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?