Question

A compound A dissociate by two parallel first order paths at certain temperature
$\mathrm{A}(\mathrm{g})\stackrel{{\mathrm{k}}_1\left({\min }^{-1}\right)}{\to }2\mathrm{B}(\mathrm{g}){\mathrm{k}}_1=6.93\times 10^{-3}{\min }^{-1}$
$\mathrm{A}(\mathrm{g})\stackrel{{\mathrm{k}}_2\left({\min }^{-1}\right)}{\to }\mathrm{C}(\mathrm{g}){\mathrm{k}}_2=6.93\times 10^{-3}{\min }^{-1}$
If reaction started with pure 'A' with 1 mole of A in 1 litre closed container with initial pressure 2 atm. What is the pressure (in atm) developed in container after 50 minutes from start of experiment ?

• A. 1.25
• B. 0.75
• C. 1.50
• D. 2.50

Answer 1

Answer: (D)

$A(g)⟶2B(g);k_1=6.93\times 10^{-3}mi{n}^{-1}$
$A(g)⟶C(g);k_2=6.93\times 10^{-3}mi{n}^{-1}$

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

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

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.