Chemical Kinetics and Nuclear Chemistry

$ \text { For } \mathrm{A} \xrightarrow{\mathrm{~K}_1} \mathrm{~B} $

$ \begin{aligned} & \ln (2)=\frac{E_{a_1}}{R}\left[\frac{1}{300}-\frac{1}{500}\right] \\ & E_{a_1}=\frac{\ln 2 \times R \times 1500}{2} \\ & E_{a_2}=\frac{E_{a_1}}{2}=\frac{\ln 2 \times R \times 1500}{4} \\ & \left(K_1\right)_{\text {at } 500 \mathrm{~K}}=\frac{\ln 2}{2} \\ & \left(K_2\right)_{\text {at } 500 \mathrm{~K}}=\ln 2 \end{aligned} $

Now for $\mathrm{C} \xrightarrow{\mathrm{K}_2} \mathrm{D}$

$ \begin{aligned} & \ln \left[\frac{\left(\mathrm{K}_2\right)_{\text {at } 500 \mathrm{~K}}}{\left(\mathrm{~K}_2\right)_{\text {at } 300 \mathrm{~K}}}\right]=\left(\frac{\ln 2 \times \mathrm{R} \times 1500}{4}\right) \times \frac{1}{\mathrm{R}} \times\left[\frac{1}{300}-\frac{1}{500}\right] \\ & \left(\mathrm{K}_2\right)_{\text {at } 300 \mathrm{~K}}=\frac{\ln 2}{\sqrt{2}}=0.49 \\ & \left(\mathrm{~K}_2\right)_{\text {at } 300 \mathrm{~K}}=4.9 \times 10^{-1} \end{aligned} $

Ans is 5 .

$ \begin{aligned} & \mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{~K}} \end{aligned} $

Here, $\mathrm{t}_{1/2}=245$ days, so we find $K$ as:

$ \begin{aligned} & \mathrm{~K}=\frac{\ln 2}{245} \end{aligned} $

Now we use the radioactive decay relation for activity:

$ \begin{aligned} & \mathrm{t}=\frac{1}{\mathrm{~K}} \ln \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}} \end{aligned} $

Given that $75\%$ of the original activity remains, so $\mathrm{a}_t = 0.75\,\mathrm{a}_0 = \frac{3}{4}\mathrm{a}_0$.

So,

$ \begin{aligned} & \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}}=\frac{\mathrm{a}_0}{\frac{3}{4}\mathrm{a}_0}=\frac{4}{3} \end{aligned} $

Therefore,

$ \begin{aligned} & \mathrm{t}_{25 \%}=\frac{1}{\mathrm{~K}} \ln \frac{4}{3} \end{aligned} $

Substitute $\mathrm{K}=\frac{\ln 2}{245}$:

$ \begin{aligned} & \mathrm{t}_{25 \%}=\frac{1}{\frac{\ln 2}{245}} \ln \frac{4}{3} \end{aligned} $

Simplifying:

$ \begin{aligned} & \mathrm{t}_{25 \%}=245 \frac{\ln \frac{4}{3}}{\ell \mathrm{n} 2}=245\left[\frac{2 \log 2-\log 3}{\log 2}\right] \end{aligned} $

Now use the given values $\log 3=0.4771$ and $\log 2=0.3010$:

$ \begin{aligned} & =245\left[\frac{2 \times 0.3010-0.4771}{0.3010}\right]=101.66 \text { day. } \end{aligned} $

The graph between concentration $[\mathrm{AB}]$ and time $t$ is a straight line that decreases with time. This type of graph shows that the reaction is zero order in $[\mathrm{AB}]$.

