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Deduce the rate law expression

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


The inversion of cane sugar proceeds with a constant half life of 500 minute at pH=5 for any concentration of sugar. When pH is changed to 4, the half life becomes 50 minutes. Deduce the rate law expression for inversion of sugar.


Homework Equations


[itex]t_{1/2}*a^{n-1}[/itex] is constant for reaction of any order


The Attempt at a Solution


Since pH=5
Therefore, [itex][H^{+}]=10^{-5}[/itex] (Initial Concentration)
Similarly Final [itex][H^{+}]=10^{-4}[/itex]
Using first equation I have
500*[itex](10^{-5})^{n-1}[/itex]=50*[itex](10^{-4})^{n-1}[/itex]
10=[itex]10^{n-1}[/itex]
n=2
So it must be second order wrt [itex][H^{+}][/itex]
Since the rate does not depend of sugar so it must be zero order wrt sugar
Therefore the rate law expression should be
Rate=k[Ester][itex][H^{+}]^{2}[/itex]
But the answer in my book says
Rate=k[Ester][itex][H^{+}][/itex]
Whats wrong?
 

Answers and Replies

  • #2
148
2
What is wrong is that the equation you have quoted relates to a changing half-life as a reaction proceeds. In this case you need a different approach because you are looking at initial rates of two separate reaction mixtures. The half-life does not change as the reaction proceeds. That in itself might tell you something.

What you can do, having decided that a(H+) is the important factor, is test how the initial half life would change if the reaction were first order, second order, zero order, minus 1 order in that activity.

When you increase the activity by a factor of 10, what happens to the half-life? What is therefore happening to the rate?

Is the rate proportional to the activity? to the square of the activity? invariant? inversely proportional to the activity?
 
Last edited:
  • #3
utkarshakash
Gold Member
855
13
What is wrong is that the equation you have quoted relates to a changing half-life as a reaction proceeds. In this case you need a different approach because you are looking at initial rates of two separate reaction mixtures. The half-life does not change as the reaction proceeds. That in itself might tell you something.

What you can do, having decided that a(H+) is the important factor, is test how the initial half life would change if the reaction were first order, second order, zero order, minus 1 order in that activity.

When you increase the activity by a factor of 10, what happens to the half-life? What is therefore happening to the rate?

Is the rate proportional to the activity? to the square of the activity? invariant? inversely proportional to the activity?
I can't understand what you are trying to say. Please make it clear.
 
  • #4
148
2
The pH4 and pH5 solutions do not count as initial and final concentrations. Both solutions are buffered, and in both cases the pH does not change as the reaction proceeds. You need some other way of getting the order with respect to the activity of H+.

There are two different reaction setups, and for each of them the pH represents both the initial and the final concentration. You simply cannot use the equation you have quoted, but there is another very simple way to get the reaction order.
 
  • #5
utkarshakash
Gold Member
855
13
The pH4 and pH5 solutions do not count as initial and final concentrations. Both solutions are buffered, and in both cases the pH does not change as the reaction proceeds. You need some other way of getting the order with respect to the activity of H+.

There are two different reaction setups, and for each of them the pH represents both the initial and the final concentration. You simply cannot use the equation you have quoted, but there is another very simple way to get the reaction order.
But what is that simple method? I have already spent days trying to figure it out.
 

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