November 24, 2024, 06:35:58 AM
Forum Rules: Read This Before Posting


Topic: Rate of reaciton expressed in ml/min  (Read 14998 times)

0 Members and 2 Guests are viewing this topic.

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Rate of reaciton expressed in ml/min
« on: February 04, 2014, 01:32:44 PM »
The problem is attached, and the question is to calculate the molarity of H2O2 at the beginning of the experiment #4 and after 4 min.
I got that the rate law is r=[H2O2][KI]. The concentration at t=4min can be calculated from the integrated rate equation of first order, but I need k. How to calculate it? The beginning concentration of peroxide is 0.294M and of KI is 0.0167M. I know that I should use the rate law r=[H2O2][KI], but r isn't expressed in concentration/min as the concentration change is the same for every experiment, because it is a gas. Never saw a case like this, when a gas is produced. Should I just put the volume/min into r=[H2O2][KI]?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #1 on: February 04, 2014, 05:49:16 PM »
To start with, surely you have to convert those volumes they've given you for the liquid-phase species to concentrations in the combined initial solution.

Generally I think it should be, like with equilibrium, activities, so you'd look at partial pressure for any gas and concentration for anything in solution (or a liquid I suppose). But here it seems that partial pressure is given (and constant).

What do you do in an equilibrium case where the pressure (total and partial) is kept constant and you have to work with volumes instead? Go back to concentration? Except it's a gas ... activity should be related to pressure, not concentration?

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #2 on: February 05, 2014, 06:28:51 AM »
I converted and wrote the concentrations for the substance in the liquid phase, but I don't know what to use for oxygen. The only two parameters that aren't constant for oxygen are the number of moles and the volume. Should I maybe put the mole increase for the gas, or the volume?

Can't find a reference where it is explained what to use for the activity of a gas at constant pressure :(.
« Last Edit: February 05, 2014, 08:30:19 AM by Raderford »

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #3 on: February 05, 2014, 09:41:55 AM »
I have an idea. Using the ideal gas law, 1.74·10-4 moles of oxygen are produced in the first minute. This means that 3.48·10-4 moles of peroxyde disappeared in the first minute, i.e. 3.48·10-4/0.15=2.32·10-3M disappeared in one minute. The rate can be expressed also as:
r=-1/2·d[H2O2]/dt=1.16·10-3M/min, now we have a valid unit for the rate. Then:
k=r/([H2O2][KI])=0.236/s. Is this correct?

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #4 on: February 05, 2014, 01:21:01 PM »
I have an idea. Using the ideal gas law, 1.74·10-4 moles of oxygen are produced in the first minute. This means that 3.48·10-4 moles of peroxyde disappeared in the first minute, i.e. 3.48·10-4/0.15=2.32·10-3M disappeared in one minute. The rate can be expressed also as:
r=-1/2·d[H2O2]/dt=1.16·10-3M/min, now we have a valid unit for the rate. Then:
k=r/([H2O2][KI])=0.236/s. Is this correct?

I will give one idea I have. If the gas is being held at constant partial pressure once it escapes, perhaps you can say that we have the reaction O2 (aq)  :rarrow: O2 (g) which is instantaneous. i.e. convert all that O2 gas into the concentration it would be in solution, just before it left the solution. Then r=d[O2 (aq)]/dt and (maybe) you can thereby express rate in terms of a concentration of oxygen, just before it leaves the solution. I think this is pretty similar to what you ended up doing.

But, I don't like the approximations. Volume of gas given off in the first minute does not have to be a good estimate of initial rate of production of gas. And then there is a more fundamental problem in the question itself: they haven't specified if the rate given refers to the average rate of production of O2 until the experiment finishes, or the initial rate. If the initial rate, why use mL/min rather than mL/s, since a minute is much larger time (so it is less precise to say initial rate wrt a minute). If the latter, we have a much harder but still solvable problem. I'm inclined to think it's the initial rate but they could have specified this.

Do you think they can ask for integrations in this IChO or in the prep problems? If not then it greatly limits the range of questions they can ask on kinetics. In any case the extent of integration I've seen on Olympiads before is still well below the complexities that could arise in kinetics, if it is allowed.

