I have studied that rate of disappearance and rate of appearance is the change in concentration of reactants and products (respectively) with respect to time. Thus,

$$\text{ROD} = -\frac{\Delta c_R}{\Delta T}$$


$$\text{ROA} = \frac{\Delta c_P}{\Delta T}$$


$$\text{rate of reaction} = % % \frac{\text{ROD}} {\text{stoichiometric coefficient}} % % = \frac{\text{ROA}} {\text{stoichiometric coefficient} }$$

Then writing rate of reaction using rate law was taught.

And according to rate law rate of reaction can be expressed as:

$$\text{rate of reaction} = k[\ce{A}]^x[\ce{B}]^y$$

here $x + y = \text{overall order of reaction}$, with $\ce{A}$ and $\ce{B}$, the reactants.

Here is my confusion: Now they wrote rate of disappearance and appearance using rate law. How using rate law they wrote expression for ROA and ROD.

For a reaction: $\ce{aA -> bB}$ (elementary reaction), it was stated that: $$\text{ROD}_\text{A} = k_d [\ce{A}]^a$$

So how ?

Using rate law we can write rate of reaction.. (OK) but how can we write ROA and ROD ??!!

Hope you understood what I meant to say.

  • 1
    $\begingroup$ How did you come to the last equation? $\endgroup$
    – Poutnik
    Commented Jan 11, 2022 at 19:59
  • $\begingroup$ @PoutnikCan we relate rate law with ROD of reactant A as this formula : Rate of reaction = $-\frac{1}{Stoichiometric\space Coefficient\space of \space reactant A}\cdot\frac{\Delta[A]}{\Delta{t}} = k[A]^x[B]^y$ $\endgroup$ Commented Jan 13, 2022 at 8:49
  • $\begingroup$ @Poutnik I have made amendment to the reaction. Sorry for caused ambiguity. Now can you please explain how we can write ROD and ROA using RATE LAW ? $\endgroup$
    – Spencer
    Commented Jan 13, 2022 at 11:24
  • $\begingroup$ IF ROD(A)=ROR . SC and if ROR= k . [A]^x . [B]^y then ....... $\endgroup$
    – Poutnik
    Commented Jan 13, 2022 at 12:15
  • 1
    $\begingroup$ If there is text in the equation, set this as text (cf. the revision). A $ROD = x $ means $R \cdot O \cdot D$ (a multiplication) yielding $x$ as a product; using $\text{ROD}$ = x$ however states $\text{ROD} = x$, i.e. the variable ROD equates to $x$. And for the use of \mhchem syntax: have a look here. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    Commented Jan 13, 2022 at 20:25

1 Answer 1


You overthink it. It is just matter of eventual multiplication or division with respective stoichiometric coefficients.

If there is an elementary reaction

$$\ce{a A -> b B}$$

then for respective reaction, appearance and disappearance rates:

$$\begin{align} R_\mathrm{r} &= k_\mathrm{r}[A]^a\\ R_\mathrm{a,B} = \frac {\mathrm{d[B]}}{\mathrm{d[t]}}=bR_\mathrm{r} &= bk_\mathrm{r}[A]^a = k_\mathrm{a,B}[A]^a\\ R_\mathrm{d,A} = -\frac {\mathrm{d[A]}}{\mathrm{d[t]}} = aR_\mathrm{r} &= ak_\mathrm{r}[A]^a = k_\mathrm{d,A}[A]^a \end{align}$$

.. and backwards:

$$\begin{align} R_\mathrm{r} = \frac{R_\mathrm{a,B} }{ b} &= \frac 1b \frac {\mathrm{d[B]}}{ \mathrm{d[t]}}\\ R_\mathrm{r} = \frac{R_\mathrm{d,A} }{ a} &= -\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}} \end{align}$$

  • $\begingroup$ thanks .According to your solution can we conclude the following expression :$-\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}}= k_\mathrm{r}[A]^a$ $\endgroup$ Commented Jan 15, 2022 at 6:42
  • 1
    $\begingroup$ @AdnanAL-Amleh If X=Y and X=Z then obviously Y=Z. :-) $\endgroup$
    – Poutnik
    Commented Jan 15, 2022 at 7:09
  • $\begingroup$ Thanks . Can we teach the following expression for 11th grade . Is it correct?: $\text{average}R_\mathrm{r} = \frac{\text{average}R_\mathrm{d,A} }{ a} = -\frac 1a \frac {\mathrm{\Delta[A]}}{\mathrm{\Delta[t]}}= k_\mathrm{r}[A]^a$ $\endgroup$ Commented Jan 15, 2022 at 7:42
  • 1
    $\begingroup$ @AdnanAL-Amleh Involving deltas and averages, you have to come with average for $\ce{[A]}^a$ too. If the delta is about 10% and a=3, it makes about 30% difference. It is a task for integration. $\endgroup$
    – Poutnik
    Commented Jan 15, 2022 at 8:19
  • $\begingroup$ thanks . Is it useful to teach the following two expressions separately without reconciliation or connection between them ,as the first expression about instantaneous rate and rate law but the second about average rate 1)$R_\mathrm{r} = -\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}}= \frac 1b \frac {\mathrm{d[B]}}{ \mathrm{d[t]}}= k_\mathrm{r}[A]^a $\ 2)$R_\mathrm{r} = -\frac 1a \frac {\mathrm{\Delta{[A]}}}{\mathrm{\Delta{t}}}= \frac 1b \frac {\Delta\mathrm{[B]}}{\Delta\mathrm{t}}$ $\endgroup$ Commented Jan 15, 2022 at 11:41

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