Question

The inversion of cane sugar proceeds with the half-life of 600 minutes at $pH=5$ for any concentration of sugar. However at $pH=6$, the half life changes to 60 minute. The rate law expression for sugar inversion can be written as: (if it is first order with respect to sugar)

• A. $r=k\left[sugar{]}^2\right[H^{+}{]}^o$
• B. $r=k\left[sugar{]}^1\right[H^{+}{]}^o$
• C. $r=k\left[sugar{]}^2\right[H^{+}{]}^1$
• D. $r=k\left[sugar{]}^1\right[H^{+}{]}^{-1}$

Answer 1

The correct option is D $r=k\left[sugar{]}^1\right[H^{+}{]}^{-1}$

When the pH is increased by 1 unit from 5 to 6, the hydrogen ion concentration decreases to one-tenth.

The half-life period also becomes one-tenth. Thus the rate of the reaction increases ten times.

Hence, the order of the reaction with respect to hydrogen ion concentration is -1.

Thus the expression for the rate of the reaction is $r=k\left[sugar{]}^1\right[H^{+}{]}^{-1}$