For the reaction A ? B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g (if the initial concentration of A was 50 g ) is _______. (Nearest integer) Given: log 2 = 0.3010

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Solution

Q.
For the reaction A → B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g $L^{- 1}$ (if the initial concentration of A was 50 g $L^{- 1}$ ) is _______.

(Nearest integer)

Given: log 2 = 0.3010 [2025]

Ans.
(43)

This question is solved by assuming the reaction to be first order. To find the rate constant, consider two values of concentration at t = 5 sec and t = 15 sec.

$k = \frac{2.303}{time difference} \log \frac{{[A]}_{t = 5}}{{[A]}_{t = 15}}$

$= \frac{2.303}{15 - 5} \log \frac{40}{20} = 0.2303 \log 2$

Let after time $t$ the concentration get reduced to $2.5 g/L$ , then

$k = \frac{2.303}{time difference} \log \frac{{[A]}_{t = 0}}{{[A]}_{t = t}}$

$0.2303 log 2 = \frac{2.303}{t - 0} \log \frac{50}{2.5}$

$0.2303 log 2 = \frac{2.303}{t} \log 20$

$0.2303 log 2 = \frac{2.303}{t} (\log 2 + \log 10)$

$0.2303 \times 0.3010 = \frac{2.303}{t} (0.3010 + 1)$

$t = 43.2 s$

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