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 interval of time
$\underset{\text{Sucrose d-}}{C_{12}{H}_{22}{O}_{11}}+H_{2}O\stackrel{H^{+}}{\to }\underset{\text{Glucose d- }}{C_6{H}_{12}{O}_6}+\underset{\text{Fructose l-}}{C_6{H}_{12}{O}_6}$
$r_0=$ angle of rotation at the start, $r_t=$ angle of rotation at time $t,r_{\infty }=$ angle of rotation at the complete reaction. There is $50\%$ inversion, when

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

Answer 1

Answer: (C)

$r_0=$ rotation at the start, $r_t=$ rotation at time $t$

$r_{\infty }=$ rotation at the end

then $\left(r_{\infty }-r_0\right)=a$ and $\left(r_{\infty }-r_t\right)=(a-x)$

When $50\%$ inversion has taken place

$a-x=\frac{a}{2}$

$\therefore 2(a-x)=a$

$2\left(r_{\infty }-r_t\right)=r_{\infty }-r_0$ or $2r_{\infty }-2r_t=r_{\infty }-r_0$

$\therefore r_0=\left(2r_t-r_{\infty }\right)$