Use the information in Table 14.1 and Figure 14.1 to estimate the instantaneous rate of appearance of NO2 at t = 350 s by calculating the average rate of appearance of NO2 over the following time intervals centered on t = 350 s. (a) 0 to 700 s (b) 100 to 600 s (c) 200 to 500 s (d) 300 to 400 s Which is the best estimate, and why?

Video transcript

Hello everyone today. We have the following problem, calculate the average rate of appearance of sulfur dioxide across the following time intervals. All centered on T. Or time equals 1 75 minutes. Identify the best estimate for the instantaneous rate of appearance of sulfur dioxide at T. Equals 1 75 minutes. And explain why. So what we're gonna do is we're gonna go through each of these answer choices and find the best estimate for the entertaining us rate. So we have our first interval are 50-300 minutes. And to calculate the rate of this, we're going to take the rate and we're going to equal that too. The change in the concentration for our sodium dioxide and divide that by the change in time. So what is that gonna look like? Well, from 50, We're gonna box this here to get a better present presentation of it from to 300 minutes At 300 minutes. Were we at 0.0163 moller. And at 50 minutes we have 0.0032 molar. And the change in the time is of course going to be minutes minus 50 minutes. And that's gonna give us 5.2 times 10 to the negative 5th moller per minute. We're gonna do the same thing with our second choice This time we'll circle it this away from 150 minutes to 200 minutes. So as before we're going to take the change in our concentration of sulfur dioxide and divide that by the change of the time. And so we're going to say that at 200 minutes, we're at 0.0116 Mohler and at 1 50 minutes we are at 0. moller. We're gonna divide that by the the time change difference which is 200 minutes -150 minutes. And this will give us 5.2 times 10, it's the negative 5th moller per minute. And we're also gonna hang on to that and for three we will do the same process This time from 100 minutes to to 50 minutes. And so with this From our 250 we're going to have 0.0141 Moeller, we're gonna subtract that by our 0.0062 molars. And the time range of course is to 50 minutes minus our 100 minutes. This is gonna give us 5.3 times 10 to the negative fifth molar per minute. If you look at our last and final one we're into the same process Which is from 0 to 350. And at 3 50 we're gonna have 0.185 moller minus zero Mohler. Of course, that's gonna be 3 50 minutes to zero minutes Which will give us 5.3 times 10. Then I get 1/5 moller per minute. So we were it was said that the shortest range. So the shortest range will give us the best estimate. So the shortest range will give us the best estimate. And so if we look out of all of these answer choices, which one has the shortest closest range of time, That is going to be great # two. That is because it is from 200 to 1 50. So there's only a 50 minute difference between these two, Which means that the best answer choice will be 5. Or 150-200 minutes as our final answer. And with that we've answered the question overall. I hope this helped. And until next time.