Following is the graph between `(a-x)` and time `t` for second order reaction `theta=tan^(-1)(0.5)OA=2L mol^(-)`

Hence, the rate at the start of the reaction is

Options

A. `1.25 mol L^(-1) mi n^(-1)`
B. `0.5 mol L^(-1)mi n^(-1)`
C. `0.125 mol L^(-1) mi ^(-1)`
D. `12.5 mol L^(-1) min^(-1)`

Answer

C

Explanation

`kt=((1)/(a-x)-(1)/(a))`
`(1)/(a-x)=kt+(1)/(a)`
Graph between `(a-x)^(-1)` and time `t` is a straight line, hence, `k= tan theta= 0.5`
`OA=(1)/(a)=2`
`:. A=(1)/(2)=0.5`
`:. R=k[A]^(2)`
`= 0.5xx(0.5)^(2)`
`=0.125Lmol^(-1)min^(-1)`