# Reaction rate ambiguity

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?

• Logically, they must proceed in sequence: first nascent oxygen is released, then O2 forms from the atomic O. – DrMoishe Pippik Feb 26 '20 at 19:10
• 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 $\ce{H2O}$ – Adnan AL-Amleh Feb 27 '20 at 1:29
• $k$ constant of disappearing $\ce{H2O2}$= $k$ constant of appearing$\ce{H2O}$=2($k$ constant of appearing$\ce{O2})$ – Adnan AL-Amleh Feb 27 '20 at 1:48

## 1 Answer

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..