Question

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 $r_{\propto },r_{\zeta }$ and $r_0$ are the rotations at $t=\propto$, t = t, and t = 0, then the first order reaction can be written as:

• A. $k=\frac{1}{t}log\frac{r_1-r_{\propto }}{r_0-r_{\propto }}$
• B. $k=\frac{1}{t}ln\frac{r_0-r_{\propto }}{r_1-r_{\propto }}$
• C. $k=\frac{1}{t}ln\frac{r_{\propto }-r_0}{r_{\propto }-r_1}$
• D. $k=\frac{1}{t}ln\frac{r_{\propto }-r_1}{r_{\propto }-r_0}$

Answer 1

Answer: (B)

$k=\frac{1}{t}ln\frac{r_0-r_{\propto }}{r_1-r_{\propto }}$

For a first order reaction $A\to P$, the expression for the rate constant is
$k=\frac{1}{t}ln\frac{a}{a-x}$
Here, A is reactant, P is product, a is the initial concentration of A and $a-x$ is the concentration of A at time t.
The inversion of a sugar follows first order rate equation which is given below.
$k=\frac{1}{t}ln\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$
Here, $a=r_0-r_{\infty }$ and $a-x=r_t-r_{\infty }$