\(\frac{1}{2} A \rightarrow 2 B\)
can be written either as
\(-2 \frac{d}{d t}[A]\) with respect to \(^{\prime} A^{\prime}\)
\(\quad \frac{1}{2} \frac{d}{d t}[B] \quad\) with respect to \(^{\prime} B^{\prime}\)
From the above, we have
\(-2 \frac{d}{d t}[A]=\frac{1}{2} \frac{d}{d t}[B]\)
\(\quad-\frac{d}{d t}[A]=\frac{1}{4} \frac{d}{d t}[B]\)