Chemistry - Chemical Kinetics Question with Solution | TestHub

$\begin{array}{ccccc}\text{A}\left(\text{g}\right)& ⟶& 2\text{B}\left(\text{g}\right)& \text{;}& {\text{k}}_1=6\text{.}93\times 10^{-3}{\text{ min}}^{-1}\\ \text{A}\left(\text{g}\right)& ⟶& \text{C}\left(\text{g}\right)& \text{;}& {\text{k}}_2=6\text{.}93\times 10^{-3}{\text{ min}}^{-1}\end{array}$

$\text{k = }\left({\text{k}}_1+{\text{k}}_2\right)$overall velocity constant = 6.93 x 2 x 10^-3min^-1

${\text{t}}_{1/2}=\frac{0\text{.}693\times 1000}{6\text{.}93\times 2}=50\text{ min}$

i.e., after = 50 min

$P_A$= 1 atm.

Since k₁= k₂

$P_B$= 0.5 x 2 = 1 atm (due to stoichio metric coefficient)

$P_C$= 0.5 x 1 = 0.5 atm.

Total pressure = 2.5 atm.