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Topic: Rates of removal/production and slope calculation  (Read 783 times)

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TheArmchairSkeptic

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Rates of removal/production and slope calculation
« on: February 06, 2019, 08:57:32 PM »
Hey there, just cramming for a test tomorrow and I'm having a couple issues with practice questions I'm hoping someone could help me with. First question:

"For the reaction of 2A + 4B → C + 3D the rate law was determined to be Rate = k[A]. The rate constant for the reaction was determined to be 0.351 min -1 and the reaction starts with [A] = 2.41 M and [ B] = 2.55 M.

After 1.2575 minutes have passed [A] would equal [?] M and the [ B] would equal [?] M.

The rate of removal of A would equal [?] mol L-1 min-1 at the initiation of the reaction.

The rate of removal of A would equal [?] mol L-1 min-1 after 1.2575 minutes have passed.

The rate of production of D would equal [?] mol L-1 min-1 after 1.2575 minutes have passed."

Because {rate = k[A]} the reaction is first order, so I've used the formula [ln([A]t/[A]o) = -kt] to solve the first 2 [?]s (1.55 and 0.83 respectively). However, I'm having a hard time with the removal/production rates. Can anyone tell me how to calculate them? I'm either using the wrong formulas or using the right formula in the wrong way...

Second question:

I've been given this data in a question about determining Ea

T (K)
750 796 850 896

1/T (10−3 K−1)
1.33 1.26 1.18 1.12

ln k
−8.62 −6.21 −3.61 −1.43

The problem is, the textbook is giving the slope of this graph as -3.27x10^4 K, but that's not the answer I'm getting at all. Slope should just be dy/dx, no? I've tried using various points on the graph to calculate but I'm not even getting within shouting range of that answer. What am I doing wrong?

Thanks very much to anyone who can help me out here, I really appreciate it.

mjc123

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Re: Rates of removal/production and slope calculation
« Reply #1 on: February 07, 2019, 05:58:35 AM »
Quote
I'm either using the wrong formulas or using the right formula in the wrong way...
Why don't you tell us what formula you're using, so we can see where you may be going wrong.
Likewise show us your working on the slope of the graph. It looks about right to me.

On a more general point, you are assuming that "Rate = k[A]" means d[A]/dt = -k[A], in which case your answers to the first part are correct. However, "rate of reaction", unless more specifically defined, usually means 1/νi*d[Ri]/dt, where νi is the stoichiometric coefficient of reagent Ri (positive for products, negative for reactants). Thus for your reaction as written
Rate = -1/2*d[A]/dt = -1/4*d[ B]/dt = d[C]/dt = 1/3*d[D]/dt
In this case [A]/[A]0 = e-2kt
The problem with this is that before 1.2575 min all the B (the limiting reagent) is gone, so the question can't be answered. (Unless that is supposed to be the answer, but I doubt it.)

TheArmchairSkeptic

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Re: Rates of removal/production and slope calculation
« Reply #2 on: February 07, 2019, 07:11:36 PM »
Thanks for the response. I was able to find the answer to the first question on my own, but the second is still stumping me. I've been using the formula dy/dx to calculate the slope, so in this case it would be -1.43-(-8.62)/1.12-1.33=7.19/-.21=-34.2380952381=(-3.42x10^1). Futhermore, the given Ea in this problem is 2.7x10^2 and because [slope=-(Ea/R)], using the slope they provided does not produce the Ea they gave. Using the slope I calculated comes very close to that Ea though, so I'm starting to think maybe I did it right and the slope given in the textbook is wrong? For the record, there are a number of confirmed-by-the-prof errors in this textbook, so I'm not trying to come off as cocky by saying the book might be wrong.

mjc123

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Re: Rates of removal/production and slope calculation
« Reply #3 on: February 08, 2019, 04:35:56 AM »
Look again at your data. What are the units of 1/T?

TheArmchairSkeptic

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Re: Rates of removal/production and slope calculation
« Reply #4 on: February 09, 2019, 04:42:14 PM »
Wow, talk about missing the forest for the trees. *eyeroll* Thanks for bringing that to my attention, but I'm still getting the answer as -3.42x10^4 and not the -3.27x10^4 given in the problem. Is there something else I'm doing wrong in addition to apparently being unable to read units? Or does that discrepancy just result from using different points on the line to measure the slope, and isn't a big deal?

mjc123

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Re: Rates of removal/production and slope calculation
« Reply #5 on: February 11, 2019, 04:00:19 AM »
Plot a regression line, don't just use the two end points.

TheArmchairSkeptic

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Re: Rates of removal/production and slope calculation
« Reply #6 on: February 11, 2019, 10:01:28 PM »
Got it. Thanks again for all the help.