The inversion of a sugar follows first order rate equation which can be followed by noting the change in the rotation of the plane of polarization of light in the polarimeter. If `r_(oo), r_(f)` and `r_(0)` are the rotations at `t = oo, t = t`, and `t = 0`, then the first order reaction can be written as

Options

A. `k=(1)/(t)In(r_(1)-r_(oo))/(r_0-r_(oo))`
B. `k=(1)/(1)In(r_(0)-r_(oo))/(rt-r_(oo))`
C. `k=(1)/(t)In(r_(o)-r_(t))/(r_oo-r_(t))`
D. `k=(1)/(1)In(r_oo-r_(t))/r_(oo-r_(o))`

Answer

B

Explanation

(b)Rate constant for first order reaction is given by
`k=(1)/(t)log"(a)/((a-x))`
`apropr_(0)-r_(prop)` ,i.e.amount of total sugar,
`(a-x)propr_(t)-r_(prop)`,i.e. left sugar at any time.
`therefore`first order rate constant for inversion of sugar may be written sa:
`k=(1)/(t)log"(r_(0)-r_(prop))/(r_(t)-r_(prop))`