Chemical Kinetics

Reaction $A \rightarrow P$, first order at $300 \mathrm{~K},[A]_0=0.5 \mathrm{~mol} / \mathrm{L}$

Rate constant, $k=0.125 \mathrm{~min}^{-1}$

for first order reaction,

$k=$ constant [ $T=$ constant]

Thus, $k=0.125 \mathrm{~min}^{-1}$

Given, $R \longrightarrow P$ (first order)

Integrated rate law for first order is given by the equation,

$ \ln [R]=\ln [R]_0-k t $

Comparing with $y=m x+c$

$ \ln [R]=-k t+\ln [R]_0 $

Slope $=-k$ (negative)

Intercept on $y$-axis $=\ln [R]_0$

This problem involves a zero-order reaction, where the rate of reaction is independent of the reactant's concentration.

The integrated rate law for a zero-order reaction is given by $[A]_t=[A]_0-k t$ where,

$t$ is time.

The half-life ( $t_{1 / 2}$ ) for a zero -order reaction is related to the initial concentration and rate constant by the expression

$ t_{1 / 2}=\frac{[A]_0}{2 k} $

Given

• $t_{1 / 2}=0.5 \mathrm{~h}$

• Initial concentration for half-life, $[A]_0=4 \mathrm{~mol} \mathrm{~L}^{-1}$.

First, calculate the rate constant ( $k$ )

$ \begin{aligned} 0.5 \mathrm{hr} & =\frac{4 \mathrm{~mol} \mathrm{~L}^{-1}}{2 k} \\ k & =\frac{4 \mathrm{~mol} \mathrm{~L}^{-1}}{2 \times 0.5 \mathrm{hr}}=\frac{4 \mathrm{~mol} \mathrm{~L}^{-1}}{1.0 \mathrm{hr}} \\ & =4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{hr}^{-1} \end{aligned} $

Next, calculate the time $(t)$ for the concentration to change from $2.0 \mathrm{~mol} \mathrm{~L}^{-1}$

to $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$;

Using the integrated rate law

$1.0 \mathrm{~mol} \mathrm{~L}^{-1}=2.0 \mathrm{~mol} \mathrm{~L}^{-1}$

$ -\left(4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{hr}^{-1}\right) t $

Rearrange to solve for $t$

$ \begin{aligned} \left(4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{hr}^{-1} t\right. & =2.0 \mathrm{~mol} \mathrm{~L}^{-1} - \left.1.0 \mathrm{~mol} \mathrm{~L}^{-1}\right) \end{aligned} $

$4 t=1.0 \mathrm{hr}$

$t=\frac{1.0}{4} \mathrm{hr}$

$t=1 / 4 \mathrm{~h}$ or 0.25 h .

Given, $\log K\left(s^{-1}\right)=7.14-1 \times 10^4 \mathrm{~K}$

$ \begin{aligned} & R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \mathrm{~T} \\ & E_a=? \end{aligned} $

Using Arrhenius equation,

$ \log k=\log A-\frac{E_a}{2.303 R T} $

Comparing $\frac{E_a}{2.303 R}=1 \times 10^4$

Calculating; $E_a=2.303 \times R \times 1 \times 10^4$

$ \begin{aligned} & =191.47 \\ & \approx 191.1 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Given reaction is, $A(g) \longrightarrow B(g)+C(g)$

Let $x$ be the decrease in pressure of $A$.

The pressure of $A$ at time $t$ is $p_A=p_0-x$

The total pressure at time $t$ is

$ \begin{aligned} & p_t=p_0-x+x+x=p_0+x \\ & 250=200+x ; x=50 \mathrm{~mm} \\ & p_A=200-50=150 \mathrm{~mm} \end{aligned} $

Use the integrated rate law,

$ k=\frac{2.303}{t} \log \left(\frac{p_0}{p_A}\right) $

Substitute the value,

$ \begin{aligned} k & =\frac{2.303}{20} \log \left(\frac{200}{150}\right) \Rightarrow k=0.0138 \mathrm{~min}^{-1} \\ t_{1 / 2} & =\frac{0.693}{k} \\ t_{1 / 2} & =\frac{0.693}{0.0138}=50.2 \text { minutes } \end{aligned} $

$R \longrightarrow P$

Half-life is independent of initial concentration, means it's a first order reaction.

Integrated rate law,

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

Thus, $\ln R v s$ time is a straight line with slope $=-k$.

Option (b) graph is incorrect because it $[R]$ vs $t$.

So, it should be exponential.

The change in concentration, $\Delta[R]$ is the final concentration minus initial concentration.