After all is said and done, I definitely want to find out, how to define activity for a gas at constant partial pressure.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Rate of reaciton expressed in ml/min
« Reply #5 on: February 05, 2014, 01:27:39 PM »
I converted and wrote the concentrations for the substance in the liquid phase, but I don't know what to use for oxygen.

Where's Oxygen in your rate expression?

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #6 on: February 05, 2014, 01:58:36 PM »
It should represent the rate itself: d[oxygen]/dt.
I have an idea. Using the ideal gas law, 1.74·10-4 moles of oxygen are produced in the first minute. This means that 3.48·10-4 moles of peroxyde disappeared in the first minute, i.e. 3.48·10-4/0.15=2.32·10-3M disappeared in one minute. The rate can be expressed also as:
r=-1/2·d[H2O2]/dt=1.16·10-3M/min, now we have a valid unit for the rate. Then:
k=r/([H2O2][KI])=0.236/s. Is this correct?

I will give one idea I have. If the gas is being held at constant partial pressure once it escapes, perhaps you can say that we have the reaction O2 (aq)  :rarrow: O2 (g) which is instantaneous. i.e. convert all that O2 gas into the concentration it would be in solution, just before it left the solution. Then r=d[O2 (aq)]/dt and (maybe) you can thereby express rate in terms of a concentration of oxygen, just before it leaves the solution. I think this is pretty similar to what you ended up doing.

But, I don't like the approximations. Volume of gas given off in the first minute does not have to be a good estimate of initial rate of production of gas. And then there is a more fundamental problem in the question itself: they haven't specified if the rate given refers to the average rate of production of O2 until the experiment finishes, or the initial rate. If the initial rate, why use mL/min rather than mL/s, since a minute is much larger time (so it is less precise to say initial rate wrt a minute). If the latter, we have a much harder but still solvable problem. I'm inclined to think it's the initial rate but they could have specified this.

Do you think they can ask for integrations in this IChO or in the prep problems? If not then it greatly limits the range of questions they can ask on kinetics. In any case the extent of integration I've seen on Olympiads before is still well below the complexities that could arise in kinetics, if it is allowed.

After all is said and done, I definitely want to find out, how to define activity for a gas at constant partial pressure.
Interesting idea. It gives exactly the same solution as my proposal, so actually it should be the same thing then, only worded differently. So I think that this is the correct solution for the problem when a gas is under constant pressure.

It has to be the initial rate, as we use the initial concentrations in the rate law expression.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #7 on: February 05, 2014, 04:17:22 PM »
Interesting idea. It gives exactly the same solution as my proposal, so actually it should be the same thing then, only worded differently. So I think that this is the correct solution for the problem when a gas is under constant pressure.

Oxygen is a gas, so I would hesitate to use d[O2]/dt as a definition of rate. Usually we would be using d(P(O2 (g)))/dt, partial pressure. Here of course we can't. But how does that legitimize using concentration?

Once rate is defined as this, it is clear how to carry forward the calculation. But now it seems to me we are modelling the reaction producing O2 (aq) rather than O2 (g). If there is a subtlety required by the calculation or units in relating what we can say for O2 (aq) to O2 (g) then I can't explain it, but we should try to get some clarification here. In general having a method for activity of gases at constant pressure would be good - surely cannot be concentration, since concentration = number of moles / volume and here we have both number of moles and volume changing (once O2 reaches gaseous phase - we can track it, by concentration, only when in solution) with time.

It has to be the initial rate, as we use the initial concentrations in the rate law expression.

Hmm. Can we say that average rate over the whole reaction procedure must be 0, since if we define average rate = final concentration / total time taken, since the concentration is asymptotic towards the final concentration and never exactly reaches it, total time taken must be ∞ and whatever the finite final concentration is (bounded by limiting reagents, etc.), average rate = 0? If we can do this then the rates must refer to initial.

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #8 on: February 06, 2014, 08:06:41 AM »
I don't see why someone would use the average rate with initial concentrations. The rate has it's biggest value at the beginning of the reaction, if you use the average rate, there would be a big error. Seems like no integration this time, you can just memorize the rate laws.

