Chemical Kinetics

Unit for rate constant of $n^{\text {th }}$ order reaction $=\left(\frac{m o l}{L}\right)^{1-n} s^{-1}$

For zero order reaction $\Rightarrow \mathrm{n}=0$; unit $\Rightarrow \mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$

For first order reaction $\Rightarrow \mathrm{n}=1$; unit $\Rightarrow \mathrm{s}^{-1}$

For second order reaction $\Rightarrow \mathrm{n}=2$; unit $\Rightarrow \mathrm{mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}$

For third order reaction $\Rightarrow n=2$; unit $\Rightarrow \mathrm{mol}^{-2} L^2 \mathrm{~s}^{-1}$

For zero order

$ [R]=\left[R_0\right]-k t $

So -ve slope shows zero order reaction

$ \begin{aligned} &\text { From Arrhenius equation }\\ &\begin{aligned} & \mathrm{k}=\mathrm{A} \mathrm{e}^{\frac{-\mathrm{E}_a}{R T}} \\ & \ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_a}{R \mathrm{~T}} \\ & \frac{\mathrm{E}_a}{R}=1.25 \times 10^4 \\ & \mathrm{E}_a=1.25 \times 10^4 \times 1.987 \\ & =24.84 \mathrm{k} \mathrm{cal} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $

$ \begin{aligned} & \operatorname{Ink}=\operatorname{In} A-\frac{E_a}{R T} \\ & \operatorname{In}\left(e^2\right)=6-\frac{6.64 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times T} \\ & 2=6-\frac{6.64 \times 10^3}{8.3 \times T} \\ & T=\frac{6.64 \times 10^3}{8.3 \times 4}=200 \mathrm{~K} \end{aligned} $

For zero order reaction

$ \begin{aligned} & -\frac{1}{2} \frac{d A}{d t}=k \\ & t=\frac{A_0-A_t}{2 k} \\ & t=\frac{2-0.5}{2}=0.75 \mathrm{~min} \end{aligned} $

To determine how long it takes for the concentration of a reactant to decrease from 7.2 mol L$^{-1}$ to 0.9 mol L$^{-1}$, we use the formula for the first-order reaction:

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

Where:

$ k = 0.03 \, \text{s}^{-1} $ is the rate constant.

$ a = 7.2 \, \text{mol L}^{-1} $ is the initial concentration.

$ a-x = 0.9 \, \text{mol L}^{-1} $ is the final concentration.

Plug these values into the equation:

$ \begin{aligned} t & = \frac{2.303}{0.03} \log \frac{7.2}{0.9} \\ & = \frac{2.303}{0.03} \log 8 \\ & = \frac{2.303}{0.03} \times 3 \times \log 2 \\ & = \frac{2.303}{0.03} \times 3 \times 0.301 \\ & = 69.3 \, \text{s} \end{aligned} $

Therefore, it takes 69.3 seconds for the concentration to decrease from 7.2 mol L$^{-1}$ to 0.9 mol L$^{-1}$.

For a first‐order reaction the half‐life and rate constant are related by

$k=\frac{\ln2}{t_{1/2}}=\frac{0.693}{1\;\text{min}}=0.693\;\text{min}^{-1}.$

We want 99.9% completion, so the fraction remaining is 0.001. Using

$\frac{[A]_t}{[A]_0}=e^{-kt}=0.001$

take natural logs:

$-kt=\ln(0.001)=-6.9078$

hence

$t=\frac{6.9078}{0.693}\approx9.97\;\text{min}\approx10\;\text{min}.$

Answer: Option B, 10 minutes.

