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:

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 equilbrium constant value is $1.39x{10}^{-16}$

Step-by-step solution

Calculation of natural logartham;

According to the given table in the above question, one would calculate natural logartham of given concendration 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	3.03x10-3	-5.80

   Graph :

Plotting $\text{In}\left[{\text{In}}^{\text{+}}\right]$ versus timeone would get the figure-1

For the first order reaction plot of $\text{In}\left[{\ln }^{+}\right]$ against time in seaconds has a stright line with slope -k.Slope of the stright line is -0.00101s^-1

Therefore valueof rate constant is -0.00101s^-1

Formulae for  calculation:

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

$$\begin{gathered} k_r=k\frac{k_{B}T}{h}k*...........\left(1\right) \\ Where \\ {k}_r=ratecons\tan t \\ k=Transmissioncoefficient \\ h=planckscons\tan t \\ {k}_B=\text{Boltzmenconstant} \\ k*=\text{equilbriumconstant} \end{gathered}$$

Calculation of  equilbrium constant value:

Substituting the values in the above equation.

$$\begin{gathered} k=0.001s^{-1} \\ h=6.625x{10}^{-34}\hspace{0.17em}\text{Js} \\ \text{T\hspace{0.17em}}={75}^{0}C \\ \text{T}=75+273\text{K} \\ \text{T}=348\text{K} \end{gathered}$$

$$\begin{gathered} \text{Assuming}k=1\text{substitutingthesevaluesineqaution-1} \\ 0.00101=\frac{1.38x{10}^{-23}\text{JK}x348\text{K}}{1.38065x{10}^{-23}{\text{JK}}^{-1}\text{x348K}}x\text{K}* \\ \text{K}*=\frac{0.00101s^{-1}x6.626x{10}^{-34}\text{Js}}{1.38065x{10}^{-23}{\text{JK}}^{-1}\text{x348}K} \\ \text{K}*=1.39x{10}^{-16} \end{gathered}$$

The equilbrium constant value is $1.39x{10}^{-16}$