Question

What will be the order of reaction for a chemical change having log $t_{1/2}$ vs log a? (where a = initial concentration of reactant; $t_{1/2}$ = half-life)

Answer 1

0

Answer 2

The half-life ($t_{1/2}$) for an n-th order reaction is generally proportional to $a^{1-n}$, where 'a' is the initial concentration. Taking the logarithm of this relationship yields $\log t_{1/2} = \text{constant} + (1-n) \log a$. This is a linear equation where the slope is $(1-n)$. From the graph, the slope is $\tan(45^\circ) = 1$. Equating the slopes, $1-n = 1$, which gives $n=0$.