xA $$ \to $$ yB

$${1 \over x}\left\{ { - {{d\left[ A \right]} \over {dt}}} \right\} = {1 \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}$$

$$ \Rightarrow $$ $$ - {{d\left[ A \right]} \over {dt}} = {x \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}$$

Taking log both sides, we get

$$ \Rightarrow $$ $${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + {\log _{10}}\left( {{x \over y}} \right)$$

Given that,

$${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010$$

So, by comparing both of them we get

$${\log _{10}}\left( {{x \over y}} \right)$$ = 0.3010 = $${\log _{10}}\left( 2 \right)$$

$$ \therefore $$ $${{x \over y}}$$ = 2

$$ \Rightarrow $$ x = 2y

If x = 2 then y = 1

The reaction is of type 2A $$ \to $$ B

So possible reaction is

2C₂H₄ $$ \to $$ C₄H₈