Question

The inversion of cane sugar proceeds with a constant half-life of $500$ minute at $pH$ $=$ 5 for any concentration of sugar. However, if $pH$ $=$ $6$, the half-life changes to $50$ minutes. Derive the rate law for inversion of cane sugar.

• A. $r=k\left[sugar\right]\left[H^{+}\right]$
• B. $r=k\left[sugar\right][H^{+}{]}^2$
• C. $r=k\left[H^{+}\right]$
• D. $r=k\left[sugar\right][H^{+}{]}^0$

Answer 1

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

At pH 5, the half-life is independent of the concentration of sugar.
Hence, the reaction is of first order in sugar.
The rate law expression is $rate=K{\left[sugar\right]}^1{\left[H^{+}\right]}^m$.
The relationship between half life and $\left[H^{+}\right]$ is $t_{1/2}\alpha {\left[H^{+}\right]}^{1-m}$
$500\alpha {\left[{10}^{-5}\right]}^{1-m}$
$50\alpha {\left[{10}^{-6}\right]}^{1-m}$
$10\alpha {\left(10\right)}^{1-m}$
$1-m=1$
$m=0$
Hence, the rate law expression is $r=k\left[sugar\right][H^{+}{]}^0.$