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

$\begin{aligned} & {[\mathrm{AB}]_0-[\mathrm{AB}]_{\mathrm{t}}=\mathrm{kt}} \\ & \mathbf{0 . 6 0 - 0 . 5 5 = k}(100) \\ & \mathrm{k}=5 \times 10^{-4} \\ & \text { Half life }\left(\mathrm{t}_{1 / 2}\right)=\frac{[\mathrm{AB}]_0}{2 \mathrm{k}} \\ & =\frac{0.60}{2 \times 5 \times 10^{-4}} \\ & =600 \mathrm{sec} \\ & =10 \mathrm{~min}\end{aligned}$

From the graph, the initial concentration is $[\mathrm{AB}]_0 = 0.60$ and after $100 \,\text{s}$ the concentration is $0.55$. Substituting these values in the zero-order equation gives the rate constant $k = 5 \times 10^{-4} \,\text{mol L}^{-1}\text{s}^{-1}$.

For a zero-order reaction, the half-life is given by $t_{1/2} = \dfrac{[\mathrm{AB}]_0}{2k}$. Putting the values, we get $t_{1/2} = 600 \,\text{s} = 10 \,\text{min}$. Therefore, $x = 10 \,\text{min}$ (nearest integer).

Given: $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$

From the Arrhenius equation in base-10 form, we write:

$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$

On comparing this with $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$, the coefficient of $\frac{1}{T}$ gives:

$\frac{\mathrm{E}_{\mathrm{a}_1}}{2.303 \mathrm{R}}=1.5 \times 10^4$

Now substitute $\mathrm{R}=8.314~\mathrm{J\,mol^{-1}\,K^{-1}}$:

$\mathrm{E}_{\mathrm{a}_1}=1.5 \times 10^4 \times 2.303 \times 8.314$

$\mathrm{E}_{\mathrm{a}_1}=28.7207 \times 10^4 \mathrm{~J}$

Convert joules to kilojoules ($1~\mathrm{kJ}=10^3~\mathrm{J}$):

$\mathrm{E}_{\mathrm{a}_1}=287.207 \mathrm{~kJ}$

Given $\mathrm{E}_{\mathrm{a}_2}=\frac{1}{5}\mathrm{E}_{\mathrm{a}_1}$, so:

$\mathrm{E}_{\mathrm{a}_2}=\frac{\mathrm{E}_{\mathrm{a}_1}}{5}=\frac{287.207}{5}=57.44 \mathrm{~kJ}$

Given,

$k_1=10^6 e^{-30000/T},\qquad k_2=10^4 e^{-24000/T}$

At the temperature $T$ when $k_1=k_2$:

$10^6 e^{-30000/T}=10^4 e^{-24000/T}$

Divide both sides by $10^4 e^{-30000/T}$:

$10^2=e^{(-24000/T)-(-30000/T)}=e^{6000/T}$

Take natural log on both sides:

$\ln(10^2)=\frac{6000}{T}$

$2\ln 10=\frac{6000}{T}$

Given $\ln 10=2.303$:

$2(2.303)=\frac{6000}{T}$

$4.606=\frac{6000}{T}$

$T=\frac{6000}{4.606}\approx 1302.65\ \text{K}$

Nearest integer:

$\boxed{1303\ \text{K}}$

Arrhenius equation:

$k=Ae^{-E_a/RT}$

Given pre-exponential factors are identical, so $A_1=A_2$.

Also, activation energy of first reaction exceeds that of second by $20\,\text{kJ mol}^{-1}$:

$E_{a1}=E_{a2}+20{,}000\ \text{J mol}^{-1}$

Now,

$ \frac{k_2}{k_1} =\frac{A e^{-E_{a2}/RT}}{A e^{-E_{a1}/RT}} =e^{-(E_{a2}-E_{a1})/RT} =e^{(E_{a1}-E_{a2})/RT} $

So,

$\ln\left(\frac{k_2}{k_1}\right)=\frac{E_{a1}-E_{a2}}{RT}=\frac{20{,}000}{8.3\times 300}$

Compute:

$8.3\times 300=2490$

$\ln\left(\frac{k_2}{k_1}\right)=\frac{20{,}000}{2490}\approx 8.03$

Nearest integer:

$\boxed{8}$

We are given the Arrhenius equation in the exponential form:

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

Comparing this with the given equation:

$ k = (5.5 \times 10^{11} \, \text{s}^{-1}) \, e^{-\frac{28000}{T}} $

we can see that:

$ \frac{E_a}{R} = 28000 $

Now, substituting the given value of the gas constant $ R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} $:

$ E_a = 28000 \times 8.3 \, \text{J mol}^{-1} $

$ E_a = 232400 \, \text{J mol}^{-1} $

Converting joules to kilojoules:

$ E_a = \frac{232400}{1000} = 232.4 \, \text{kJ mol}^{-1} $

Rounding to the nearest integer:

$ \boxed{E_a = 232 \, \text{kJ mol}^{-1}} $

• The reaction is of the first order.

• The half-life of the reaction, $t_{1/2}$, is 3 hours.

• We need to find the percentage of sucrose remaining after a total time, $t$, of 6 hours.

Let's begin.

Step 1: Understand the concept of half-life ($t_{1/2}$)

For a chemical reaction, the half-life is the time it takes for the concentration of a reactant to reduce to exactly half of its initial value.

In this case, it means that every 3 hours, the amount of sucrose will become half of what it was at the start of that 3-hour period.

Step 2: Determine the number of half-lives passed

We are given a total time of 6 hours and a half-life of 3 hours. We can find the number of half-lives that have occurred, which we can call 'n'.

$ n = \frac{\text{Total time}}{\text{Half-life}} = \frac{t}{t_{1/2}} $

Substituting the given values:

$ n = \frac{6 \text{ hours}}{3 \text{ hours}} = 2 $

So, a total of 2 half-lives have passed.

Step 3: Calculate the amount of sucrose remaining

Let's assume the initial percentage of sucrose is 100%.

• After the first half-life (at t = 3 hours):

  The amount of sucrose remaining will be half of the initial amount.

  $ \text{Sucrose remaining} = \frac{1}{2} \times 100\% = 50\% $

• After the second half-life (at t = 6 hours):

  The amount of sucrose will be half of the amount that was present at the end of the first half-life (which was 50%).

  $ \text{Sucrose remaining} = \frac{1}{2} \times 50\% = 25\% $

We can also use a general formula for the fraction of reactant remaining after 'n' half-lives:

$ \text{Fraction remaining} = \left(\frac{1}{2}\right)^n $

In our case, $n=2$, so:

$ \text{Fraction remaining} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $

To find the percentage remaining, we multiply this fraction by 100:

$ \text{Percentage remaining} = \frac{1}{4} \times 100 = 25\% $

The question asks for the answer as the nearest integer.

Therefore, the percentage of sucrose remaining after 6 hours is 25.

From the graph, it is clear that the half-life $ t_{1/2} $ increases directly with the initial concentration $[A]_0$.

We know that for a reaction of order $ N $:

$ t_{1/2} \propto [A]_0^{1-N} $

Here, since $ t_{1/2} \propto [A]_0 $, we can write:

$ 1 - N = 1 \Rightarrow N = 0 $

This means the reaction is of zero order. For a zero-order reaction, the half-life is given by:

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

From the data in the graph: For the first point, $ t_{1/2} = 240 \, s $ and $[A]_0 = 4 \times 10^{-3}$

So,

$ 240 = \frac{4 \times 10^{-3}}{2k} \tag{1} $

For the second point, $[A]_0 = 1.5 \times 10^{-3}$ and $ t_{1/2} = x \, s $

$ x = \frac{1.5 \times 10^{-3}}{2k} \tag{2} $

Now, divide equation (1) by (2):

$ \frac{240}{x} = \frac{4}{1.5} $

Therefore,

$ x = 90 \, s $

For a first order reaction,

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

Given:

$ t_{1/2}=6.93 \text{ min} $

So,

$ k=\frac{0.693}{6.93}=0.1\ \text{min}^{-1} $

Now, $99\%$ completion means only $1\%$ of the reactant is left.

