Question

The inversion of cane sugar proceeds with a half-life of $600$ minutes at $pH=5$ for any concentration of sugar. However, if $pH=6$, the half-life changes to $60$ minutes. The rate law expression for the sugar inversion can be written as:
A. $r=K{{[sugar]}^{2}}{{[{{H}^{+}}]}^{0}}$
B. $r=K{{[sugar]}^{1}}{{[{{H}^{+}}]}^{0}}$
C. $r=K{{[sugar]}^{1}}{{[{{H}^{+}}]}^{1}}$
D. $r=K{{[sugar]}^{0}}{{[{{H}^{+}}]}^{1}}$

Answer 1

Hint: In the given question, see whether the half-life is dependent upon the concentration of the reactants and products or not, then it will be easy to figure out about the order of the reaction. Remember, the half-life is inversely proportional to the concentration of the reactants.

Complete answer:
In the question given that,
At pH$=5$, the inversion of sugar is proceeding with a half-life (${{[{{t}_{1/2}}]}_{1}}$) of $600$minutes for any concentration of sugar.
While, at pH$=6$ the half-life changes (${{[{{t}_{1/2}}]}_{2}}$) to $60$ minutes. We have to find out the rate law expression for the sugar inversion.
So, here we have reactants as sugar (i.e. sucrose) and acid (in hydrogen ion form). And looking at the options, we see that only reactants are used in the rate law expression.
The rate law is expressed as the product of the rate constant (K) with the concentration of the reactants to the power of their stoichiometry.
The half-life period is inversely proportional to that of the concentration of the reactants or products and it can be expressed as:
${{t}_{1/2}}=\dfrac{1}{conc{{.}^{n-1}}}$, where n refers to the order of the reaction.
As, pH is a negative log of hydrogen concentration.
So at pH$=5$, the concentration of hydrogen will be ${{10}^{-5}}$ while at pH$=6$ , the concentration of the hydrogen will be ${{10}^{-6}}$.
So, the half-life period for the hydrogen at different time periods will be:
$\dfrac{{{[{{t}_{1/2}}]}_{1}}}{{{[{{t}_{1/2}}]}_{2}}}={{\left( \dfrac{{{10}^{-5}}}{{{10}^{-6}}} \right)}^{1-n}}$
Then,
$\dfrac{600}{60}={{10}^{1-n}}$
Thus, ${{10}^{1-n}}=10$
And, $1-n=1$
So, $n=0$
So, the order of the hydrogen concentration will be zero order. As, the half-life does not depend upon the concentration of the sugar, so the order will be a first order.
So, the rate law will be $rate=K{{[sugar]}^{1}}{{[{{H}^{+}}]}^{0}}$.

Hence, the correct option is B.

Note:
Sucrose (which is also called as cane sugar) in presence of dilute acid (such as hydrochloric acid, etc.) gets converted to glucose and fructose. Generally, sucrose is dextrorotatory, but while hydrolysis it gets converted to laevorotatory and this process of conversion is the inversion of sugar.