A reaction with respect to X is zero order till the concentration is reduced to half of initial concentration.Then this reaction become first order with respect to it.If the value of rate constants for the zero order and first order are equal to k (in magnitude), then find the time taken to reduce the concentration of X to `1/16th` of its original concentration `X_0`.

Options

A. `1/(2k)+(3ln2)/k`
B. `x_0/(2k)+(3ln2)/k`
C. `x_0/(2k)+(4ln2)/k`
D. `1/(2k)+(4ln2)/k`

Answer

B

Explanation

`xto` product (zero order)
Till the concentration of X changes from `X_0` to `X_0/2` , it is zero order with rate constant k.
Time taken for concentration to becomes `X_0/2=(t_(1//2)` for zero order )=`X_0/(2k)`
After reaching `X_0/2` , the reaction becomes first order
`Xto`product (I order)
`X_0/2underset((ln2)/k)toX_0/4underset((ln2)/k)toX_0/8underset((ln2)/k)toX_0/16implies ` total time of reach `X_0/16` from `X_0` is `=X_0/(2k)+(3ln2)/k`