Question

The inversion of cane sugar proceeds with the half-life of 500 min at pH 5 for any concentration of sugar. However, if pH = 6, the half-life changes to 50 min. The rate law expression for the sugar inversion can be written as:

• A. $r=k\left[sugar{]}^1\right[H{]}^0$
• B. $r=k\left[sugar{]}^2\right[H{]}^6$
• C. $r=k\left[sugar{]}^0\right[H^{\oplus }{]}^6$
• D. None of these

Answer 1

The correct option is A $r=k\left[sugar{]}^1\right[H{]}^0$

Since $t_{1/2}$ does not depend upon the sugar concentration, it is first order w.r.t [sugar]
$\therefore t_{1/2}\propto [sugar{]}^1$
$t_{1/2}\times a^{n-1}=k$
$\frac{(t_1/2{)}_1}{(t_{1/2{)}_2}}=\frac{[H^{\oplus }{]}_1^{1-n}}{[H^{\oplus }{]}_2^{1-n}}$
$\frac{500}{50}={\left(\frac{{10}^{-5}}{{10}^{-6}}\right)}^{1-n}$
$10=$ $(10{)}^{1-n}\Rightarrow n=0$