Let the rate equation is

$\text { Rate }=k[A] \times[B]^y$

Therefore, we can write

$\begin{aligned} & 2 \times 10^{-3}=k[0.1]^x[0.1]^y \quad \text{..... (i)}\\ & 4 \times 10^{-3}=k[0.2]^x[0.1]^y \quad \text{..... (ii)}\\ & 1.6 \times 10^{-2}=k[0.2]^x[0.2]^y \quad \text{..... (iii)} \end{aligned}$

$\begin{aligned} & \text { (ii) } \div \text { (i); } \\ & \frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y} \\ & \Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x \\ & \therefore \quad x=1 \end{aligned}$

$\begin{aligned} & \text { (ii) } \div \text { (iii); } \\ & \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^x[0.1]^y}{k[0.2]^x[0.2]^y} \\ & \Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y \\ & \therefore \quad y=2 \\ & \therefore \quad \text { Rate }=k[A]^1[B]^2 \end{aligned}$

First order with respect to A while second order with respect to B.

Using Arrhenius equation, $ \begin{aligned} & k=A e^{-E_a / R T} \\ & \ln k=\ln A-\frac{E_a}{R T} \\ & y=c+m x, \quad \text { slope }(m)=-\frac{E_a}{R} \\ & \text { Intercept }=\ln A \end{aligned} $

$K=A e^{-E_a / R T}$

After taking In both side

$\ln \mathrm{K}=\ln \mathrm{A}-\frac{E_a}{R T}$

$\ln K_1=\ln A-\frac{E_a}{R T_1}$ at temp. $T_1$ .... (i)

$\ln K_2=\ln A-\frac{E_a}{R T_2}$ at temp. $T_2$ .... (ii)

$\begin{aligned} & \text { (ii) }- \text { (i) } \\ & \ln K_2-\ln K_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ & \ln \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{1}{500}-\frac{1}{700}\right] \\ & \ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500 \times 700}\right] \\ & \ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500 \times 700}\right] \\ & \log 3.5=\frac{E_a}{2.303 \times R}\left[\frac{1}{250 \times 7}\right] \\ & 0.5441=\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{250 \times 7}\right] \\ & E_a=0.5441 \times 8.31 \times 250 \times 7 \times 2.303 \\ & =0.5441 \times 83.1 \times 25 \times 7 \times 2.303 \\ & =18222.65 \\ & \approx 18219 \mathrm{~J} \end{aligned}$

To calculate value of $\mathrm{E_a}$

Equation used is

$\mathrm{\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

Hence $\mathrm{E_a}$ can be calculated if value of rate constant k is known at two different temperatures $\mathrm{T_1}$ and $\mathrm{T}_2$.

The Arrhenius equation is given as

$\begin{aligned} & \mathrm{k}=\mathrm{Ae}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ & \therefore \quad \ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}} \end{aligned}$

$\ln \mathrm{k~} \text{v/s} ~\frac{1}{\mathrm{~T}} \text { gives a straight line graph with slope }=-\frac{E_a}{R} \text { and intercept }=\ln A$

To solve this problem, we will use the Arrhenius equation, which relates the rate constant ($k$) of a chemical reaction to the temperature ($T$) and the activation energy ($E_a$). The equation in its logarithmic form is:

where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor (frequency factor),
• $E_a$ is the activation energy,
• $R$ is the universal gas constant ($8.314 \, \text{J mol}^{-1} \text{K}^{-1}$),
• $T$ is the temperature in Kelvin,
• $2.303$ is the factor to convert from natural log to common log.

According to the problem, the rate of the reaction quadruples ($k_2 = 4k_1$) when the temperature increases from $27^{\circ}C$ (which equals $27 + 273.15 = 300.15 \, K$) to $57^{\circ}C$ (which equals $57 + 273.15 = 330.15 \, K$). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have:

With $k_2 = 4k_1$, substituting in the values and taking their difference:

Given that $\log 4 = 0.6021$, we now solve for $E_a$:

First, calculate $\frac{1}{300.15} - \frac{1}{330.15}$:

So,

Solve for $E_a$:

Therefore, the activation energy $E_a$ is $38.07 \, \text{kJ/mol}$, which best matches Option A:

$ \begin{aligned} & 3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\ & \mathrm{r}=-\frac{1}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}=+\frac{1}{2} \frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}} \\ &+\frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}}=-\frac{2}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}} \end{aligned}$

$\begin{aligned} & \mathrm{r}=\mathrm{k}[\mathrm{A}]^{-1 / 2}[\mathrm{~B}]^{3 / 2} \\ & \text { order }=-\frac{1}{2}+\frac{3}{2} \\ & =\frac{2}{2} \\ & =1 \end{aligned}$

Rate $=k[A]^{2}[\mathrm{~B}]$

If $[A]$ is tripled and $[B]$ is kept constant.

