Question

what is t(0.5)in terms of t(0.75) for 1st order reaction​

Answer 1

rate of first order reaction is given as $t=\frac{2.303}{k}log_{10}\frac{a}{a_t}$

where k is rate constant, a is initial amount of substance and $a_t$ is final amount of substance.

now, t(0.5) = (2.303/k) log{a/(a - 0.5a)}

= 2.303log2/k

= 2.303log2/k .....(1)

similarly, t(0.75) = (2.303/k)log{a/(a - 0.75a)}

= (2.303/k)log(a/0.25a)

= (2.303/k) log(4)

= 2 × 2.303log2/k.....(2)

from equations (1) and (2),

we get, t(0.75) = 2 × t(0.5)