# Can I express the rate of a reaction as percent per second (%/s)?

Rate is calculated as change in concentration over change in time, so usually $$\pu{M/s}$$ $$(\pu{mol L-1 s-1}).$$

Say, I'm calculating it for $$20\%$$ solution of ammonium hydroxide, so my final rate is in $$\pu{\%/s}.$$ Is this a proper format, or should I calculate for $$\pu{mol L-1}?$$

According to IUPAC Recommendations for chemical kinetics , rates $$\nu$$ are defined for reactant $$\ce{B}$$ and product $$\ce{Y}$$ as a time $$(t)$$ derivative of the amount $$n$$ (in general) or per unit volume $$V$$ (for the closed system):

$$\begin{array}{lll} \hline \text{System} & \text{Consumption} & \text{Formation} \\ \hline \text{Open} & \displaystyle\nu(n_\ce{B}) = -\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(n_\ce{Y}) = \frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\ \hline \text{Closed} & \displaystyle\nu(c_\ce{B}) = -\frac{1}{V}\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(c_\ce{Y}) = \frac{1}{V}\frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\ & \displaystyle\nu(c_\ce{B})|_{V = \mathrm{const}} = -\frac{\mathrm d[\ce{B}]}{\mathrm dt} & \displaystyle\nu(c_\ce{Y})|_{V = \mathrm{const}} = \frac{\mathrm d[\ce{Y}]}{\mathrm dt} \\ \hline \end{array}$$

Rates of consumption and formation that are measured in $$\pu{mol L-1 s-1}$$ are defined for the closed system at constant volume, and molar concentration arises from its definition. Whether you can or cannot substitute molar concentration $$c_i$$ with fractions can be answered by looking at the relationship between the two.

For example, molar concentration $$c_i$$ expressed in terms of mass fraction $$\omega_i$$ via the density of solution $$\rho$$ and the molar mass $$M_i$$ of the $$i$$th component:

$$c_i = \frac{\omega_i \rho}{M_i}\tag{1}$$

Similarly, molar concentration $$c_i$$ expressed in terms of mole fraction $$x_i$$ via the density of solution $$\rho$$ and the average molar mass $$\bar{M}$$ of the solution:

$$c_i = \frac{x_i\rho}{\bar{M}}\tag{2}$$

As you can see, there is a problem: as the reaction progresses, not only the amounts (or fractions) would change, but also the density and the molar mass of the solution may and will change in most cases, which leads to inapplicability of percent fractions as time derivatives. All in all, in chemistry it's just easier to manipulate with the quantities based on amount of substance.

### References

1. Laidler, K. J. A Glossary of Terms Used in Chemical Kinetics, Including Reaction Dynamics (IUPAC Recommendations 1996). Pure and Applied Chemistry 2009, 68 (1), 149–192. DOI: 10.1351/pac199668010149. (Free Access)
• I don't fully understand. I get that % values change over the reaction (first order btw) , but if I have a rate that's in %/s, isn't that expressing the change in concentration in and of itself? – Raymo111 Jan 1 at 1:12

To summarize, and to answer the question, it is correct to say that the rate of a chemical reaction can be expressed in % per second if and only if the reaction is first order. This percentage is the first order rate constant $$k$$.

• Can I get a source? That % is first order? – Raymo111 Dec 31 '19 at 20:43

For a reaction like this: $$\ce{aA +bB⟶cC +dD}$$, the rate is given by

Rate=$$−a\frac{d[A]}{dt}=−b\frac{d[B]}{dt}=c\frac{d[C]}{dt}=d\frac{d[D]}{dt}$$ where the unit is usually $$\ce{mol L^{-1}{s^{-1}}}$$.

The rate of a reaction is different for different orders and has different units of the same. It's unit usually has the terms $$\ce{mol, L, and s}$$.

• For a zero order reaction,

Rate= k, unit= $$\ce{mol L^{-1}{s^{-1}}}$$

• For a first order reaction,

Rate= $$\ce{k[A]}$$, unit=$$\ce{s^{-1}}$$

• For a second order reaction,

Rate=$$\ce{k[A]^2}$$, unit=$$\ce{L mol^{-1}{s^{-1}}}$$