# 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 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 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 at 1:48

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