For a sequential reaction,
$\begin{array}{l} A \overset{k = 0.1386 \min^{- 1}}{\to} B \overset{k = 1.386 \times 10^{3} \min^{- 1}}{\to} C \\ \overset{\overset{}{k = 1.386 \times 10^{7} \min^{- 1}}}{\to} D \overset{k = 6.93 \times 10^{10} \min^{- 1}}{\to} E \end{array}$
starting with pure A initially (concentration= 1 M). What will be concentration of D after 10 min. assuming steady state is obtained ?

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$(1 - \frac{1}{4} - \frac{1}{8}) M$

$2 \times 10^{- 11} M$

$5 \times 10^{- 13} M$

$10^{- 5} M$

answer is A.

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Detailed Solution

$A \overset{K_{1}}{⟶} B \overset{K_{2}}{⟶} C \overset{K_{8}}{⟶} D \overset{K_{4}}{⟶} E$

At steady state, $k_{3} [C] = k_{4} [D]$ ;

$\begin{array}{l} k_{2} [B] = k_{3} [C]; k_{1} [A] = k_{2} [B] \\ \Rightarrow k_{1} [A] = k_{4} [D] \end{array}$

$\Rightarrow [D] = \frac{0.1386}{6.93 \times 10^{10}} \times e^{- 0.1386 \times 10} = 5 \times 10^{- 13}$ .

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