Let’s suppose the following reaction:

$$\ce{ 2 H2O2 —> 2 H2O + O2}\tag{1}$$ with reaction rate $v$

Now let’s suppose the same equation with different balance:

$$\ce{ H2O2 —> H2O + 1/2O2 }\tag{2}$$

I assume that the reaction rate must be the same ($v$), right? It should not change with stoichometry.

However, depending on the balancing, I get different rates for each substance as shown in the equations below, different $k$ constant and so on.

$$\begin{align}v_1 &= -\frac12 \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac12 \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\\[1em] v_2 &= \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = 2\cdot \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\end{align}$$

Any help?

  • $\begingroup$ Logically, they must proceed in sequence: first nascent oxygen is released, then O2 forms from the atomic O. $\endgroup$ Feb 26, 2020 at 19:10
  • $\begingroup$ According the picture , the first relation represent the rate of appearing $\ce{O2}$ , the second relation represent the rate of disappearing $\ce{H2O2}$ or the rate of appearing $\ce{H2O}$ $\endgroup$ Feb 27, 2020 at 1:29
  • $\begingroup$ $k$ constant of disappearing $\ce{H2O2}$= $k$ constant of appearing$\ce{H2O}$=2($k$ constant of appearing$\ce{O2})$ $\endgroup$ Feb 27, 2020 at 1:48

1 Answer 1


I assume that the reaction rate must be the same (v), right? It should not change with stoichometry.

No, the reaction rates changes when you double all coefficients. Nothing real changes though (i.e. the rate of disappearing $\ce{H2O2}$ etc.).

You have a similar case with equilibrium constants: when you double or halve all the coefficients, the new equilibrium constant will be the square or the square root of the original one. The molar reaction enthalpy will also change when you change coefficients. It is a choice, and it still correctly describes the reaction.

Any help?

There is no need for help. If you use either chemical equation format to calculate a measurable quantity (changes in concentration, equilibrium concentrations, enthalpy change for a specific amount that reacts, for the three topics respectively), you will get the same result. It is almost like the choice of describing a volume using liters or milliliters - it does not change the result even though it might look different to someone not used to the milli prefix..


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.