Chemistry - Chemical Kinetics Question with Solution | TestHub

[B]/[C]=K₁/K₂= 1.303= where A_t = concentration after 1 hr.

$-\frac{d[A]}{dt}=K_1[A]+K_2[A]$

$-\frac{d[A]}{dt}=(K_1+K_2)[A]$

$\left(K_1+K_2\right)=\frac{2.303}{t}{\log }_{10}\frac{[A_t{]}_{}+\left[B\right]+\left[C\right]}{[A_t{]}_{}}$

$2.303K_2=\frac{2.303}{1}{\log }_{10}\left[\frac{1+2.303\left[c\right]}{[A_t{]}_{}}\right]$

$K_2={\log }_{10}\left[\frac{1+2.303\left[c\right]}{[A_t{]}_{}}\right]$

$10^2=1+\frac{2.303\left[c\right]}{A}$

$\frac{2.303[c]}{[A_t]}=99$

$\frac{\left[c\right]}{\left[A_t]\right]}=\frac{99}{2.303}$

So,$\frac{\left[c\right]}{[A_t]}=42.987\approx 43$