay the company the thank you like that will simplify this. So we have the natural log of point a one Is -4.605. Okay And then we'll do minus the natural log of .451 is negative 3.099, Divided by negative 0.02-8. Then I gave signs are going to cancel out. Okay, I'm not going to break. And this comes out 231.3 seconds from 30.5 to our desired final point. So then we're just going to add 30.5. Mhm. And we get 161.8. Okay. Mhm. I'm very proud. Total seconds will elapse to get to the desired concentration of 0.1.

Okay here's the integrated um equation for first order. We're told that at 30.5 seconds the concentration is .0451. Mhm. Yeah. And then in um what is it? 45 seconds is .0321. Yeah. Yeah appreciate it. Mhm. So if you compare these two times the time if you say this is like the initial time and this is the time after um the reaction is progress. It's A total time difference of um 14.5 seconds. Yeah and then we can just plug in our numbers. So the we have the natural log of the concentration at time T which is point okay 0321 is equal to the natural log of the initial concentration. Mhm. Okay -K. Which we're solving for because we don't know times the time that has elapsed which is 14.5 seconds. So K is going to be an inverse seconds. Um if we rearrange this equation we have the natural log of 0.3 to one minus the natural log of .0451 it's going to be equal to negative Katie. So we're gonna divide by negative t. Which is negative 14.5. And that's going to give us Okay? Yeah. All right so we have 2.321. We're going to take the natural log of that shit Is equal to negative 3.43 you're a minus sign and then point 0451. Take the natural log of that. It's negative three points 099. Why do I negative 14.55. Mhm. All right. And then we can simplify that. Yeah anything like that And it comes out 2.02-8 inverse seconds. Now we want the half life which is just the Ln of two divided by that K. Value. You know And that will give us half life in seconds. So let's eat the natural life too. Divided by .02-8 comes out to 30.4 seconds. All right. Um Okay So we calculated the half life and then we want how many seconds after the whole reaction starts does it take for the reacting constitution to decrease 2.01. So we'll go back to this equation. Let's use this as the initial concentration but then we'll just have to add 3.5 seconds. Um which already progressed to get to this point to get the total time. So we have natural log of .01 is our final concentration. Yeah Equals the natural log of. Our initial 1. .0451. Yeah I'm going to I mean thank you minus K. Which we solved for In the 1st part. Okay and then times the time. So then the time is just gonna be we'll rearrange to solve like we did before. Okay the company the thank you like that will simplify this. So we have the natural log of point a one Is -4.605. Okay And then we'll do minus the natural log of .451 is negative 3.099, Divided by negative 0.02-8. Then I gave signs are going to cancel out. Okay, I'm not going to break. And this comes out 231.3 seconds from 30.5 to our desired final point. So then we're just going to add 30.5. Mhm. And we get 161.8. Okay. Mhm. I'm very proud. Total seconds will elapse to get to the desired concentration of 0.1.

In this problem, A second order decomposition reaction and then I did 550°C has a rate constant. Mhm Equal to 3.1 into changes to the power -2 later. Permal per second having initial concentration. Mm hmm. Mhm. Mhm or not is equal to 0.10 more left. We have to calculate the concentration of reactant after 1.500 tender. Step up to the power two seconds. D. Is alleged time. Mhm. Which is equal to 1.5- 10 days. To the power two seconds. Yeah. Okay. Yeah. Let final concentration is denoted by 80 with the constitution for a second order. The accent one by 80 is equal to Katie Plus one by initial concentration or not. Now substitute the value of care. T. And a Not. In this expression. To evaluate the value of 80. Mhm. Value 40 is equal to 1.5 in two 10 days. To the power two seconds Plus one by a note is equal to 0.10 Mueller. With the help of calculator, we can obtain this value equals to 14.65. Formula. Therefore 80 is equal to One divided by 14.65 smaller university, which is equal to 6.8 into 10 days. To the power -2 moller. Okay This is concentration after time. $1.500 to the power two seconds. Now we have to calculate half life for this reaction. Half life is denoted by the half, which is equal to one bite k into concentration of or not. Now substitute the value of K. And a note in this expression. Mr. And we will obtain the value of the have which is equal to 3.2 into 10 days to the power to second. This is an answer for the health. Thank you. Mhm.


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