For the reaction $2 A \longrightarrow B+3 C$; if $-\frac{d[A]}{d t}=k_1[A]^2 ; \quad \frac{d[B]}{d t

Question

For the reaction $2 A \longrightarrow B+3 C$; if $-\frac{d[A]}{d t}=k_1[A]^2 ; \quad \frac{d[B]}{d t}=k_2[A]^2 ; \quad \frac{d[C]}{d t}=k_3[A]^2$ the correct relation between $k_1, k_2$ and $k_3$ is:

Professor PC Thomas Classes · Answer

(d) Rate of reaction
$
\begin{gathered}
=-\frac{1}{2} \frac{d[A]}{d t}=\frac{d[B]}{d t}=\frac{1}{3} \frac{d[C]}{d t} \
	ext { or } \frac{k_1[A]^2}{2}=\frac{k_2[A]^2}{1}=\frac{k_3[A]^3}{3} \
\frac{k_1}{2}=k_2=\frac{k_3}{3}
\end{gathered}
$