Question

For a first order reaction derive the relationship \(t_{99\%} = 2 t_{90\%}\)

Answer 1

For 1^st order reaction k.t = \(ln\left(\frac{a_0}{a_t}\right)\) 

For t₉₉, if a₀ = 100 then a_t = 1

\(\therefore \ t_{99} = \frac{1}{k} ln \left(\frac{100}{1}\right)\) ...(1)

For t₉₀, if a₀ = 100 then a_t = 10

\(\therefore \ t_{90} = \frac{1}{k} ln \left(\frac{100}{10}\right)\)  ...(2)

On dividing (1) \(\div\)(2)

\(\frac{t_{99}}{t_{90}} = 2\) 

\(\therefore \ t_{99} = 2t_{90}\)