According to my Chemistry textbook (Chemistry by Raymond Chang 10th edition, pg 570), for a first-order reaction of the type $$\ce{A->product}$$ the rate is $$rate=-\frac{[\Delta A]}{\Delta t}$$ From the rate law we also know that $$rate=k[A]..... (13.2)$$ Combining the two equations for the rate we write $$-\frac{[\Delta A]}{\Delta t}=k[A]$$ And finally, using calculus, $$\ln\frac{[A]_t}{[A]_0}=-kt...... (13.3)$$ All of this feels okay to me. However, later on, the book uses the same equation (13.3) to solve a first-order reaction of the type $$\ce{2A->product}$$ and I don't understand why. Because the rate of reaction would now be $$rate=-\frac12\frac{[\Delta A]}{\Delta t}$$ and equating this relation with (13.2) gives $$-\frac12\frac{[\Delta A]}{\Delta t}=k[A]$$ and finally $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ I know I'm probably misunderstanding something. Please can someone explain this to me. I've been trying to figure it out for days now.

[Edit] The problem that I'm referring to is a Practice Exercise, the book only gives the problem and the answer, not the solution. However, based on the answer is obvious that equation 13.3 was used to solve it.

Problem: The reaction $\ce{2A->B}$ is first order in A with a rate constant of $\pu{2.8 \times 10^{-2}s^{-1}}$ at $\pu{80 ^{\circ}C}$. How long (in seconds) will it take for A to decrease from $\pu{0.88M}$ to $\pu{0.14M}$?

Answer: $\pu{66s}$

If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$.

  • $\begingroup$ It could be that the 2k here be changed to a, which one could designate as the rate constant. As far as I can tell, it just seems as a change of variable. $\endgroup$ – user79161 May 4 at 19:57
  • $\begingroup$ Hi! I added (as an edit) the practice excercise where the book uses equation 13.3 to solve a reaction of the type 2A --> B $\endgroup$ – Manuel1299 May 4 at 20:09
  • $\begingroup$ @user79161 But in the case of the practice excercise above, how would you know that the rate constant that was given is 2k or a? $\endgroup$ – Manuel1299 May 4 at 20:33
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    $\begingroup$ The genuine of your question based on the idea which confused a lot of chemistry teacher may be this link help you: chemistry.stackexchange.com/questions/38167/… $\endgroup$ – Adnan AL-Amleh May 4 at 23:36
  • $\begingroup$ Thanks @AdnanAL-Amleh !! but I unable to see how that thread answers my doubt $\endgroup$ – Manuel1299 May 5 at 1:38

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients.

Definition of elementary reaction: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state (more details on elementary reactions).

Rate of reactions are determined by experiments. We never write the rate law based on the balanced equation. Compare this with equilibrium constant where coefficients are important. Even $\ce{2 A -> B}$ may be proceeding in several elementary steps. The observed rate may be the effect of several combined elementary reactions.

In those kinetic problems like yours, the writer will tell you the order of the reaction. Don't proceed to write the rate law on the basis of balanced equation.

Reaction rate and writing the rate law are two different things. For a reaction

$$\ce{a A + b B → p P + q Q}$$

Reaction is written as

$$r = -\frac{1}{a}\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -\frac{1}{b}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p}\frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q}\frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$

but we cannot write the rate law i.e. you do not know what would be the order of the reaction in general.

The relation above only shows how the rate of disappearance of reactants and rate of appearance of products are related. It should not be used anywhere else. Your book answer is correct.

  • $\begingroup$ If given a problem with no other restrictions, then an elementary reaction should be assumed. However for the particular problem under consideration it was explicitly stated that the reaction was first order. $\endgroup$ – MaxW May 5 at 0:32
  • $\begingroup$ Exactly! @MaxW If it is first ordered and the stoichiometric coefficient of A is 2, aren't I supposed to include the number 2 in there? Since, according to the IUPAC definition, the reaction rate depends on the stoichiometric coefficients goldbook.iupac.org/html/R/R05156.html $\endgroup$ – Manuel1299 May 5 at 1:36
  • $\begingroup$ @Manuel1299 - First order $r = k\ce{[A]}$ is first order. $r = k\ce{[A]^2}$ would be a second order reaction. $\endgroup$ – MaxW May 5 at 2:13
  • $\begingroup$ The goal is not to write the rate law. The rate law is already given since it is stated in the problem that is a first-ordered reaction. Hence the rate is $r=k[A]$. Also, based on IUPAC definition of rate, $r=-\frac12\frac{[\Delta A]}{\Delta t}$. Aren't these two expressions equal to each other? $\endgroup$ – Manuel1299 May 5 at 2:52
  • $\begingroup$ You are missing the punchline of the answer above. The IUPAC definition of the rate just shows the inter-relationship of rates among reactants and products. That is all. It is not the definition of rate. Rate is just [conc]/time- no coefficients. IUPAC relation only helps when you know the rate of a reactant but now you wish to know rate of appearance of a product. The expression r=k[A] is not equivalent to the relationship shown by IUPAC. $\endgroup$ – M. Farooq May 5 at 2:59

If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$.

You cannot substitute $k=2.8\cdot{10^{-8}}$ in the above equation because of the value of $k$ in the last equation of your question equal half the value of k in equation(3.2).

Why the values of k in the two equations are different?

Because they are different in defining the rate:

  • The equation(3.2) defines the rate as the rate of disappearance of the reactant A: $$r_1 = -\frac{\Delta[A]}{\Delta t}= k_1[A]$$
  • The last equation in the question defines the rate as the rate of appearance of the product P: $$r_2 = \frac{\Delta[P]}{\Delta t}= k_2[A]$$
  • But , the rate of appearance of the product $P=\frac{1}{2}$disappearance of the reactant$A$ $$\frac{\Delta[P]}{\Delta t}= \frac{1}{2}\frac{\Delta[A]}{\Delta t}$$ $$ k_2[A] = \frac{1}{2} k_1[A]$$ So : $$ k_2= \frac{1}{2} k_1$$

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