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) ${\text{r}} = {\text{k}}{[{\text{sugar}}]^2}{[{\text{H}}]^6}$
B) ${\text{r}} = {\text{k}}{[{\text{sugar}}]^1}{[{\text{H}}]^0}$
C) ${\text{r}} = {\text{k}}{[{\text{sugar}}]^0}{[{\text{H}}]^6}$
D) ${\text{r}} = {\text{k}}{[{\text{sugar}}]^0}{[{\text{H}}]^1}$

Answer 1

If half life remains unaffected by the concentration then it is a second order reaction. If half life increases with increase in concentration then it is zero or less than zero order reaction.

Formula used:
${t_{1/2}} = \dfrac{{\text{k}}}{{{{\text{a}}^{{\text{n}} - 1}}}}$ here ${t_{1/2}}$ is the half life, ${\text{k}}$ is the rate constant, ‘a’ is the concentration and n is the order of reaction.

Complete step by step solution:
It is given to us that the half life of sugar at a pH of 5 is always 500 min and is independent of the concentration of sugar. In the formula if we put the value of n equals to 1 then we will get,
${t_{1/2}} = \dfrac{{\text{k}}}{{{{\text{a}}^{{\text{n}} - 1}}}}$
Substituting the value of n = 1
${t_{1/2}} = \dfrac{{\text{k}}}{{{{\text{a}}^{1 - 1}}}}$
$\Rightarrow {t_{1/2}} = \dfrac{{\text{k}}}{{{{\text{a}}^0}}}$
Anything raised to a power of zero is 1. Hence for a first order reaction the half life and concentration are independent of each other. So the order with respect to sugar in the reaction should also be first order.
Now if we put n equals to zero we will get.
${t_{1/2}} = \dfrac{{\text{k}}}{{{{\text{a}}^{0 - 1}}}} = \dfrac{{\text{k}}}{{{{\text{a}}^{ - 1}}}}$
The negative power will become positive if we bring it to the numerator,
${t_{1/2}} = {\text{k}} \times {\text{a}}$
So for a zero order reaction the half life is directly proportional to the concentration.
As the pH increases from 5 to 6 then the concentration of hydrogen ion decreases and the half life also decreases form 500 to 50 and hence it follows the zero order with respect to the concentration of hydrogen ion,

Thus, the only option that fits above criteria is option B.

Note:
${\text{pH}} = - \log [{{\text{H}}^ + }]$ . The higher the concentration of hydrogen ion lower will be its pH and hence the reverse is also true. The order with respect to hydrogen ion can also be calculated as:
$\dfrac{{{{({t_{1/2}})}_1}}}{{{{({t_{1/2}})}_2}}} = \dfrac{{{{{\text{(}}{{\text{a}}_2})}^{{\text{n}} - 1}}}}{{{{{\text{(}}{{\text{a}}_1})}^{{\text{n}} - 1}}}}$. By substituting the respective concentration and half life will give the value of n.