$ \Delta[R]=0.03-0.04=-0.01 \mathrm{~mol} / \mathrm{L} $

$ \begin{aligned} \text { Average velocity } & =-\frac{\Delta[R]}{\Delta t}=-\left(-\frac{0.01}{40}\right) \\ & =0.00025 \mathrm{~mol}^{-1} \mathrm{~min}^{-1} \end{aligned} $

or $2.5 \times 10^{-4} \mathrm{~mol}^{-1} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

In $\mathrm{Mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=4.167 \times 10^{-6} \mathrm{~mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

The slope of the $\ln K$ vs $\frac{1}{T}$ plot is given by

$ \begin{aligned} \text { Slope }= & -\frac{E_a}{R} \Rightarrow-2 \times 10^4=\frac{E_a}{8.314} \\ E_a & =166280 \mathrm{~J} / \mathrm{mol} \\ & =166.28 \mathrm{~kJ} / \mathrm{mol} \approx 166 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

From the graph,

Energy of reactant $A=10 \mathrm{~kJ} / \mathrm{mol}$

Energy of product $P=5 \mathrm{~kJ} / \mathrm{mol}$

Energy of transition state (peak) $=25 \mathrm{~kJ} / \mathrm{mol}$.

Activation energy $E_a=$ Energy of transition sate - Energy of reactant

Heat of reaction $\Delta H=$ Energy of product

- Energy of reactant

$\Delta H=5-10=-5 \mathrm{~kJ} / \mathrm{mol}$

$|\Delta H|=5 \mathrm{~kJ} / \mathrm{mol}$

For half completion $[A]_t=\frac{[A]_0}{2}$

$ t_{1 / 2}=\frac{2.303}{K} \log \frac{[A]_0}{[A]_{0 / 2}} $

Simplify $t_{1 / 2}=\frac{0.693}{K}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

for $3 / 4$ completion

$ \begin{aligned} {[A]_t } & =\frac{1}{4}[A]_0 \\ t_{3 / 4} & =2 \times t_{1 / 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Taking ratio of Eqs. (i) and (ii),

$ \frac{t_{3 / 4}}{t_{1 / 2}}=\frac{2 \times t_{1 / 2}}{t_{1 / 2}}=2: 1 $

The given equation is

$ \log _{10} \frac{k}{A}=0.00174 $

Multiply 2.303 on both side

$ \begin{aligned} \quad \ln \left(\frac{k}{A}\right) & =2.303 \times 0.00714 \\ \Rightarrow \quad \ln \left(\frac{k}{A}\right) & =0.00400722\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

The Arrhenius equation is

$ k=A e^{-E_e / R T} \Rightarrow \frac{k}{A}=e^{-E a / R T} $

Taking natural log

$ \ln \left(\frac{k}{A}\right)=-\frac{E_a}{R T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Compare Eqs. (i) and (ii)

$ -\frac{E_a}{R T}=0.00400722 $

Substituting all values

$ E_a \simeq 10.0 \mathrm{~J} / \mathrm{mol} $

We want to find the average velocity (rate) for the reaction as $A$ goes to $B$ when the amount of $A$ drops from $x$ to $y$ mol/L in 100 minutes.

The change in the concentration of $A$ is: $ \Delta [A] = y - x \text{ mol/L} $ This tells us how much $A$ has changed in value.

The average velocity means how much $A$ changes per minute. We can find it by dividing the total change in $A$ by the time taken: $ \text{Average velocity} = \frac{\Delta [A]}{\Delta t} $ where $\Delta t = 100$ minutes.

So, $ \text{Average velocity} = \frac{y - x}{100} \text{ mol L}^{-1} \text{ min}^{-1} $ If you need the answer as a positive value (since rates are usually positive), write it as $ \frac{|x - y|}{100} \text{ mol L}^{-1} \text{ min}^{-1} $

The concentration becomes $\frac{1}{8}$ of what it was at the start.

$ \frac{1}{8} = \left(\frac{1}{2}\right)^3 $

This means it takes 3 half-lives for the concentration to become $\frac{1}{8}$.

$ \begin{array}{r} t_{1 / 2} = \frac{75}{3} \\ t_{1 / 2} = 25\ \text{minutes} \end{array} $

Given equation,

$ \log \left(\frac{K}{A}\right)=-\frac{x}{T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using Arrhenius equation

$ k=A e^{-E_q / R T} $

Taking log on both side

$ \begin{aligned} & \log k=\log A-\frac{E_a}{2.303 R T} \\ & \log \left(\frac{k}{A}\right)=-\frac{E_a}{2.303 R T}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Compare Eqs. (i) and (ii),

$ \frac{x}{T}=\frac{E_a}{2.303 R T} \Rightarrow E_a=2.303 R \cdot x $

Given reaction is,

$ A+2 B \rightarrow 3 C+2 D $

The rate of reaction is, $r=-\frac{1}{2} \frac{d[B]}{d t}$

Given, $\frac{d[B]}{d t}=-x \times 10^{-2}$

Thus, $r=\frac{1}{2} \times x \times 10^{-2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Rate of appearance of $C$

$ \begin{aligned} & =\frac{1}{3} \frac{d[C]}{d t}=-\frac{1}{2} \frac{[d B]}{d t} \\ \frac{d[C]}{d t} & =\frac{3}{2} \times x \times 10^{-2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Taking the ratio of (i) and (ii)

$ \frac{\text { rate of reaction }}{\text { rate of appearance of } C}=1: 3 $

The activation energy for acid hydrolysis of sucrose is $6.22 \mathrm{~kJ} / \mathrm{mol}$. The activation energy for sucrose by sucrase is $2.15 \mathrm{~kJ} / \mathrm{mol}$.

