Question

Inversion of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$ is a first order reaction and is studied by measuring angle of rotation at different intervals of time $\underset{\text{d-Sucrose}}{C_{12}{H}_{22}{O}_{11}}+H_{2}O\stackrel{H^{+}}{\to }\underset{\text{d -Glucose}}{C_6{H}_{12}{O}_6}+\underset{\text{l -Fructose}}{C_6{H}_{12}{O}_6}$
$r_0$ = angle of rotation at the start, $r_t$ = angle of rotation at time ti, and $r_{\infty }$ = angle of rotation at the complete reaction, if 50% of sucrose undergoes inversion, then

• A. $r_0=r_t-2r_{\infty }$
• B. $r_0=2r_t-r_{\infty }$
• C. $r_0=r_t+r_{\infty }$
• D. $r_0=r_t-r_{\infty }$

Answer 1

The correct option is B $r_0=2r_t-r_{\infty }$

Lets assume that $a_0$ mole per litre of sucrose is present at $t=0$
$sucrose\to glucose+fructose$
$att=0a_{0}00r_o$
Now at 50% inversion
$att=t{a}_0/2a_0/2a_0/2r_t$
Now at $t=\infty$
$att=\infty a_0/2a_0/2a_0/2r_{\infty }$
$r_0={\alpha }_s\left[a_0\right]\left(i\right)$
$r_t={\alpha }_s\left[a_0/2\right]+{\alpha }_g\left[a_0/2\right]+{\alpha }_f\left[a_0/2\right]\left(ii\right)$
$r_{\infty }={\alpha }_g\left[a_0\right]+{\alpha }_f\left[a_0\right]\left(iii\right)$
by $e{q}^n$ i ,ii and iii
$\frac{r_0+r_{\infty }}{2}=r_t$