Question

Consider a certain reaction \(A\) \(→\) \(Products\) with \(k = 2.0 \times 10^{-2 }s^{-1}\) . Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.

Answer 1

k= 2.0 × 10^-2 s^-1 

T= 100 s

[A]_o= 1.0 moL^-1

Since the unit of k is \(s^{-1}\) , the given reaction is a first order reaction.

Therefore, \(k = \frac {2.303}{t} log\ \frac {[A]_0}{[A]}\)

⇒ \(2.0 \times 10^{-2} s^{-1} = \frac {2.303}{100\  s} log\ \frac {1.0}{[A]}\)

⇒ \(2.0 \times 10^{-2} s^{-1} = \frac {2.303}{100 \ s} (-log\ A)\)

⇒ \(-log [A] = \frac {2.0\times 10^{-2} \times  100}{2.303}\)

⇒ \([A] = Antilog\  (-\frac {2.0 \times 10^{-2}\times 100}{2.303})\)

⇒ \([A]= 0.135\  mol \ L^{-1} (approximately)\)

Hence, the remaining concentration of A is \(0.135\  mol \ L^{-1}\).