So,

$ \frac{[A]_0}{[A]}=\frac{100}{1}=100 $

For a first order reaction,

$ t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} $

Substituting the values:

$ t=\frac{2.303}{0.1}\log 100 $

Since,

$ \log 100=2 $

therefore,

$ t=\frac{2.303}{0.1}\times 2 $

$ t=23.03\times 2=46.06 \text{ min} $

Hence, the time required for $99\%$ completion of the reaction is

$ \boxed{46.06\ \text{minutes}} $

For a first order reaction, the integrated rate law is

$ k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t} $

We use the data at $t=20$ min:

• Initial concentration, $[A]_0=0.6500\ \text{M}$

• Concentration at $20$ min, $[A]_{20}=0.00065\ \text{M}$

So,

$ k=\frac{2.303}{20}\log\frac{0.6500}{0.00065} $

Now,

$ \frac{0.6500}{0.00065}=1000 $

and

$ \log 1000 = 3 $

Therefore,

$ k=\frac{2.303}{20}\times 3 $

$ k=\frac{6.909}{20} $

$ k=0.34545\ \text{min}^{-1} $

Now we use the concentration at time $x$:

• $[A]_x=0.0650\ \text{M}$

Again,

$ k=\frac{2.303}{x}\log\frac{[A]_0}{[A]_x} $

$ 0.34545=\frac{2.303}{x}\log\frac{0.6500}{0.0650} $

Now,

$ \frac{0.6500}{0.0650}=10 $

and

$ \log 10 = 1 $

So,

$ 0.34545=\frac{2.303}{x} $

Hence,

$ x=\frac{2.303}{0.34545} $

$ x\approx 6.67\ \text{min} $

Nearest integer:

$ \boxed{7\ \text{min}} $

For the reaction

$2A(g) \longrightarrow 4B(g) + C(g)$

let the initial pressure of $A$ be $P_0$.

Since the vessel is closed and temperature is constant, pressure is proportional to number of moles.

At $t=\infty$, reaction is complete.

From the stoichiometry:

• $2$ moles of $A$ produce $4+1=5$ moles of products

So if initially only $A$ is present, total moles increase by a factor

$\frac{5}{2}$

Hence final total pressure is

$P_\infty = \frac{5}{2}P_0$

Given:

$P_\infty = 600 \text{ mm Hg}$

So,

$\frac{5}{2}P_0 = 600$

$P_0 = 600 \times \frac{2}{5} = 240 \text{ mm Hg}$

Now at $30$ min, let the extent of reaction correspond to consumption of $x$ pressure units of $A$ according to:

$2A \to 4B + C$

Then pressure changes are:

• $A$: decreases by $x$

• $B$: increases by $2x$

• $C$: increases by $\frac{x}{2}$

This is because for $2A \to 4B + C$,

if $2$ units of $A$ react, $4$ units of $B$ and $1$ unit of $C$ are formed.

So total pressure at time $t$ is

$P_t = (240 - x) + 2x + \frac{x}{2}$

$P_t = 240 + \frac{3x}{2}$

Given at $30$ min,

$240 + \frac{3x}{2} = 300$

$\frac{3x}{2} = 60$

$x = 40$

Therefore pressure of $C$ at $30$ min is

$P_C = \frac{x}{2} = 20 \text{ mm Hg}$

So, the required pressure is

$\boxed{20 \text{ mm Hg}}$

Use the Arrhenius equation in its two-temperature form:

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

Given:

$ k_1=1.5\times 10^3\ \mathrm{s}^{-1}, \quad T_1=27^\circ\mathrm{C}=300\ \mathrm{K} $

