Question

8.23 × 1023 mol of InCl(s) is placed in 1.00 L of 0.010 M HCl(aq) at 75°C. The InCl(s) dissolves quite quickly, and then the following reaction occurs:

$\mathbf{3}\mathit{I}{\mathit{n}}^{\mathbf{+}}\left(aq\right)\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{2}\mathit{I}\mathit{n}\left(s\right)\mathbf{}\mathbf{+}\mathbf{ }\mathbf{}\mathit{I}{\mathit{n}}^{\mathbf{+}\mathbf{3}}\left(aq\right)$

As this disproportionation proceeds, the solution is analyzed at intervals to determine the concentration of In⁺(aq) that remains.

(a) Plot ln [In1] versus time, and determine the apparent rate constant for this first-order reaction.

(b) Determine the half-life of this reaction.

(c) Determine the equilibrium constant K for the reaction under the experimental conditions.

Short answer

The equilibrium constant value is $1.39\times {10}^{-16}$.

Step-by-step solution

Step-1: Calculation of natural logarithm

According to the given table in the above question, one would calculate natural logarithm of given concentration of and finally plot .

Time(s)	$\left[I{n}^{+}\right]\left(mol{L}^{-1}\right)$	$In\left[I{n}^{+}\right]$
0	$8.23\times {10}^{-3}$	-4.80
240	$6.41\times {10}^{-3}$	-5.50
480	$5.00\times {10}^{-3}$	-5.30
720	$3.89\times {10}^{-3}$	-5.55
1000	$3.03\times {10}^{-3}$	-5.80
1200	$3.03\times {10}^{-3}$	-5.80
10,000	$3.03\times {10}^{-3}$	-5.80

Step-2 :   Graph :

Plotting $\ln \left[I{n}^{+}\right]versustime$one would get the figure-1

The first order reaction plot of against time in seconds has a straight line with slope -k. The slope of the straight line is -0.00101s^-1

Therefore the value of the rate constant is -0.00101s^-1

Step-3 :  Formulae for   calculation:

(C) Transition rate theory predicts the reaction rate of the reaction and given by eyring formulae

$k_r=k\frac{k_{B}T}{h}K*.........\left(1\right)$

Where

$\begin{array}{rcl}{k}_{{}_r}& =& ratecons\tan t\\ K*& =& Equilbriumcons\tan t\\ k& =& Transmissioncoefficient\\ h& =& planckcons\tan t\\ K_B& =& Boltzmencons\tan t\end{array}$

STEP-4:  Calculation of  equilibrium constant value:

Substituting the values in the above equation.

$\begin{array}{rcl}{k}_r& =& 0.00101s^{-1}\\ h& =& 6.626\times {10}^{-34}Js\\ k_B& =& 1.38065\times {10}^{-23}J{K}^{-1}\\ T& =& {75}^{0}C\\ T& =& \left(75+273\right)K\\ T& =& 348K\end{array}$

Assuming k = 1 ,substituting these values in equation-1,

$\begin{array}{rcl}0.00101& =& 1\times \frac{1.38\times {10}^{-23}J{K}^{-1}\times 348K}{6.626\times {10}^{-34}Js}{K}^{*}\\ K^{*}& =& \frac{0.00101s^{-1}\times 6.626\times {10}^{-34}Js}{1.38065\times {10}^{-23}J{K}^{-1}\times 348K}\\ K^{*}& =& 1.39\times {10}^{-16}\end{array}$

The equilbrium constant value is $1.39\times {10}^{-16}$.