Kinetics question regarding Rate Law

Chemistry Asked by Tony Stark on December 5, 2020

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 $$text{Rate} = k [ce{A}][ce{H+}]^xtag1$$
Now let:

And then:
$$k’ t_frac 12 = ln2$$

Is it conceptually correct?


  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.

One Answer

We can do this question in two steps, finding the rate law with respect to both A and $ce{H+}$ individually.


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:


Correct answer by Safdar on December 5, 2020

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