1. Tardigrade
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  4. A compound A dissociate by two parallel first order paths at certain temperature mathrmA( mathrmg) stackrel mathrmk1( mathrm~min-1)arrow 2 mathrm~B( mathrm~g) mathrmk1=6.93 × 10-3 mathrm~min-1 mathrmA( mathrmg) stackrel mathrmk2( mathrm~min-1)arrow mathrmC( mathrmg) mathrmk2=6.93 × 10-3 mathrm~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 ?

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Q. A compound A dissociate by two parallel first order paths at certain temperature
$\mathrm{A}(\mathrm{g}) \stackrel{\mathrm{k}_1\left(\mathrm{~min}^{-1}\right)}{\rightarrow} 2 \mathrm{~B}(\mathrm{~g}) \mathrm{k}_1=6.93 \times 10^{-3} \mathrm{~min}^{-1}$
$\mathrm{A}(\mathrm{g}) \stackrel{\mathrm{k}_2\left(\mathrm{~min}^{-1}\right)}{\rightarrow} \mathrm{C}(\mathrm{g}) \mathrm{k}_2=6.93 \times 10^{-3} \mathrm{~min}^{-1}$
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 ?

NTA AbhyasNTA Abhyas 2020Chemical Kinetics

Solution:

$A ( g ) \longrightarrow 2 B ( g ) ; k _{1}=6.93 \times 10^{-3} min ^{-1}$
$A ( g ) \longrightarrow C ( g ) \quad ; k _{2}=6.93 \times 10^{-3} min ^{-1}$

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

$\text{t}_{1 / 2} = \frac{0 \text{.} 6 9 3 \times 1 0 0 0}{6 \text{.} 9 3 \times 2} = 5 0 \text{ min}$

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.