Question

The inversion of cane sugar proceeds with the half-life of 500 min at any concentration of sugar. However, if , the half-life changes to $50$ min. The rate law expression for sugar inversion can be written as:
Option:
A) $r = k{[sugar]^2}{[H]^6}$
B) $r = k{[sugar]^1}{[H]^0}$
C) $r = k{[sugar]^0}{[{H^ \oplus }]^6}$
D) $r = k{[sugar]^0}{[{H^ \oplus }]^1}$

Answer 1

Basically, rate law is a law which gives the relationship between the rate of reaction and concentration of reactants which are involved in it. For a reaction $aA + bB \to cC + dD$
Where $a,b,c,d$ are stoichiometric coefficient of products and reactant in the reaction then the rate law is given as $Rate \propto {[{\rm A}]^x}{[{\rm B}]^y}$ where $x,y$ denote partial reaction orders for reactants.

Complete step by step solution:
Now, we know the relation between the rate of reaction and concentration of reactants.
Lets assume the half-life rate to be {t_{{\raise0.7ex\hbox{ 1 } \\!\mathord{\left/ {\vphantom {1 2}}\right.} \\!\lower0.7ex\hbox{ 2 }}}} , half-life is independent of sugar concentration which indicates its first order with respect to sugar concentration
$\therefore {t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}} \propto {[sugar]^1}$ …….eq $(1)$
Now, in general $rate = k{[A]^x}{[B]^y}$ or we can also represent it as k = {\raise0.7ex\hbox{ {rate} } \\!\mathord{\left/ {\vphantom {{rate} {{{[A]}^x}{{[B]}^y}}}}\right.} \\!\lower0.7ex\hbox{ {{{[A]}^x}{{[B]}^y}} }}
So for this problem \dfrac{{{t_{{\raise0.7ex\hbox{ 1 } \\!\mathord{\left/ {\vphantom {1 2}}\right.} \\!\lower0.7ex\hbox{ 2 }}}}}}{{{{[{H^ \oplus }]}^n}}} = k
Where we assume $n$ as partial reaction order for hydrogen ion
Now there are two conditions given for half time, let us try to find out by putting both condition in one equation
${t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}} = \dfrac{1}{{{{[{H^ \oplus }]}^n}}} \\\ \therefore {t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}} = {[{H^ \oplus }]^{1 - n}} \\\ \dfrac{{{{({t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}})}_1}}}{{{{({t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}})}_2}}} = \dfrac{{{{[{H^ \oplus }]}_1}^{1 - n}}}{{{{[{H^ \oplus }]}_2}^{1 - n}}} \\\$
Let's put the values in the equation
$\dfrac{{500}}{{50}} = {\left( {\dfrac{{{{10}^{ - 5}}}}{{{{10}^{ - 6}}}}} \right)^{1 - n}} \\\ 10 = {10^{1 - n}} \\\ \therefore n = 0 \\\$ …………….eq $(2)$
So, by concluding eq $(1)$ and $(2)$ the final rate equation will be
$r = k{[sugar]^1}{[H]^0}$

Note:
For a few reactions, we can only determine rate law only by performing it. We cannot obtain a rate law equation from a balanced chemical equation as many times partial orders of the reactants are not equal to the stoichiometric coefficients.