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In my book it was written that

The differential rate law applies to only one single species, so unless we have equimolar amounts of each reactant or make an approximation using an excess of one, we cannot apply the differential rate law.

What does this statement mean? Can anyone please explain?

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  • $\begingroup$ Do not confuse the applicability of the differential rate law and ability (easily or at all) to solve respective differential equations. There are many cases in chemistry and physics where the law and it's equation fit well, but it is not possible to solve it analytically. $\endgroup$
    – Poutnik
    Commented Nov 12, 2022 at 7:50
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    $\begingroup$ Which book is that? $\endgroup$ Commented Nov 12, 2022 at 10:04

2 Answers 2

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For a reaction at constant volume of the form:

$$\ce{A +B->C}$$

If the rate of this reaction follows a power law relationship, then:

$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}=f(C_\pu{A},C_\pu{B})$$

Since both $C_\pu{A}$ and $C_\pu{B}$ are functions of time $t$, you won't be able to integrate -$\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}$.

However, if we use an excess of $\pu{B}$, then the change in $C_\pu{B}$ will be negligible compared to the change in $C_\pu{A}$ after the reaction is finalized, and the following approximation becomes valid:

$$C_\pu{B}^{\beta}≈constant$$

This constant can be grouped up with $k$, which is another constant:

$$k'=k\;C_\pu{B}^{\beta}$$

So we would get:

$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}\pu{t}}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}≈k'C_\pu{A}^{\alpha}≈f(C_\pu{A})$$

So the rate would only depend on $C_\pu{A}$, and you can integrate it:

$$\frac{-\mathrm{d}C_\pu{A}}{C_\pu{A}^{\alpha}}=k'\;\mathrm{d}\pu{t}$$

Conversely, when no excess of either $\pu{A}$ or $\pu{B}$ are used, but equimolar initial amounts are, we can integrate regardless.

Assuming $\pu{A}$ is the limiting reagent, and constant volume, pressure, and temperature conditions, the concentration of $\pu{A}$ and $\pu{B}$ in terms of conversion are defined as:

$$C_\pu{A}=C_{\pu{Ao}}\;(1-X)$$

$$C_\pu{B}=C_{\pu{Ao}}\left(\frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}-X\right)$$

Since we have equimolar amounts, it follows that:

$$C_{\pu{Ao}}=C_{\pu{Bo}}\implies \frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}=1\implies C_\pu{B}=C_{\pu{Ao}}\;(1-X)\implies C_\pu{A}=C_\pu{B}$$

So the rate would be:

$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}=k\;C_\pu{A}^{\alpha+\beta}$$

And you could integrate the resulting expression if orders are known or reaction is elemental:

$$\frac{-\mathrm{d}C_\pu{A}}{C_\pu{A}^{\alpha+\beta}}=k\;\mathrm{d}t$$

In conclusion, if either of the two conditions are used, and you know the values of $\alpha$ and $\beta$, you can integrate these differential equations to obtain a function of $C_\pu{A}$ vs $t$.

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    $\begingroup$ for further reference: upright vs italic $\endgroup$
    – Poutnik
    Commented Nov 12, 2022 at 7:40
  • $\begingroup$ There is possible the parameterization , expressing both concentrations of A and B as functions of the single parameter. Still, more complicated solving than the simple case of pseudo first order. $\endgroup$
    – Poutnik
    Commented Nov 12, 2022 at 7:57
  • $\begingroup$ That is true. I had initially expressed both concentrations as functions of conversion $X$, but I thought that could confuse OP, so I just left it as is. $\endgroup$
    – Sam202
    Commented Nov 12, 2022 at 7:59
  • $\begingroup$ I see, I can agree with that. $\endgroup$
    – Poutnik
    Commented Nov 12, 2022 at 8:00
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If the reaction has the second order rate law $dA/dt=-kAB$ with initial concentrations $A_0,B_0$ then by stoichiometry $A-A_0=B-B_0$ and then

$$\frac{dA}{A(B_0-A_0+A)}=-kdt$$

which can be integrated after expanding as partial fractions as it is a function only of one variable, $A$.

Alternatively using the reaction variable method let $x$ be the amount reacted at time $t$ then $\displaystyle \frac{dx}{dt}=k(a-x)(b-x)$, which can be integrated when $a,b$ are initial amounts, to give an equivalent result to that above.

These solutions would seem to indicate that the statement in your book is not quite correct.

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