Rate constant k = 2.3 x 10-4 s-1
Initial concentration of reactant [A0] = 0.01 M
Concentration of reactant after one hour = [A]
\(\therefore k = \frac{2.303}t \log \frac{[A_0]}{[A]}\)
⇒ \(2.3 \times 10^{-4} = \frac{2.303}{3600} \log \frac{[A]_0}{[A]}\)
⇒ \(\frac{2.3 \times10^{-4}\times 3600}{2.303} = \log 0.01 - \log [A]\)
⇒ \(0.36 = \log 0.01 - \log[A]\)
⇒ \(\log[A] =-2 - 0.36
\)
\(\log[A] =-1.64
\)
\([A] = 10^{(-1.64)}\)
\([A] = 0.023\) M
