A plot of log `t_(1//2)` versus log `C_(0)` is given in the adjoining fig. The conclusion that can be drawn from this graph is

Options

A. Order `=1,t_(1//2)=(1)/(ka)`
B. Order `=1,t_(1//2)=(2.303)/(k)log^(2)`
C. Order `=0,t_(1//2)=(1)/(2" ka")`
D. Order `=2,t_(1//2)=(1)/(a)`

Answer

B

Explanation

`t_(1//2) proph (C_(0))^(1-n) or t_(1//2) = KC_(0)^(1-n)` (K = constant)
`therefore log t_(1//2) = log K + (1-n) log C_(0) ( y = mx + c)`
Thus, Plot of log `t_(1//2) "vs" log C_(0)` will be linear with slope = 1 - n. But as is parallel to log `C_(0)` axis, slope = 0, i.e., 1 - n = 0 or n = 1. Hence, it is a reaction of 1st order and for 1st order reactions,
`t_(1//2) = (2*303)/(k) log2`.