Question

The complex of ${\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{5}}}{\text{F}}} \right]^{{\text{2 + }}}}$ reacts with water according to the following equation.
${\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{5}}}{\text{F}}} \right]^{{\text{2 + }}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{5}}}\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)} \right]^{{\text{3 + }}}} + {{\text{F}}^ - }$
The rate of the reaction is = ${\text{rate constant}} \times {\left[ {{\text{Complex}}} \right]^{{\text{ax}}}}{\left[ {{{\text{H}}^{\text{ + }}}} \right]^{\text{b}}}$. The reaction is acid catalysed i.e. $\left[ {{{\text{H}}^{\text{ + }}}} \right]$does not change during the reaction. Thus rate = ${{\text{k}}_{\text{1}}}{\left[ {{\text{complex}}} \right]^{\text{a}}}$where ${{\text{k}}_{\text{1}}} = {\text{k}}{\left[ {{{\text{H}}^{\text{ + }}}} \right]^{\text{b}}}$, calculate ‘a’ and ‘b’ given the following data at ${\text{2}}{{\text{5}}^{\text{0}}}{\text{C}}$

[Complex]M	$\left[ {{{\text{H}}^{\text{ + }}}} \right]$M	${T_{1/2}}$ hr	${T_{3/4}}$ hr
$0.1$	$0.01$	1	2
$0.2$	$0.02$	$0.5$	1

A. a =b =$0.5$
B. a =b = 1
C. a =b = 2
D. a =b = 3

Answer 1

The reaction between the cobalt ammine fluoride and water is a substitution reaction. Substitution reactions can be of two types: bimolecular and unimolecular. We shall find the ratios of the time given and the hydrogen ion concentration to find the order of reaction.

Complete step-by-step answer: From the values of the half-lives and the three-fourth lives of the concentration of the reactants, we know that the half-life of a reaction and the time required for the decay of three-fourth of the species is equal to half.
Hence, $\dfrac{{{T_{1/2}}}}{{{T_{3/4}}}} = \dfrac{1}{2}$,
when the molarity of the reactants is $0.1$ and when the molarity of the reactants is $0.2$, the ratio is $\dfrac{{{T_{1/2}}}}{{{T_{3/4}}}} = \dfrac{{0.5}}{1}$.
Hence, in both the cases the ratio is the same. This is only possible if the order of the reaction is 1. Hence the value of ‘a’ is 1 and the reaction is first-order with respect to the complex.
For the hydrogen ion concentration, let us consider that the time –period of the reaction is directly proportional to the concentration of the hydrogen ions raise to the power “b”, according to the following equation, ${{\text{k}}_{\text{1}}} = {\text{k}}{\left[ {{{\text{H}}^{\text{ + }}}} \right]^{\text{b}}}$.
Hence it can be said that $\dfrac{{{{\left( {{T_{1/2}}} \right)}_1}}}{{{{\left( {{T_{3/4}}} \right)}_2}}} = {\left( {\dfrac{{0.01}}{{0.02}}} \right)^b}$= $\dfrac{{0.5}}{1} = {\left( {\dfrac{1}{2}} \right)^b}$
Hence, the value of b is also 1.

Thus, the correct answer is option B.

Note: In the above reaction, the water molecules behave as the nucleophile to replace the fluoride anions from the complex. Since the rate of the reaction is dependent on the concentration of the complex and as there is only one molecule of the complex that takes in the reaction, so the reaction should be independent of the concentration of the water molecules and only dependent on the concentration of the complex.