Question

For the reaction $2A⟶B+2C$; if $-\frac{d\left[A\right]}{dt}=k_1[A{]}^2$; $-\frac{d\left[B\right]}{dt}=k_2[A{]}^2$;$-\frac{d\left[C\right]}{dt}=k_3[A{]}^2$ the correct relation between $k_1$, $k_2$ and $k_3$ is :

• A. $k_1=2k_2=k_3$
• B. $2k_1=k_2=3k_2$
• C. $4k_1=k_2=3k_2$
• D. $\frac{k_1}{2}=k_2=\frac{k_3}{3}$

Answer 1

The correct option is A $k_1=2k_2=k_3$

$2A\to B+2C$

Law of chemical equilibrium :

$\Rightarrow \frac{-1}{2}\frac{dA}{dt}=\frac{dB}{dt}=\frac{1}{2}\frac{dC}{dt}$

Given $\frac{-dA}{dt}=K_1{\left[A\right]}^2\frac{dC}{dt}=K_3{\left[A\right]}^2$

$\frac{dB}{dt}=K_2{\left[A\right]}^2$

$\frac{1}{2}{K}_1{\left[A\right]}^2=K_2{\left[A\right]}^2=\frac{1}{2}{K}_3{\left[A\right]}^2$

$K_1=2K_2=K_3$ is the correct answer.