$ k_2=4.5\times 10^3\ \mathrm{s}^{-1} $

$ E_a=60\ \mathrm{kJ\,mol^{-1}}=60000\ \mathrm{J\,mol^{-1}} $

$ R=8.3\ \mathrm{J\,K^{-1}\,mol^{-1}} $

First, find the ratio:

$ \frac{k_2}{k_1}=\frac{4.5\times 10^3}{1.5\times 10^3}=3 $

So,

$ \ln 3=\frac{60000}{8.3}\left(\frac{1}{300}-\frac{1}{T_2}\right) $

Now convert $\ln 3$ using the given values:

$ \ln 3 = 2.3 \log 3 = 2.3 \times 0.48 = 1.104 $

Also,

$ \frac{60000}{8.3} \approx 7229 $

Hence,

$ 1.104 = 7229\left(\frac{1}{300}-\frac{1}{T_2}\right) $

Therefore,

$ \frac{1}{300}-\frac{1}{T_2}=\frac{1.104}{7229} $

$ \frac{1.104}{7229}\approx 1.53\times 10^{-4} $

So,

$ \frac{1}{T_2}=\frac{1}{300}-1.53\times 10^{-4} $

$ \frac{1}{300}=3.33\times 10^{-3} $

Thus,

$ \frac{1}{T_2}=3.33\times 10^{-3}-0.153\times 10^{-3}=3.18\times 10^{-3} $

$ T_2 \approx \frac{1}{3.18\times 10^{-3}} \approx 314\ \mathrm{K} $

Now convert into Celsius:

$ t = 314 - 273 = 41^\circ\mathrm{C} $

Hence, the required temperature is

$ \boxed{41^\circ\mathrm{C}} $

To determine the time required for the concentration of A to decrease from an initial value of $50 \, \mathrm{g} \, \mathrm{L}^{-1}$ to $2.5 \, \mathrm{g} \, \mathrm{L}^{-1}$ in the reaction $ \mathrm{A} \rightarrow \mathrm{B} $, we assume first-order kinetics. Although the graph does not provide a clear indication of the reaction order over the intervals $0-5$, $5-10$, and $10-15$ seconds, where the order appears to be zero, we'll proceed with the assumption of first-order kinetics, since the graph is not a straight line.

The rate constant $ \mathrm{K} $ for first-order reactions can be calculated using the formula:

$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{\mathrm{A}_0}{\mathrm{A}_{\mathrm{t}}} $

For the interval where $\mathrm{A}_0 = 40 \, \mathrm{g/L}$ and $\mathrm{A}_{\mathrm{t}} = 20 \, \mathrm{g/L}$ after 10 seconds:

$ \mathrm{K} = \frac{1}{10} \ln \frac{40}{20} $

Now, to find the time $ \mathrm{t} $ required for the concentration to reduce to $2.5 \, \mathrm{g/L}$:

$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{50}{2.5} $

Equating the two expressions for $\mathrm{K}$:

$ \frac{1}{10} \ln 2 = \frac{1}{\mathrm{t}} \ln 20 $

Solving for $\mathrm{t}$:

$ \mathrm{t} = \frac{1.3010 \times 10}{0.3010} = 43.3 \, \mathrm{sec} $

Therefore, the time required for the concentration to decrease to $2.5 \, \mathrm{g/L}$ is approximately 43 seconds.

Order of Reaction:

Since $ t_{1/2} \propto [A]_0 $, the reaction follows a zero-order kinetics.

Half-life Equation for Zero-order Reaction:

The half-life ($ t_{1/2} $) is calculated as:

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

Given the slope from the graph is $76.92$, which equals $\frac{1}{2K}$. This implies:

$ K = \frac{1}{2 \times 76.92} $

Concentration of A at 10 Minutes:

Apply the zero-order kinetics equation:

$ [A]_{10} = -Kt + [A]_0 $

Substituting the values:

$ [A]_{10} = -\left(\frac{1}{2 \times 76.92}\right) \times 10 + 2.5 = 2.435 \ \text{mol L}^{-1} $

Final Concentration:

Convert to scientific notation:

$ [A]_{10} = 2435 \times 10^{-3} \ \text{mol L}^{-1} $

Thus, the concentration of A at 10 minutes is approximately $ 2435 \times 10^{-3} \ \text{mol L}^{-1} $.