$r^{1}=k[3 A]^{2}[B]$

$r^{1}=9 k[A]^{2}[B]$

$r^{1}=9 r$

Increased by a factor of nine

A reaction cannot have zero activation energy.

$\mathrm{E_{a}}$ is minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value.

4A + 3B $\to$ 6C + 9D

Rate of reaction = ${{ - d[A]} \over {dt}} \times {1 \over 4} = {{ - d[B]} \over {dt}} \times {1 \over 3} = {{ + d[C]} \over {dt}} \times {1 \over 6} = {{ + d[D]} \over {dt}} \times {1 \over 9}$

Rate of reaction $ = {{ + d[C]} \over {dt}} \times {1 \over 6} = {{6 \times {{10}^{ - 2}}} \over 6} = {10^{ - 2}}$ mol L^$-$1 s^$-$1

Rate of reaction $ = {{ - 1} \over 3}{{d[B]} \over {dt}}$

${{ - d[B]} \over {dt}} = 3 \times $ rate of reaction $ = 3 \times {10^{ - 2}}$ mol L^$-$1 s^$-$1

After interval of 10 sec. $ = 3 \times {10^{ - 2}} \times 10$

$ = 30 \times {10^{ - 2}}$ mol L^$-$1

$\bullet$ For zero order reaction

r = k[A]⁰

r = k (constant)

hence, 'y' as 'rate' and 'x' as concentration will give desired graph.

$\bullet$ For first order reaction

${t_{1/2}} = {{0.693} \over k}$ (constant)

hence, 'y' as 't_1/2' and 'x' as concentration will given desired graph.

For first order reaction,

$K = {{2.303} \over t}\log {{[{A_0}]} \over {[A]}}$; where A₀ is the initial concentration of reactant A.

A₀ = 0.1 M

A = 0.001 M

t = 5 minute

$K = {{2.303} \over 5}\log {{0.1} \over {0.001}} = {{2.303} \over 5}\log {10^2}$

$ = {{2.303} \over 5} \times 2$

K = 0.9212 min^$-$1

The correct answer to this question is collision frequency. Here's an explanation for each option to understand why:

Option A - Heat of Reaction: The heat of reaction, also known as the enthalpy change $(\Delta H)$, is determined by the difference in the energy levels of the reactants and products. It is a characteristic feature of a particular chemical reaction and is not typically affected by the concentration of reactants except in cases of extremely high or non-ideal concentrations where volume changes might somewhat influence pressure and therefore the heat of reaction in gases. Under normal conditions, however, changing the concentration of reactants does not alter the heat of reaction.

Option B - Threshold Energy: Threshold energy is the minimum energy that reactant molecules must possess in order to undergo a successful collision, leading to a chemical reaction. This value is inherent to the specific reaction and is related to the strength of the bonds in the reactants as well as the energy required to form the activated complex during the reaction. The threshold energy does not change simply because the concentration of reactants is altered.

Option C - Collision Frequency: Collision frequency corresponds to how often reacting particles collide in a given time interval. When the concentration of reactants is increased, there are more reactant particles per unit volume. This statistically leads to an increased number of collisions per unit time, thereby raising the collision frequency. This is in accordance with the collision theory of chemical kinetics.

Option D - Activation Energy: Activation energy is the minimum energy barrier that must be overcome for reactants to be converted into products. It is a property of a particular reaction and is related to the nature of the reactants and the reaction pathway. Altering the concentration of reactants does not change the activation energy for that reaction.

Thus, the only factor among the provided options that changes with an increase in the concentration of reactants is indeed the collision frequency.

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.