For 1st order rate is, Rate $=k[A]$

$ \frac{\text { Rate }_1}{\text { Rate }_2}=\frac{\left[A_1\right]}{\left[A_2\right]} $

For first two data points,

$ \begin{aligned} & & \frac{0.2}{0.4} & =\frac{0.02}{x} \\ \Rightarrow & & x & =0.04 \mathrm{~m} \end{aligned} $

For second and third data point,

$ \begin{aligned} & \frac{0.4}{1}=\frac{0.04}{y} \Rightarrow y=0.1 \mathrm{M} \\ & x: y \Rightarrow 0.04: 0.1 \Rightarrow 2: 5 \end{aligned} $

The given reaction is:

$ \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2} + \frac{1}{2} \mathrm{O}_{2} $

For this reaction, the rates of change of concentrations are expressed as:

$ -\frac{d[\mathrm{N}_{2} \mathrm{O}_{5}]}{dt} = K_{1}[\mathrm{N}_{2} \mathrm{O}_{5}] $

$ \frac{d[\mathrm{NO}_{2}]}{dt} = K_{2}[\mathrm{N}_{2} \mathrm{O}_{5}] $

$ \frac{d[\mathrm{O}_{2}]}{dt} = K_{3}[\mathrm{N}_{2} \mathrm{O}_{5}] $

From the stoichiometry of the reaction:

$ -\frac{d[\mathrm{N}_{2} \mathrm{O}_{5}]}{dt} = \frac{1}{2} \frac{d[\mathrm{NO}_{2}]}{dt} = 2 \frac{d[\mathrm{O}_{2}]}{dt} $

Therefore, by equating the expressions, we get:

$ K_{1}[\mathrm{N}_{2} \mathrm{O}_{5}] = \frac{1}{2} K_{2}[\mathrm{N}_{2} \mathrm{O}_{5}] = 2 K_{3}[\mathrm{N}_{2} \mathrm{O}_{5}] $

Multiplying through by 2 to eliminate fractions gives:

$ 2K_{1}[\mathrm{N}_{2} \mathrm{O}_{5}] = K_{2}[\mathrm{N}_{2} \mathrm{O}_{5}] = 4K_{3}[\mathrm{N}_{2} \mathrm{O}_{5}] $

Thus, the relationship between the rate constants is:

$ 2K_{1} = K_{2} = 4K_{3} $

The first order reaction equation is

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

graph is drawn between $ \log \frac{(a)}{(a-x)} $ on

$ y $-axis and $ t $ on $ x $-axis.

Slope from graph is

$ \frac{\log \frac{a}{a-x}}{t}= $ slope

From Eq. (i), $ \frac{\log \frac{a}{a-x}}{t}=\frac{k}{2303} $

Slope $ =\frac{k}{2303} $

$ 2 \times 10^{-3}=\frac{k}{2303} $

$ k=4.606 \times 10^{-3} / \mathrm{min} $

For a first-order reaction, the time required for the reaction to reach a certain fraction of its initial concentration is given by the equation:

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

where:

$a$ is the initial concentration,

$(a-x)$ is the concentration after time $t$,

$k$ is the rate constant.

Case I: $\frac{1}{4}$ of the initial concentration remains after decomposition.

For this scenario, we have:

$ t_{\frac{1}{4}} = \frac{2.303}{k} \log \frac{1}{\frac{1}{4}} $

$ \Rightarrow t_{\frac{1}{4}} = \frac{2.303}{k} \times 0.6 $

Case II: $\frac{1}{10}$ of the initial concentration remains after decomposition.

For this situation:

$ t_{\frac{1}{10}} = \frac{2.303}{k} \log \frac{1}{\frac{1}{10}} $

$ \Rightarrow t_{\frac{1}{10}} = \frac{2.303}{k} $

Calculating the ratio and scaling it by 100:

The task requires finding the ratio $ \frac{t_{\frac{1}{4}}}{t_{\frac{1}{10}}} $ and multiplying it by 100.

$ \begin{aligned} \frac{t_{\frac{1}{4}}}{t_{\frac{1}{10}}} \times 100 & = \left( \frac{\frac{2.303}{k} \times 0.6}{\frac{2.303}{k}} \right) \times 100 \\ & = 0.6 \times 100 \\ & = 60 \end{aligned} $

Thus, the calculated value is 60.