To determine the overall energy of activation for the given complex reaction with rate constants $k_1$, $k_2$, and $k_3$, we start with the expression for the overall rate constant $k$:

$ k = \sqrt{\frac{k_1 k_3}{k_2}} $

The rate constant can also be expressed in terms of the Arrhenius equation:

$ A \cdot e^{-E_a / RT} = \sqrt{\frac{A_1 e^{-E_{a_1} / RT} \cdot A_3 e^{-E_{a_3} / RT}}{A_2 e^{-E_{a_2} / RT}}} $

By comparing the exponential terms from both sides, we have:

$ \frac{E_a}{RT} = \frac{1}{2} \left( \frac{E_{a_1}}{RT} + \frac{E_{a_3}}{RT} - \frac{E_{a_2}}{RT} \right) $

This simplifies to:

$ E_a = \frac{E_{a_1} + E_{a_3} - E_{a_2}}{2} $

Substituting the given activation energies—$E_{a_1} = 60 \, \text{kJ mol}^{-1}$, $E_{a_2} = 30 \, \text{kJ mol}^{-1}$, and $E_{a_3} = 10 \, \text{kJ mol}^{-1}$:

$ E_a = \frac{60 + 10 - 30}{2} = \frac{40}{2} = 20 \, \text{kJ mol}^{-1} $

Therefore, the overall energy of activation is 20 kJ/mol.

$\begin{aligned} & \mathrm{K}_{\mathrm{N}_2 \mathrm{O}_5}=2 \times 4.606 \times 10^{-2} \mathrm{~S}^{-1} \\ & 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \end{aligned}$

$\matrix{ {{P_i}} & {0.6} & 0 & 0 \cr {{P_f}} & {0.6 - P} & P & {{P \over 2}} \cr } $

$\begin{aligned} & 2 \times 4.606 \times 10^{-2}=\frac{2.303}{100} \log \frac{0.6}{0.6-\mathrm{P}} \\ & \quad 4 \log _{10} \frac{0.6}{0.6-\mathrm{P}} \\ & \quad 10^4=\frac{0.6}{0.6-\mathrm{P}} \\ & \Rightarrow 0.6 \times 10^4-10^4 \mathrm{P}=0.6 \end{aligned}$

$\begin{aligned} \begin{aligned} & \Rightarrow 10^4 \mathrm{P}=0.6\left(10^4-1\right) \\ & \mathrm{P}=(6000-0.6) \times 10^{-4} \\ &=5999 . \times 10^{-4} \\ &=0.59994 \\ & \mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{P}}{2} \\ &= 0.6+0.29997 \\ &= 0.89997 \\ &=899.97 \times 10^{-3} \\ & \text { Ans. } 900 \end{aligned} \end{aligned}$

$\mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{x}}{2}$

As given in equation

$\mathrm{K}_{\mathrm{r}}=4.606 \times 10^{-2} \mathrm{sec}^{-1}$

(Here language conflict in question)

($\mathrm{K}_{\mathrm{r}}=\frac{\mathrm{KA}}{2}$ not considered)

$\begin{aligned} \mathrm{K}_{\mathrm{r}} \mathrm{t} & =\ln \frac{0.6}{0.6-\mathrm{x}} \\ 4.606 & \times 10^{-2} \times 100=2.303 \log \frac{0.6}{0.6-\mathrm{x}} \\ \mathrm{P}_{\text {Total }} & =0.6+\frac{0.594}{2}=0.897 \mathrm{~atm} \\ \quad & =897 \times 10^{-3} \mathrm{~atm} \end{aligned}$

