The reaction A(g) arrow 2B(g)+C(g) follows first order kinetics. The reaction is started with pure ' A ' in a rigid closed vessel maintained at constant temperature. After 10s , a pin hole is developed in the vessel. If the molar ratio of gases ' A ' and ' B ' coming out initially is 1:2 the rate constant of reaction(in s- 1 ) is X . Calculate 100X ? (Molar masses of A,B and C are 16,4 and 8g/mol respectively, ln 2=0.7 , ln 3=1.1 )

Question

The reaction A(g) arrow 2B(g)+C(g) follows first order kinetics. The reaction is started with pure ' A ' in a rigid closed vessel maintained at constant temperature. After 10s , a pin hole is developed in the vessel. If the molar ratio of gases ' A ' and ' B ' coming out initially is 1:2 the rate constant of reaction(in s- 1 ) is X . Calculate 100X ? (Molar masses of A,B and C are 16,4 and 8g/mol respectively, ln 2=0.7 , ln 3=1.1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/75787e7933199aaee5443aba021df511-.png" /> Now, (rA/rB)=(PA/PB)√(MB/MA)⇒ (1/2)=(a - x/2 x)⋅ √(4/16) ⇒ x=(a/3) Now, K=(1/t)⋅ ln(a/a - x)=(1/10 sec)⋅ ln(a/(2 a)3) =0.04sec- 1100X=4

Now, rB​rA​​=PB​PA​​MA​MB​​​⇒21​=2xa−x​⋅164​​ ⇒x=3a​ Now, K=t1​⋅lna−xa​=10sec1​⋅ln32a​a​ =0.04sec−1100X=4

Now, $\frac{r _{A}}{r _{B}} = \frac{P _{A}}{P _{B}} \frac{M _{B}}{M _{A}} \Rightarrow \frac{1}{2} = \frac{a - x}{2 x} \cdot \frac{4}{16}$
$\Rightarrow x = \frac{a}{3}$
Now, $K = \frac{1}{t} \cdot l n \frac{a}{a - x} = \frac{1}{10 sec} \cdot l n \frac{a}{\frac{2 a}{3}}$
$= 0.04 se c^{- 1} 100 X = 4$