$ -\frac{d A}{d t}=k[A]^x[B]^y \quad \ldots \text { (i) } $

On doubling [ $A$ ] rate is doubled

Hence, $2^x=2$

$ x=1 \quad \ldots \text { (ii) } $

Similarly, $y=1$

$ \therefore \quad-\frac{d[A]}{d t}=k[A][B] \quad \ldots \text { (iii) } $

Substituting the given values,

$ \begin{aligned} k & =\frac{2.5 \times 10^{-4}}{0.1 \times 0.1} \\\\ & =2.5 \times 10^{-2} \end{aligned} $

The Arrhenius equation describes the temperature dependence of the rate constant $k$ for a reaction, and is given by:

$ k = A e^{-\frac{E_a}{RT}} $

Taking the natural logarithm of both sides gives:

$ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} $

This equation represents a linear plot of $\ln k$ vs. $\frac{1}{T}$, where the slope is $-\frac{E_a}{R}$ and the intercept is $\ln A$.

Given:

Slope $= -10^3 \, \mathrm{K}$

Intercept $= 2.303$

$R = 8314 \, \mathrm{J \, K^{-1} \, mol^{-1}}$

The slope is related to the activation energy $E_a$ by:

$ -\frac{E_a}{R} = -10^3 $

Solving for $E_a$, we have:

$ E_a = 10^3 \cdot R $

Substituting the value of $R$:

$ E_a = 10^3 \cdot 8314 \, \mathrm{J \, mol^{-1}} $

To convert to kilojoules per mole, divide by 1000:

$ E_a = \frac{8314 \times 10^3}{1000} \, \mathrm{kJ \, mol^{-1}} = 8314 \, \mathrm{kJ \, mol^{-1}} $

Therefore, the correct answer is:

Option A: 8.314

The instaneous lst order reaction rate is given as

$ \text { rate }=\frac{-d[A]}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On $X$-axis $=$ time, $Y$-axis $=$ concentration

From graph slope $=-m$

i.e., concentration is decreasing with time.

$ -m=\frac{d[A]}{d t} \text { from graph }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Equating (i) and (ii)

Rate of reaction $=m$

To find the approximate activation energy ($E_a$) for a first-order reaction where the rate constant doubles when the temperature changes from 300 K to 310 K, we can use the following relationship:

$ \log \left(\frac{K_2}{K_1}\right) = \frac{E_a}{2.303 R} \frac{T_2 - T_1}{T_2 \times T_1} $

For $T_1 = 300 \, \text{K}$, the rate constant is $K_1 = K$.

For $T_2 = 310 \, \text{K}$, the rate constant becomes $K_2 = 2K$.

By substituting the known values into the equation:

$ \log \left(\frac{2K}{K}\right) = \frac{E_a}{2.303 \times 8.314} \times \frac{310 - 300}{310 \times 300} $

Simplifying and solving the equation:

$ \log 2 = \frac{E_a}{2.303 \times 8.3} \times \frac{10}{310 \times 300} $

Given that $\log 2 = 0.3$, we rearrange to find $E_a$:

$ E_a = 5333 \, \text{J/mol} = 53.33 \, \text{kJ/mol} $

Thus, the approximate activation energy is $53.33 \, \text{kJ/mol}$.

For a first-order reaction, the integrated rate law is:

$ k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t} \quad \text{...(i)} $

Where:

$ k $ is the rate constant,

$ [A]_0 $ is the initial concentration,

$ [A]_t $ is the concentration at time $ t $.

Given:

$ k = 3.3 \times 10^{-4} \, \mathrm{s}^{-1} $,

$ [A]_0 = 100 $ (assuming initial concentration as 100 for calculation),

$ [A]_t = 10 $ (since 90% of the reaction is complete, only 10% remains).

Substitute these values into equation (i):

$ t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t} = \frac{1}{3.3 \times 10^{-4}} \ln \left(\frac{100}{10}\right) $

This simplifies to:

$ t = \frac{10^4 \times 2.303}{3.3} = 6977.53 \, \mathrm{s} $

Converting seconds to minutes:

$ \frac{6977.5}{60} = 116.67 \, \mathrm{min} $

So, the time required to complete 90% of the reaction is 116.67 minutes.

For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. Given the rate constant $ k = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} $, the initial concentration of $ A $ is $ [A]_0 = 0.1 \, \text{mol L}^{-1} $, and the time $ t = 10 $ seconds, we use the following formula to calculate the concentration of $ A $ after 10 seconds:

$ k = \frac{[A]_0 - [A]_t}{t} $

Substituting the known values into the formula, we get:

$ 1 \times 10^{-3} = \frac{0.1 - [A]_t}{10} $

Solving for $ [A]_t $:

$ 0.01 = 0.1 - [A]_t $

$ [A]_t = 0.1 - 0.01 $

$ [A]_t = 0.09 \, \text{mol L}^{-1} $

Therefore, the concentration of $ A $ after 10 seconds is $ 0.09 \, \text{mol L}^{-1} $.

