Chemical Kinetics

To determine the time period of a first-order reaction when it is two-thirds complete, and given a rate constant of $ 4.3 \times 10^{-4} \, \text{s}^{-1} $, use the following approach :

For a first-order reaction :

$ t = \frac{2.303}{k} \log \frac{a}{a - x} $

Where :

$ a $ is the initial concentration

$ a - x $ is the final concentration

Given :

Rate constant, $ k = 4.3 \times 10^{-4} \, \text{s}^{-1} $

Initial concentration, $ a = a $

When the reaction is two-thirds complete :

$ a - x = a - \frac{2}{3}a = \frac{a}{3} $

Now, substitute into the first-order reaction formula :

$ t = \frac{2.303}{k} \log \frac{a}{a - x} = \frac{2.303}{4.3 \times 10^{-4}} \log \frac{a}{\frac{a}{3}} $

Solving this :

$ t = \frac{2.303}{4.3 \times 10^{-4}} \log 3 = 2.5 \times 10^3 \, \text{seconds} $

Thus, the time period for the reaction to be two-thirds complete is $ 2.5 \times 10^3 \, \text{seconds} $.

Therefore, the correct option is (D).

$\text { Key idea } \log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

where $T_1$ and $T_2$ are initial and final temperature and $K_1$ and $K_2$ are initial and final rate constant $E_a=$ activation energy and $R=$ gas constant.

$\begin{aligned} \text { Given, } T_1 & =300 \mathrm{~K}, T_2=400 \mathrm{~K} \\ R & =8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ \frac{K_2}{K_1} & =\frac{2}{1} \\ \log \frac{K_2}{K_1} & =\frac{E_a}{2303 \times R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \end{aligned}$

Thus,

$\begin{aligned} \log \frac{2}{1} & =\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{300}-\frac{1}{400}\right] \\ E_a & =0.30 \times 2.303 \times 8.31 \times 3 \times 400 \\ E_a & =6.88 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

For a first order reaction, the integrated rate law is written as,

$k=\frac{2.303}{t} \log \frac{a}{a-x}$

Where, $k=$ Rate constant

$a=$ initial concentration

$a-x=$ concentration after time '$t$'

For half-life, $t=t_{1 / 2}, x=\frac{a}{2}$

On substituting the values, we get

$\begin{aligned} k=\frac{2.303}{t_{1 / 2}} \log \frac{a}{a-\frac{a}{2}} & =\frac{2.303}{t_{1 / 2}} \log 2 \\ & =\frac{0.693}{t_{1 / 2}} \\ k & =\frac{0.693}{t_{1 / 2}} \end{aligned}$

Thus, $t_{1 / 2}$ of a first order reaction does not depend upon the concentration.

Given that,

$\ln k\left(\mathrm{~s}^{-1}\right)=14.34-\frac{1.25 \times 10^4}{T} \mathrm{~K}$ ..... (i)

We know that,

$\ln k\left(\mathrm{~s}^{-1}\right)=\ln A-\frac{E_a}{R T}$ ...... (ii)

On comparing equation (i) and (ii), we get

$\begin{aligned} & \frac{E_a}{R}=1.25 \times 10^4 \mathrm{~K} \\ \therefore \quad & E_a=1.25 \times 10^4 \mathrm{~K} \times R \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ & =1.25 \times 10^4 \mathrm{~K} \times 2 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ & \quad\left[\because R=2 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right] \\ & =2.50 \times 10^4 \mathrm{~cal} \mathrm{~mol}^{-1} \end{aligned}$

Both assertion and reason are false.

The reaction $2 \mathrm{NO}+\mathrm{O}_2 \longrightarrow 2 \mathrm{NO}_2$ and $2 \mathrm{CO}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2$ proceeds at the different rates. Both the reactions have different activation energies.