Question

Inversion of sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in the polarimeter. If r_∞, r_t and r₀ are the rotations at t = ∞, t = t and t = 0, then first order reaction can be written as ______.

Options

• `k =1/t log_e  (r_t - r_infty)/(r_0 - r_infty)`

• `k =1/t log_e  (r_0 - r_infty)/(r_t - r_infty)`

• `k =1/t log_e  (r_infty - r_0)/(r_infty - r_t)`

• `k =1/t log_e  (r_infty - r_t)/(r_infty - r_0)`

Answer 1

Inversion of sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in the polarimeter. If r_∞, r_t and r₀ are the rotations at t = ∞, t = t and t = 0, then first order reaction can be written as `bbunderline(k =1/t log_e  (r_0 - r_infty)/(r_t - r_infty))`.

Explanation:

In the inversion of cane sugar (a first-order reaction), the progress of the reaction is followed using optical rotation.

r₀ = initial optical rotation at t = 0

r_t = rotation at time t

r_∞ = final rotation at \[\ce{t → \infty}\]

The concentration of the reactant is proportional to (r_t − r_∞), and the integrated rate law becomes:

`k = 1/t ln ([R]_0/[R]_t)`

= `1/t ln ((r_0 - r_infty)/(r_t - r_infty))`