To determine the rate constant of a reaction at a different temperature using the Arrhenius equation, follow this process:

The formula for calculating the change in the rate constant with temperature is:

$ \log \frac{K_{T_2}}{K_{T_1}} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) $

Where:

$ K_{T_1} $ is the rate constant at temperature $ T_1 $.

$ K_{T_2} $ is the rate constant at temperature $ T_2 $.

$ E_a $ is the activation energy.

$ R $ is the universal gas constant.

Given Values:

Initial rate constant, $ K_{T_1} = 3.46 \times 10^{-2} \, \mathrm{s}^{-1} $ at $ T_1 = 298 \, \mathrm{K} $

Activation energy, $ E_a = 50.1 \, \mathrm{kJ/mol} = 50.1 \times 10^3 \, \mathrm{J/mol} $

Universal gas constant, $ R = 8.314 \, \mathrm{J/K \cdot mol} $

Temperature, $ T_2 = 350 \, \mathrm{K} $

Calculation Steps:

Substitute the given values into the formula:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = \frac{50.1 \times 10^3}{2.303 \times 8.314} \times \left( \frac{350 - 298}{298 \times 350} \right) $

Evaluate the expression:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = 1.30 $

Solve for $ K_{T_2} $:

$ \frac{K_{T_2}}{3.46 \times 10^{-2}} = 10^{1.30} = 19.95 $

Calculate $ K_{T_2} $:

$ K_{T_2} = 3.46 \times 10^{-2} \times 19.95 = 0.6903 \, \mathrm{s}^{-1} $

Thus, the rate constant at 350 K is approximately $ 0.692 \, \mathrm{s}^{-1} $.

For the given reaction:

$ 5 \mathrm{Br}^{-}(aq) + \mathrm{BrO}_3^{-}(aq) + 6 \mathrm{H}^{+}(aq) \longrightarrow 3 \mathrm{Br}_2(aq) + 3 \mathrm{H}_2 \mathrm{O}(l) $

The rate of change for the concentration of $\mathrm{Br}^{-}$ ions is given as:

$ -\frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = x \, \text{mol L}^{-1} \text{min}^{-1} $

The overall rate of the reaction can be calculated by considering the stoichiometry of $\mathrm{Br}^{-}$. Since 5 moles of $\mathrm{Br}^{-}$ are consumed per mole of reaction, the rate of the reaction is:

$ \frac{\Delta R}{\Delta t} = -\frac{1}{5} \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} $

$ = -\frac{1}{5} \times (-x) = \frac{x}{5} \, \text{mol L}^{-1} \text{min}^{-1} $

Therefore, the rate of the reaction is $\frac{x}{5} \, \text{mol L}^{-1} \text{min}^{-1}$.

The average rate of a first-order reaction is calculated using the formula:

$ \frac{\Delta R}{\Delta t} = -\left[\frac{R_2 - R_1}{t_2 - t_1}\right] $

In this context:

$ R_1 $ is the initial concentration:

$ R_1 = 0.03 \, \text{mol/L} $

$ R_2 $ is the final concentration:

$ R_2 = 0.02 \, \text{mol/L} $

$ t_2 - t_1 $ is the time interval over which the change occurs, given as 25 minutes. To convert this time into seconds:

$ t_2 - t_1 = 25 \times 60 \, \text{seconds} $

By substituting these values into the rate formula, we obtain:

$ \frac{\Delta R}{\Delta t} = -\frac{0.02 - 0.03}{25 \times 60} $

Calculating the above expression provides:

$ \frac{\Delta R}{\Delta t} = 6.667 \times 10^{-6} \, \text{mol/L/s} $

For a first-order reaction,

$ A(g) \rightarrow B(g) + 2C(g) $

After 24 minutes, the ratio of the concentration of the products to the reactant is $1:3$.

The rate constant $k$ for a first-order reaction is given by the formula:

$ k = \frac{2.303}{t} \log \left[\frac{a}{a-x}\right] $

Where:

$ t = 24 $ minutes

$ a = 4 $ (initial concentration)

$ a-x = 3 $

Substituting these into the equation, we have:

$ k = \frac{2.303}{24} \log 1.11 $

Using $\log 1.11 = 0.046$, we get:

$ k = \frac{2.303}{24} \times 0.046 \approx 4.40 \times 10^{-3} \, \text{min}^{-1} $

The half-life ($t_{1/2}$) for a first-order reaction is calculated using:

$ t_{1/2} = \frac{0.693}{k} $

Substitute $k = 4.40 \times 10^{-3}$ min$^{-1}$:

$ t_{1/2} = \frac{0.693}{4.40 \times 10^{-3}} \approx 157.8 \, \text{minutes} $

A reaction is described as autocatalytic when one of its products also serves as a catalyst for the same reaction.

For the given reaction:

$ \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5 + \mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOH} + \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} $

Here, $\mathrm{CH}_3 \mathrm{COOH}$ is both a product and a catalyst for the reaction, illustrating the autocatalytic nature.

