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Q.

A reaction : $A + B ⟶ C$ follows the rate law as shown $R = 10^{- 2} [A] [B] M / \sec$ . Calculate time required for concentration of [B] to reduce to one fourth of its original value if initially A and B are taken at 3 M and 3.2 M respectively.

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a

$(\ln \frac{5}{4}) \times \frac{1}{2 \times 10^{- 3}} \sec$

b

$(\ln 5) \times \frac{1}{2} \sec$

c

$\ln 4 \times \frac{1}{2 \times 10^{- 3}} \sec$

d

$\ln 5 \times \frac{1}{2 \times 10^{- 2}} \sec$

answer is A.

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Detailed Solution

$A + B \to C \begin{array}{cccc} t = 0 & 3 & 3.2 \\ t = t & 3 - x & 3.2 - x & x \\ = 0.6 & = 0.8 & = 2.4 \end{array} \begin{array}{l} t = \frac{- 1}{k (b - a)} \ln [(\frac{b}{b - x}) \times \frac{(a - x)}{a}] \\ = \frac{- 1}{10^{- 2} \times 0.2} \ln (\frac{3.2}{0.8} \times \frac{0.6}{3}) = \frac{1}{2 \times 10^{- 3}} \ln (\frac{5}{4}) \end{array}$

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