Following is the graph between `(a-x)^(-1)` and time for second order reaction `Ato2B,Q=tan^(-1)(0.5),OA=2Lmol^(-1)`. Here is the initial concentration of reactant and x is its amount reactedc at any time t. Hence rate at the start of the reaction is:

Options

A. `0.125Lmol^(-1)mol^(-1)`
B. `0.25molL^(-1)min^(-1)`
C. `0.125molL^(-1)min^(-1)`
D. `0.25Lmol^(-1)min^(-1)`

Answer

C

Explanation

For second order reaction,
`because(a-x)^(-1)=(1)/(Kt)+(I)/(a)`
`becausek=tantheta(becausetantheta=0.5` from question)
`becauseK=0.5` lit `mol^(-1)min^(-1)`
and `(I)/(a)=2` or `a=0.5M`
`r=Ka^(2)`
`=0.5xx0.5^(2)`
`=0.125molL^(-1)min^(-1)`