Given, first order reaction graph between $\ln (a-x)$ on $Y$-axis and $t$ on $X$-axis.

According to first order reaction, $\ln (a-x)$

Rate constant is given by

$ k=\frac{\ln }{t} \frac{[a]}{[a-x]} \quad \ldots \text { (i) }$

or $ \begin{aligned} k t & =\ln \frac{[a]}{[a-x]} \\ k t & =\ln [a]-\ln [a-x] \end{aligned} $

or $\ln [a-x]=\ln [a]-k t \quad \ldots \text { (ii) } $

Compare Eq. (ii) by straight line equation.

$ \begin{aligned} & Y=\ln [a-x], x=t, m(\text { slope })=-k \\ & C \text { (intercent })=\ln \text { al } \end{aligned} $

$C$ (intercept) $=\ln [a]$

Therefore, slope $=$ rate constant

$ =-\left(-10^{-2}\right)=10^{-2} \quad \ldots \text { (iii) } $

Now initial concentration (time $t=0$ ) Eq. (ii) becomes

$ \ln [a-x]=\ln [a] $

From graph at $t=0, \ln [a-x]=-2.303$

$ -2.303=\ln [a] $

Initial concentration, $[a]=10^{-1}$

The rate of reaction, in a multistep reaction is determined by slowest step. So,

$ R=k_1\left[\mathrm{~N}_2 \mathrm{O}_2\right]\left[\mathrm{H}_2\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$\mathrm{N}_2 \mathrm{O}_2$ is not the part of overall reaction. Hence, its concentration can be calculated from step 1 i.e.

$ \left[\mathrm{N}_2 \mathrm{O}_2\right]=k_2[\mathrm{NO}]^2 $

[where, $k_2$ is rate constant for step 1]

By substituting concentration of $\left[\mathrm{N}_2 \mathrm{O}_2\right]$ in (i). We get,

$ R=k_1 k_2[\mathrm{NO}]^2\left[\mathrm{H}_2\right] \text { or } R=k[\mathrm{NO}]^2\left[\mathrm{H}_2\right] $

where, $k=k_1 \times k_2$

Rate law $=-\frac{d(\mathrm{HI})}{d t}=-k(\mathrm{HI})^2$

Since, it is a $2^{\circ}$ reaction.

∴ Putting the value of ( $n$ ) in general expression,

$ \begin{aligned} & \left(\mathrm{mol} \mathrm{~L}^{-1}\right)^{1-n} \mathrm{~s}^{-1} \quad(n=2) \\ & \left(\mathrm{molL}^{-1}\right)^{1-2} \mathrm{~s}^{-1} \\ & \text { or } \mathrm{mol}^{-1} \mathrm{Ls}^{-1} \end{aligned} $

Given, for zero order reaciton, $A \rightarrow$ product

$ \begin{aligned} & \text { Slope }=-3 \times 10^{-3} \mathrm{M} \mathrm{~min}^{-1},(x \text {-axis }) \\ & \text { Intercept }=2 \times 10^{-2} \mathrm{M} \text { (on } y \text {-axis) } \end{aligned} $

For zero order reaction,

$ [A]=-k t+[A]_0 $

If we plot $[A]$ against $t$, we get a straight line with slope equal to $-k$ and intercept equal to $[A]_0$ as shown below.

$ \therefore \quad k=3 \times 10^{-3} \mathrm{M} \mathrm{~min}^{-1} $

According to Arrhenius equation,

$ \log \frac{k_1}{k_2}=\frac{-E_a}{2303}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] $

Given rate of 1st order reaction will be double where temperature changes from 300 K to 310 K that means $2 K_1=K_2$

$ \begin{aligned} \log \frac{K_1}{2 K_1} & =\frac{-E_a}{2.303 R}\left[\frac{1}{300}-\frac{1}{310}\right] \\ \log \left(\frac{1}{2}\right) & =\frac{-E_a}{2.303 R} \times \frac{10}{300 \times 310} \\ -0.3 & =\frac{-E_a}{2.303 \times 300 \times 31 \times 8.3} \\ E_a & =6425.37 \times 8.3 \\ \Rightarrow & =53.330 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

The graph between $\ln k$ (where $k$ is the rate constant) on the $y$-axis and $1/T$ on the $x$-axis is a straight line. The slope of this line is $-4 \times 10^4 \text{ K}$. We need to determine the activation energy of the reaction in $\text{kJ mol}^{-1}$ given that $R = 831 \text{ J K}^{-1} \text{ mol}^{-1}$.

According to the Arrhenius equation:

$ k = A e^{-E_a / RT} $

Taking the natural logarithm:

$ \ln k = \ln A - \frac{E_a}{RT} $

This can be compared to the equation of a straight line $y = mx + c$, where:

The slope (m) is $-\frac{E_a}{R}$.