I have another question. As the concentration of peroxide is first order and the reaction has a second order overall, is it correct to use the k from: r=k[H2O2][KI] in this equation: ln[H2O2]=ln[H2O2]0-kt? I think not, as I found a equation for the second order reaction here http://en.wikipedia.org/wiki/Rate_equation#Zero-order_reactions (the one with A and B), but there are two unknowns.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #9 on: February 06, 2014, 10:46:27 AM »
I don't see why someone would use the average rate with initial concentrations. The rate has it's biggest value at the beginning of the reaction, if you use the average rate, there would be a big error. Seems like no integration this time, you can just memorize the rate laws.

Good. So then average rate over the whole reaction must indeed be 0 M/s (because the reaction never completely finishes until t=∞)?

0, 1st and 2nd order aren't very difficult to integrate anyway. A+B :rarrow: Products though is worth memorizing.

I have another question. As the concentration of peroxide is first order and the reaction has a second order overall, is it correct to use the k from: r=k[H2O2][KI] in this equation: ln[H2O2]=ln[H2O2]0-kt? I think not, as I found a equation for the second order reaction here http://en.wikipedia.org/wiki/Rate_equation#Zero-order_reactions (the one with A and B), but there are two unknowns.

I haven't actually done the problem yet but I'm sure you got r=k[H2O2][I-] right.

Of course you can't treat it as first-order, only pseudo first-order. You will want to look this up if you haven't already. But to do this, you will have to forget about modelling [H2O2] as a function of time, and model [I-] instead. This makes sense because even if all the I- were consumed (and it never quite gets there) most of H2O2 will be left over - i.e. its concentration is said to be unchanging and can be taken as constant throughout the experiment. So you will use r=kmod[I-] where kmod = k*[H2O2]0. This is an ugly approximation as I'm sure you can see, because H2O2 is not that huge this time - I will test it using exact integral, after my first round is done.

If you have learnt partial fractions you can integrate r=k[A][B] exactly. This is what is done in the section "Second-order reactions" on your Wikipedia page. As for the presence of two variables, you can easily express [A] in terms of [B] and substitute.

However, looking at the initial concentrations in the question it seems that this is not expected and all they expect is you to use pseudo-first order. Many times in IChO I have seen the same problem - you either use constant-concentration approximation, or you can do exact integration. I have seen r=k[A]2[B] which is a more challenging integration but still can be done. (Don't ask for derivation, I haven't yet, just seen the result in my book)

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #10 on: February 06, 2014, 11:03:47 AM »
I know about pseudo-first order. You mean that I should calculate the concentration change of KI after 4 min, and the concentration change of peroxyde would be the same? The concentration of peroxyde is not so big, but I can't think of something else to try.
« Last Edit: February 06, 2014, 11:25:45 AM by Raderford »

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #11 on: February 06, 2014, 11:23:07 AM »
I know about pseudo-first order. You mean that I should calculate the concentration change of KI after 4 min, and the concentration change of peroxyde would be the same? k would have a different value, and the concentration of peroxyde is not so big, but I can't think of something else to try.

Yeah, calculate the concentration change of I- assuming H2O2 concentration to be constant and then using this approximate concentration of I- at the end back-calculate to the approximate concentration of H2O2. If this isn't very close to the initial concentration of H2O2, you will have to scrap the whole method and go for some other way.

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #12 on: February 06, 2014, 11:31:35 AM »
Okay, I got a relative change of concentration of the peroxyde to be 1.38%. Seems okay.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Rate of reaciton expressed in ml/min
« Reply #13 on: February 06, 2014, 11:42:20 AM »
BTW, you spoke about concentration of KI. Here we have I- and K+, but you raise an interesting point with calling it KI. How do we define concentration of a dissolved salt? [KI]=min(I-,K+) - maybe in general then for a salt AaBb, [AaBb] = min([A]/a,[B]/b) ? Seems to be what you used so far (since I- is decreasing and K+ is staying constant, [KI]=[I-]), but I don't feel sure this is good to define [KI], or even if there can be a decent definition for [KI] once it is dissolved. After all imagine if Na+ were also present in this solution in high concentration - then we would have definitions [KI]=[I-] and [NaI]=[I-], but if you say there is 0.200 M KI and 0.200 M NaI (for example) you make out there is 0.400 M of I- which there is not ...

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: Rate of reaciton expressed in ml/min
« Reply #14 on: February 06, 2014, 11:48:09 AM »
I meant I-.

Sponsored Links