The reaction is initiated by taking 1 M concentration of A and B each. If the rate constant (k) is , then the time taken for A to become 0.1 M is _____ sec. (nearest integer)

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Solution

Q.
Consider the following reaction, the rate expression of which is given below

$A + B \to C$

$rate = k {[A]}^{1 / 2} {[B]}^{1 / 2}$

The reaction is initiated by taking 1 M concentration of A and B each. If the rate constant (k) is $4.6 \times 10^{- 2}$ $s^{- 1}$ , then the time taken for A to become 0.1 M is _____ sec. (nearest integer) [2024]

Ans.
(50)

$A$    $+$    $B$    $\to$     $C$

   $t = 0$    $a_{o}$ $a_{o}$ 0

   $t = t$ $a_{o} - x$    $a_{o} - x$    $x$

   $r = k {[A]}^{1 / 2} {[B]}^{1 / 2} = k {(a_{o} - x)}^{1 / 2} {(a_{o} - x)}^{1 / 2}$

   $\frac{d x}{d t} = k (a_{o} - x)$

   $\frac{d x}{(a_{o} - x)} = k d t$

Integrate both sides

$\int_{0}^{x} \frac{dx}{(a_{o} - x)} = \int_{0}^{t} k dt$

   ${[\frac{\ln (a_{o} - x)}{- 1}]}_{0}^{x} = {[k t]}_{0}^{t}$

$[\frac{\ln (a_{o} - x)}{- 1} - \frac{\ln (a_{o})}{- 1}] = k t$

$\ln (\frac{a_{o}}{a_{o} - x}) = k t$

$2.303 \log_{10} (\frac{a_{o}}{a_{o} - x}) = k t$

$In the question a_{0} = 1 M and (a_{0} - x) = 0.1 M$

$2.303 \log_{10} (\frac{1}{0.1}) = 4.6 \times 10^{- 2} t$

   $2.303 \log_{10} 10 = 4.6 \times 10^{- 2} t$

   $2.303 \times 1 = 4.6 \times 10^{- 2} t$

$t = 50 s$

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