Given the slope $m = -4 \times 10^4$, we equate and solve for $E_a$:

$ -4 \times 10^4 = -\frac{E_a}{831} $

Solving for $E_a$, we get:

$ E_a = 4 \times 10^4 \times 831 $

$ E_a = 33.2 \times 10^4 \text{ J/mol} $

Converting joules to kilojoules:

$ E_a = 332 \text{ kJ/mol} $

To solve the problem, we have the following information:

The reaction follows first-order kinetics: $ n = 1 $

The rate constant is $ k = 6.93 \times 10^{-2} \, \mathrm{min}^{-1} $

Initial concentration, $ a = 100 $

Final concentration after 90% completion, $ a-x = 10 $

We need to find the time $ t $ it takes for 90% of the reaction to complete.

Using the formula for a first-order reaction rate constant:

$ t = \frac{2.303}{k} \log \frac{[A_0]}{[A]} $

Substituting the given values:

$ t = \frac{2.303}{6.93 \times 10^{-2}} \log \left(\frac{100}{10}\right) $

Simplifying:

$ t = \frac{2.303}{6.93 \times 10^{-2}} \times \log(10) $

Since $\log(10) = 1$:

$ t = \frac{2.303}{6.93 \times 10^{-2}} $

Calculating this gives:

$ t \approx 33.23 \, \text{min} $

Therefore, rounding to the nearest integer, the time for 90% completion of the reaction is approximately 33 minutes.

Given, Threshold energy $=75$

$\mathrm{kJ} / \mathrm{mol}$

(Internal energy) $=20 \mathrm{~kJ} / \mathrm{mol}$

Activation energy $=$ ?

Activation energy

= Threshold energy - internal energy

$=75-20$

$=55 \mathrm{~kJ} / \mathrm{mol}$.

Rate of reaction, $1=K_1[A]$

Rate of reaction, $2=K_2[A]$

Rate of reaction, $3=K_3[A]$

$ \begin{aligned} K_1 & =1 \mathrm{~s}^{-1}, K_2=0.1 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} \\ K_3 & =0.01 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~s}^{-1} \\ r_1 & =1 \times[A], r_2=0.1 \times[A] \\ r_3 & =0.01 \times[A] \\ \frac{r_1}{1} & =\frac{r_2}{0.1}=\frac{r_3}{0.01} \\ \therefore \quad r_1 & =10 \times r_2 \\ r_3 & =\frac{r_2}{10} . \end{aligned} $

The photochemical reaction, $\mathrm{H}_2+\mathrm{Cl}_2 \xrightarrow{h \nu} 2 \mathrm{HCl}$ is a zero order reaction as rate of reaction doesn't depend on concentration of reactants ( $\mathrm{H}_2$ and $\mathrm{Cl}_2$ ). It depends on the incident light.

For the given reaction,

$ 2 A \longrightarrow 2 B+C $

Rate constant, $K=1.2 \times 10^{-2} \mathrm{~s}^{-1}$

For $n^{\text {th }}$ order reaction, unit of rate constant, $K$ is $\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{1-n} \mathrm{~s}^{-1}$

Unit of given $K$ is $\mathrm{s}^{-1}$. Thus,

$ 1-n=0 \Rightarrow n=1 $

Order of the reaction is 1 .

For a first order reaction, the plot of ln [A] vs $t$ is a straight line with slope $=-k$.

According to Arrhenius theory,

$ k=A e^{-\frac{E_a}{R T}} $

Constant temperature, $T=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ For initial reaction, $k=A e^{\left(-E_{a_1} / R T\right)}$ Taking $\log$ on both sides, we get

$ \ln k=\ln A-\frac{E_{a_1}}{R T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For final reaction, $4 k=A e^{-\frac{E_{a_2}}{R T}}$ Taking $\log$ on both sides, we get

$ \ln 4 k=\ln A-\frac{E_{a_2}}{R T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On subtract Eq. (i) from Eq. (ii), we get

$ \begin{aligned} & \therefore \ln k-\ln 4 k=\frac{E a_2}{R T}-\frac{E a_1}{R T} \\ & \ln \left(\frac{k}{4 k}\right)=\frac{\Delta E_a}{R T}-1.386=\frac{\Delta E_a}{8.314 \times 300} \\ & \begin{aligned}\Rightarrow \Delta E_a & =-1.386 \times 300 \times 8.314 \\ & =-3456.96 \mathrm{~J} / \mathrm{mol} \end{aligned} \end{aligned} $

Or $\Delta E_a=-3.45 \mathrm{~kJ} \mathrm{~mol}^{-1}$

∴ Change in activation energy of this reaction $=-3.456 \mathrm{~kJ} / \mathrm{mol}$.

