Question

Chloroethane decomposes at elevated temperatures according to the reaction

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}\to {\mathrm{C}}_2{\mathrm{H}}_4+\mathrm{HCl}$

This reaction obeys first-order kinetics. After 340 s at 800 K, a measurement shows that the concentration of C₂H₅Cl has decreased from 0.0098 mol L21 to 0.0016 mol L^-1. Calculate the rate constant k at 800 K

Short answer

The rate constant of the reaction is 0.0053 S

Step-by-step solution

Rate expression

The expression for the first order reaction when initial and final concendration is given as below

$$\begin{gathered} \mathrm{In}\left[\frac{{\mathrm{c}}_{\mathrm{t}}}{{\mathrm{c}}_0}\right]=-\mathrm{kt}.......\left(1\right) \\ 2.303\log \left[\frac{{\mathrm{c}}_{\mathrm{t}}}{{\mathrm{c}}_0}\right]=-\mathrm{kt} \end{gathered}$$

Calculation of rate constant

Initial , final concentration and temperature is given as

$$\begin{gathered} C\hspace{0.17em}=0.0098\mathrm{mol}{\mathrm{L}}^{-1} \\ {C}_t=0.0016\mathrm{mol}{\mathrm{L}}^{-1} \\ \mathrm{T}=800\mathrm{k} \\ \mathrm{K}=? \end{gathered}$$

Substituting all the above values in equation 1 one would get

$$\begin{gathered} 2.303\log \left[\frac{{\mathrm{c}}_{\mathrm{t}}}{{\mathrm{c}}_0}\right]=-\mathrm{kt} \\ 2.303\log \left[\frac{0.0016}{0.0098}\right]=-\mathrm{k}\times 340 \\ \mathrm{K}=\frac{2.303}{340}\log (6.125)s^{-1} \\ \mathrm{K}=0.0053s^{-1} \end{gathered}$$

Hence the rate constant at 800 K is 0.0053S^-1