In a certain reaction `B^(n+)` is getting converted to `B^((n+4)^(+))` in solution. The rate constant of this reaction is measured by titrating a volume with a reducing agent which reacts only with `B^(n+)` and `B^((n+4)^(+))`. In the process it converts `B^(n+)` to `B^((n-2)^(+))`and `B^((n+4)^(+))` to `B^((n-1)^(+))`. At, t = 0 the volume of reagent consumed is 25 mL and at t = 10 min, the volume used is 32 mL. Calculate the rate constant for the conservation of `B^(n+)` to `B^((n+4)^(+))` assuming it to be a first order reaction.

Options

A. `2.07 xx 10^(-2) "min"^(-1)`
B. `2.07 xx 10^(-4) "min"^(-1)`
C. `5.36 xx 10^(-2)"min"^(-1)`
D. `5.36 xx 10^(-4) "min"^(-1)`

Answer

A

Explanation

`B^(n+1) rarr B^((n+4)+)`
`{:("Mimimole at = 0",a,0,2e+ B^(n+) ,rarr,B^((n-2)^(+))),(t=t,(a-x),x,5e + B^((n+4)x),rarr,B^((n-1)+)):}`
Let normality be N for reducing agent.
Thus, at `t = a xx 2 = N xx 25 therefore a = (25)/(2)N`
`{:("at t = t",(a-x) xx 2,+,x xx 5,=N xx 32),(,"for " B^(n+),,"for " B^((n+4)+) ,):}`
`therefore 3x + 7 N` or `x = (7)/(3)N` Now, `K = (2.303)/(10)log""(25//2N)/(((25)/(2) -(7)/(3))N) = (2.303)/(10)log""(25 xx 6)/(2 xx 61) = 2.07 xx 10^(-2) "min"^(-1)`