For the reaction A(g)⟶2B(g)d[B]dt=6×10−4Msec−1 when [A]=0.1Md[B]dt=2.4×10−3Msec−1 when [A]=0.4MFind the time taken (in seconds) for concentration of A to change from 0.6 M to 0.15 M.

Clearly order =1 w.r.t. A−d[A]dt=12d[B]dt=k[A]⇒ 3×10−4=k(0.1)k=3×10−3sec−1t1/2=0.6933×10−3=6933=321 sect3/4=462 sec

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For the reaction $A (g) ⟶ 2 B (g)$
$\begin{aligned} \frac{d [B]}{d t} = 6 \times 10^{- 4} Msec - 1 when [A] = 0.1 M \\ \frac{d [B]}{d t} = 2.4 \times 10^{- 3} Msec - 1 when [A] = 0.4 M \end{aligned}$
Find the time taken (in seconds) for concentration of A to change from 0.6 M to 0.15 M.

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answer is 462.

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Detailed Solution

Clearly order $= 1 w.r.t. A$

$\begin{array}{l} - \frac{d [A]}{dt} = \frac{1}{2} \frac{d [B]}{dt} = k [A] \\ \Rightarrow 3 \times 10^{- 4} = k (0 .1) \\ k = 3 \times 10^{- 3} \sec^{- 1} \\ t_{1 / 2} = \frac{0 .693}{3 \times 10^{- 3}} \\ = \frac{693}{3} = 321 \sec \\ t_{3 / 4} = 462 \sec \end{array}$