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The acid catalysed hydrolysis of an compound $\ce{A}$ at $\pu{303K}$: $$\ce {A ->[$k$] B}$$ has a half-life of $\pu{100 min}$ when carried out in a buffer solution of $\mathrm{pH=5}$ and $\pu{10 min}$ when carried out at $\mathrm{pH=4}$. Both the times the half-life are independent of the initial concentration of $\ce{A}$. Find its Rate Law.


I am planning something like this:

Let $$\text{Rate} = k [\ce{A}][\ce{H+}]^x\tag1$$ Now let: $$k'=k[\ce{H+}]^x\tag{constant}$$

And then: $$k' t_\frac 12 = \ln2$$


Is it conceptually correct?


Note:

  1. I have found tons of solutions over the internet all with different answers.

  2. The answer given in my book is x=1 (x is same as is used in eq(1)) and I found the same myself but some sites give answer of the same question as x=2.

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  • $\begingroup$ Umm, you have a rate law in first eq. so... $\endgroup$ – Mithoron Sep 18 '20 at 19:12
  • $\begingroup$ could you share the different answers given? just a link to contradictory answers would be enough.. $\endgroup$ – Safdar Sep 19 '20 at 7:09
  • $\begingroup$ @Safdar Check the note section in my question. $\endgroup$ – Tony Stark Sep 19 '20 at 8:12
  • $\begingroup$ @ Tony Stark. As the rate is independent of the concentration of A, it means that the rate is only dependent on the concentration of $\ce{H+}$ and first order. $\endgroup$ – Maurice Sep 19 '20 at 9:33
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We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.

$$\text{Rate}=[A]^y[\ce{H+}]^x$$


The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.

Therefore as you've found, $$\text{Rate}=[A][\ce{H+}]^x$$


Now, onto finding the order with respect to $\ce{H+}$.

From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:

$$t_{1/2}\, \alpha\, [A]^{1-n}$$

where $n$ is the order with respect to A in a reaction.

Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.

$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$

Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.

Therefore, what you've done is conceptually sound and is the correct method.

$$k'=k[\ce{H+}]^x = \text{constant}$$

\begin{align} \frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\ \implies \frac{100}{10}&=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\ \implies x &=1 \end{align}

Therefore, the rate would be:

$$\text{Rate}=k[A][\ce{H+}]$$

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  • $\begingroup$ There is a similar question -- Rate Law for inversion of cane sugar -- chemistry.stackexchange.com/questions/104543/… . Applying the same method does not give me the answer. Please look into it. $\endgroup$ – Tony Stark Sep 19 '20 at 10:57
  • $\begingroup$ @TonyStark It is the exact same question, poorly formatted.. except A and B are given (conditons seem similar as well). Did you read the comments? this is same.. Also, I seem to get the correct answer there as well. $\endgroup$ – Safdar Sep 19 '20 at 11:00
  • $\begingroup$ Check this : toppr.com/ask/question/… . This is the same question as chemistry.stackexchange.com/questions/104543/… but well formatted. $\endgroup$ – Tony Stark Sep 19 '20 at 11:48
  • $\begingroup$ @TonyStark What part of my above statement did you not understand? "Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture," the other answers are wrong.. the internet isn't always right.. one group adds a solution, others plagiarise. nobody cares if they are correct since they aren't community vetted.. $\endgroup$ – Safdar Sep 19 '20 at 11:50
  • $\begingroup$ Actually I was looking for this statement clearly stated by some experienced person - " the other answers are wrong.. the internet isn't always right.. one group adds a solution, others plagiarise." $\endgroup$ – Tony Stark Sep 19 '20 at 12:29

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