A gaseous compound A reacts by three independent first order reactions (as shown in figure) with the rate constant $2 \times 10^{- 3}, 3 \times 10^{- 3}$ and $1.93 \times 10^{- 3}$ sec^-1for products B, C and D respectively. If initially pure A was taken in a closed container with P=2atm, and the partial pressure of B (in atm) , among the products, after 100 sec from the start of experiment is found to be $[\frac{25 y}{693}]$ then find out the value of y?

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answer is 8.

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Detailed Solution

Overall rate constant $(K) = K_{1} + K_{2} + K_{3}$
$= 6.93 \times 10^{- 3}$
After one $t_{1 / 2}$
$\begin{array}{l} P_{B} + P_{C} + P_{D} = 1 atm \\ \frac{P_{B}}{P_{B} + P_{C} + P_{D}} = \frac{K_{1}}{K_{1} + K_{2} + K_{3}} = \frac{200}{693} \\ P_{B} = \frac{200}{693} \times (P_{B} + P_{C} + P_{D}) \\ P_{B} = \frac{200}{693} \times 1 \\ P_{B} = \frac{200}{693} atm \\ So, 2 5 y = 200 \\ ∴ y = 8 \end{array}$