Chemistry - Chemical Kinetics Question with Solution | TestHub

(1) 1ˢᵗ order:

$$\begin{array}{rl}& t=18sec,C_A=1-\frac{1}{3}=\frac{2}{3}\\ & t=36sec,C_A=\frac{2}{3}\left(1-\frac{1}{3}\right)=\frac{4}{9}\\ & t=54sec,C_A=\frac{4}{9}\left(1-\frac{1}{3}\right)=\frac{8}{27}\\ & \therefore 1-\frac{8}{27}=\frac{19}{27}\\ & \therefore t_{19/27}=54sec.\end{array}$$

(2) 1ˢᵗ order:

$$\begin{array}{rl}& t=16sec,C_A=1-\frac{1}{4}=\frac{3}{4}\\ & t=32sec,C_A=\frac{3}{4}\left(1-\frac{1}{4}\right)=\frac{9}{16}\\ & 1-\frac{9}{16}=\frac{7}{16}\\ & \therefore t_{7/16}=32sec.\end{array}$$

(3) Zero order reaction:

$$\begin{array}{rl}& \frac{1}{C_A}=\frac{1}{C_{A_0}}+kt\\ & \frac{3}{2}=\frac{1}{1}+k\times 4\Rightarrow k=\frac{1}{8}\\ & \frac{1}{C_A}=1+\frac{1}{8}\times t=\frac{8+t}{8}\\ & C_A=\frac{8}{(8+t)}\end{array}$$

When $t=56sec$, $C_A=\frac{8}{8+56}=\frac{1}{8}$.

$$\therefore 1-\frac{1}{8}=\frac{7}{8}\therefore t_{7/8}=56sec.$$

(4) Zero order:

$$\begin{array}{rl}& t=18sec,\therefore 1-\frac{1}{3}=\frac{2}{3}\\ & C_A=C_{A_0}-kt\\ & \frac{2}{3}=1-k\times 18\Rightarrow k=\frac{1}{54}\\ & \therefore C_A=1-\frac{1}{54}\times t\end{array}$$

When $t=30sec$:

$$\begin{array}{rl}& C_A=1-\frac{30}{54}=1-\frac{5}{9}=\frac{4}{9}\\ & 1-\frac{4}{9}=\frac{5}{9}\\ & \therefore 30sec=t_{5/9}\end{array}$$

(5) Zero order