1. For the reaction

$xA \to yB$

the rates of disappearance and appearance are related by stoichiometry:

$-\frac{1}{x}\frac{d[A]}{dt}=\frac{1}{y}\frac{d[B]}{dt}$

So,

$-\frac{d[A]}{dt}=\frac{x}{y}\frac{d[B]}{dt}$

2. The question gives:

${\log_{10}}\left[-\frac{d[A]}{dt}\right]={\log_{10}}\left[\frac{d[B]}{dt}\right]+0.3010$

Using the log property,

$\log_{10}\left[-\frac{d[A]}{dt}\right]-\log_{10}\left[\frac{d[B]}{dt}\right]=0.3010$

$\log_{10}\left(\frac{-d[A]/dt}{d[B]/dt}\right)=0.3010$

Since

$0.3010=\log_{10}2$

we get

$\frac{-d[A]/dt}{d[B]/dt}=2$

Thus,

$-\frac{d[A]}{dt}=2\frac{d[B]}{dt}$

Comparing with

$-\frac{d[A]}{dt}=\frac{x}{y}\frac{d[B]}{dt}$

we obtain

$\frac{x}{y}=2$

So the stoichiometric ratio must be

$x:y=2:1$

3. Now check the options.

Option A: n-Butane and Iso-butane

This is isomerization:

$\text{n-Butane} \to \text{Iso-butane}$

Stoichiometry is $1:1$, not $2:1$.

So, A is incorrect.

Option B: $C_2H_4$ and $C_4H_8$

Dimerization:

$2C_2H_4 \to C_4H_8$

Here $x:y=2:1$, which matches the condition.

So, B is correct.

Option C: $C_2H_4$ and $C_6H_6$

There is no simple reaction of the form

$2C_2H_4 \to C_6H_6$

and stoichiometry does not match the required $2:1$ in a valid direct form.

So, C is incorrect.

Option D: $N_2O_4$ and $NO_2$

Dissociation is:

$N_2O_4 \to 2NO_2$

Here $x:y=1:2$, not $2:1$.

So, D is incorrect.

4. Therefore, the correct option is:

$\boxed{B}$