Given,

$ \begin{aligned} \left(t_{1 / 2}\right)_1 & =5 \text { minutes } \\ \left(t_{1 / 2}\right)_2 & =50 \text { minutes } \\ a_1 & =0.1 \mathrm{M} \\ a_2 & =0.01 \mathrm{M} \\ \frac{\left(t_{1 / 2}\right)_1}{\left(t_{1 / 2}\right)_2} & =\left(\frac{a_2}{a_1}\right)^{n-1} \Rightarrow \frac{5}{50}=\left(\frac{0.01}{0.1}\right)^{n-1} \\ (10)^1 & =(10)^{n-1} \text { or } 1=n-1 \text { or } n=2 \end{aligned} $

∴ It is second order reaction.

The temperature coefficient is given by the ratio $\frac{k_2}{k_1}$, where $k_1$ and $k_2$ are the rate constants at temperature $T_1$ and $T_2$ when $T_2=T_1+10$. This is approximately equal to 2 .

$ \begin{gathered} \log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ \log 2=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{310}\right] \\ E_a=53.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{gathered} $

Given,

$\begin{aligned} & k_1=0.02 \mathrm{~s}^{-1} \\ & k_2=0.2 \mathrm{~s}^{-1} \\ & E_d=? \\ & R=8.3 \\ & T_1=500 \mathrm{~K} \\ & T_2=700 \mathrm{~K} \end{aligned}$

$\begin{aligned} & \text { In } \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{r_2}-\frac{1}{T_1}\right) \\ & \text { In } \frac{0.2}{0.02}=-\frac{E_a}{8.3}\left[\frac{1}{700}-\frac{1}{500}\right] \\ & \text { In } 10=\frac{-E_a}{8.3}\left(\frac{500-700}{700 \times 500}\right) \\ & 2.302 \times 8.3 \times 700 \times 500=200 E_a \\ & \Rightarrow E_a=\frac{2.302 \times 8.3 \times 700 \times 500}{200} \\ & =33436 \mathrm{~J} / \mathrm{mol} \\ & \approx 33.44 \mathrm{~kJ} / \mathrm{mol} \\ \end{aligned}$

Conc. at time, $t, A=A_0-\frac{93.75}{100} A_0=\frac{6.25}{100} A_0$

$A_0 \rightarrow$ initial conc. $k \rightarrow$ rate constant, time, $t=x \mathrm{~min}$ Using integrated rate law for lst order reaction,

$\begin{aligned} k & =\frac{2.303}{t} \log \frac{A_0}{A} \\ & =\frac{2.303}{x} \log \frac{100 A_0}{6.25 A_0} \end{aligned}$

$\text { So, } \quad k=\frac{2.303}{x} \log \frac{100}{6.25}$

$\begin{aligned} & =\frac{2.303 \times 1.20}{x} \\ \Rightarrow \quad k & =\frac{2.303 \times 1.20}{x} \min ^{-1} \end{aligned}$

We know,

$\begin{aligned} t_{1 / 2} & =\frac{0.693}{k} \\ & =\frac{0.693 x}{2.303 \times 1.20} \end{aligned}$

So, $\quad t_{1 / 2}=\frac{x}{4} \min$

Given, rate constant, $k_0=0.003 \mathrm{molL}^{-1} \mathrm{~s}^{-1}$

Initial conc., $A_0=0.1 \mathrm{M}$

Final conc., $A=0.075 \mathrm{M}$

Using integrated rate law for zeroth order,

$\begin{aligned} A_0-A & =k t \\ 0.1-0.075 & =0.003 \times t \\ t & =\frac{0.025}{0.003}=8.33 \mathrm{~s} \end{aligned}$

Given:

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

Step 1: Write dimensions of each term

• Rate of reaction has units:

  $ \text{Rate} = \mathrm{mol\,L^{-1}\,s^{-1}} $

• $[A]$ and $[B]$ (concentrations) both have units:

  $ \mathrm{mol\,L^{-1}} $

Step 2: Substitute into the rate equation

$ \mathrm{mol\,L^{-1}\,s^{-1}} = k \times (\mathrm{mol\,L^{-1}}) \times (\mathrm{mol\,L^{-1}})^2 $

Simplify right-hand side:

$ k \times \mathrm{mol^3\,L^{-3}} $

Step 3: Solve for units of $ k $

$ k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol^3\,L^{-3}}} $

$ k = \mathrm{mol^{-2}\,L^{2}\,s^{-1}} $

✅ Final Answer:

$ \boxed{k = \mathrm{mol^{-2}\,L^{2}\,s^{-1}}} $

Correct option: D

$\begin{aligned} & X(g) \longrightarrow Y(g)+Z(g) \\ & t_{1 / 2}=10 \mathrm{~min} \quad \text{(Given)}\\ & A=A_0 e^{-k t} \quad \text{....... (i)}\\ & {\left[A_0\right.}=\text { Initial concentration }] \\ & k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{10}=0.0693 \end{aligned}$

From eq. (i),

$\begin{aligned} 0.1 A_0 & =A_0 e^{-0.0693 \times t} \\ \ln 0.1 & =-0.0693 \times t \\ -2.303 & =-0.0693 t \Rightarrow t=33.23 \mathrm{~min} . \end{aligned}$