Chemical Kinetics / Rate Law Question - Kaplan Mess-up??

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

mdk2121

New Member
10+ Year Member
Joined
Dec 8, 2009
Messages
3
Reaction score
0
I am reading Kaplan's General Chemistry MCAT Review 2018-2019, and think that one of its chemical kinetic questions is wrong. (Or perhaps I am simply missing something). I have attached an image of the question, which is easier than typing out as it includes the standard table of concentrations at different trials.

The answer reads: "In the first two trials, the concentration of XH4 is held constant while the concentration of O2 is multiplied by 4, and the rate of the reaction also increases by a factor of approximately 4..." It ultimately comes up with a rate of k[XH4]^2[O2]

Am I completely missing something? My first question is, is the book's answer right? If it is not, is the problem solvable (i.e. does the table make any sense at all)? I've tried to come up with rates using the table and encountered the following issues. Please let me know if I am right in approaching the question in this way:

1. If you try to find the rate with respect to XH4 using trials 1 and 3, I believe you get an order of 3. However, if you try to find the same rate using trials 1 and 4, you get 4. This isn't normal, right? Maybe an example of a mixed order reaction? Or just a mess up?

2. How do we interpret the data from trials 3 and 4? The concentration of neither reactant changes and yet the rate doubled. If we ever see something like this in a table, does it suggest that the reaction is 0 order? Or can we not make that leap? And is it even possible to have this data alongside the other data in the table?

3. It seems to me that, based on Trials 1 and 2, the rate is 0 order with respect to [O2]. Is that a correct interpretation?

Thank you so much!

Members don't see this ad.
 

Attachments

  • RATEQUESTION.JPG
    RATEQUESTION.JPG
    116.7 KB · Views: 92
I am reading Kaplan's General Chemistry MCAT Review 2018-2019, and think that one of its chemical kinetic questions is wrong. (Or perhaps I am simply missing something). I have attached an image of the question, which is easier than typing out as it includes the standard table of concentrations at different trials.

The answer reads: "In the first two trials, the concentration of XH4 is held constant while the concentration of O2 is multiplied by 4, and the rate of the reaction also increases by a factor of approximately 4..." It ultimately comes up with a rate of k[XH4]^2[O2]

Am I completely missing something? My first question is, is the book's answer right? If it is not, is the problem solvable (i.e. does the table make any sense at all)? I've tried to come up with rates using the table and encountered the following issues. Please let me know if I am right in approaching the question in this way:

1. If you try to find the rate with respect to XH4 using trials 1 and 3, I believe you get an order of 3. However, if you try to find the same rate using trials 1 and 4, you get 4. This isn't normal, right? Maybe an example of a mixed order reaction? Or just a mess up?

2. How do we interpret the data from trials 3 and 4? The concentration of neither reactant changes and yet the rate doubled. If we ever see something like this in a table, does it suggest that the reaction is 0 order? Or can we not make that leap? And is it even possible to have this data alongside the other data in the table?

3. It seems to me that, based on Trials 1 and 2, the rate is 0 order with respect to [O2]. Is that a correct interpretation?

Thank you so much!

I also was confused. Can anyone clarify this?
 
No, this question does not make sense to me and must be a mess up on Kaplan's part. If it's second-order in XH4 and first-order in O2, then the rate must quadruple when XH4 doubles and double when O2 doubles. This is not what the rate data shows.
 
I am reading Kaplan's General Chemistry MCAT Review 2018-2019, and think that one of its chemical kinetic questions is wrong. (Or perhaps I am simply missing something). I have attached an image of the question, which is easier than typing out as it includes the standard table of concentrations at different trials.

The answer reads: "In the first two trials, the concentration of XH4 is held constant while the concentration of O2 is multiplied by 4, and the rate of the reaction also increases by a factor of approximately 4..." It ultimately comes up with a rate of k[XH4]^2[O2]

Am I completely missing something? My first question is, is the book's answer right? If it is not, is the problem solvable (i.e. does the table make any sense at all)? I've tried to come up with rates using the table and encountered the following issues. Please let me know if I am right in approaching the question in this way:

1. If you try to find the rate with respect to XH4 using trials 1 and 3, I believe you get an order of 3. However, if you try to find the same rate using trials 1 and 4, you get 4. This isn't normal, right? Maybe an example of a mixed order reaction? Or just a mess up?

2. How do we interpret the data from trials 3 and 4? The concentration of neither reactant changes and yet the rate doubled. If we ever see something like this in a table, does it suggest that the reaction is 0 order? Or can we not make that leap? And is it even possible to have this data alongside the other data in the table?

3. It seems to me that, based on Trials 1 and 2, the rate is 0 order with respect to [O2]. Is that a correct interpretation?

Thank you so much!


I'm working on this right now and also didn't see a correct answer option. Also the explanation seems to be describing the results of a different question, since it says that during the first two trials the concentration of O2 is multiplied by 4, which is clearly not the case...

I guess Kaplan messed up! Lost some credibility :(
 
Members don't see this ad :)
No, this question does not make sense to me and must be a mess up on Kaplan's part. If it's second-order in XH4 and first-order in O2, then the rate must quadruple when XH4 doubles and double when O2 doubles. This is not what the rate data shows.
I would argue that XH4 is raised to the 4th power while O2 isn't even in the rate law for the answer to be correct with the given table.
 
I would argue that XH4 is raised to the 4th power while O2 isn't even in the rate law for the answer to be correct with the given table.

Read my post again. I'm saying what the data should show if their answer was correct.

In fact, the data shows that [XH4] is raised to the third power. When you double XH4 while holding O2 constant, the rate is multiplied by a facfor of eight.
 
  • Like
Reactions: 1 user
Top