To determine the time required for 50% of molecule A to change into its isomeric form B, follow these steps:

Half-life Formula for First Order Kinetics:

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

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

Calculate the Rate Constant (K):

The rate constant $ K $ can be calculated using the Arrhenius equation:

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

Given:

$ A $ (frequency factor) = $ 10^{20} $

$ E_a $ (activation energy) = $ 191.48 \, \text{kJ/mol} = 191.48 \times 10^3 \, \text{J/mol} $

$ R $ (universal gas constant) = $ 8.314 \, \text{J/mol} \cdot \text{K} $

$ T = 1000 \, \text{K} $

Substitute the values into the Arrhenius equation:

$ K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} $

Simplify this calculation:

$ K = 10^{20} \times e^{-23.031} $

Simplifying further by recognizing that $ e^{-23.031} $ is a very small number, gives:

$ K \approx \frac{10^{20}}{10^{10}} = 10^{10} \, \text{sec}^{-1} $

Calculate the Half-life:

Using the calculated value of $ K $:

$ t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{seconds} $

Convert to Picoseconds:

Since $ 1 \, \text{second} = 10^{12} \, \text{picoseconds} $:

$ t_{1/2} = 6.93 \times 10^{-11} \times 10^{12} \, \text{picoseconds} = 69.3 \, \text{picoseconds} $

Therefore, the time required for 50% of the molecules of A to become B is approximately 69 picoseconds (nearest integer).

The relative error between the actual rate and the pseudo-first order rate is computed by:

$ \text{Relative Error (%)} = \frac{\text{Rate}_2 - \text{Rate}_1}{\text{Rate}_1} \times 100 $

Plugging the calculations:

$ \text{Relative Error (%)} = \frac{(0.6 \times 10 - 0.6 \times 9.6)}{0.6 \times 9.6} \times 100 = 4.1666\% $

To find the rate constant of the first-order reaction indicated, let us first understand the reaction dynamics based on the total pressure change over time. The reaction is:

$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$

Initially, only A is present. As the reaction progresses, A decreases, and B and C are formed. For each mole of A that reacts, a total of 3 moles of gas are produced (2 moles of B and 1 mole of C), leading to an increase in the total pressure of the system if the volume remains constant.

At the start: Total pressure due to A only.

Finally, when A has completely decomposed: Total pressure is due to 3 times the initial amount of A, as no A is left and for every mole of A decomposed, 3 moles of gas (2B + 1C) are produced.

The total pressure change thus directly relates to the extent of reaction, i.e., how much A has decomposed into B and C.

The total pressure at 23 sec is 200 torr, and the final total pressure is 300 torr after a very long time, indicating that the pressure increases by 100 torr due to the complete decomposition of A.

To use this information, we need to understand the relationship between pressure change and concentration in a closed system, especially for a first-order reaction. For a first-order reaction:

$\ln\left(\frac{[A]_0}{[A]}\right) = kt$

Where: $[A]_0$ is the initial concentration (or in this context, partial pressure), $[A]$ is the concentration (or partial pressure) at time t, $k$ is the rate constant, and $t$ is the time.

Let's denote the initial total pressure due to A as $P_0$. At complete decomposition, the pressure increase is from the conversion of A to 2B + C which increases the total pressure to 300 torr, indicating that initially, $P_0$ must have been 100 torr since the pressure increase is due to the tripling effect of the reaction, completing to 300 torr.

At 23 sec, the total pressure is at 200 torr. This means 100 torr is due to the unreacted A, and the additional 100 torr is from the formation of 2B + C. Given the nature of the reaction (1:3 stoichiometry of A to the total gas), we can deduce that at 23 seconds, half of A has reacted because the pressure due to the products matches the pressure due to the unreacted A.

Hence, at 23 sec, $[A] = \frac{1}{2}[A]_0$. Plugging this into the first-order reaction formula:

$\ln\left(\frac{[A]_0}{\frac{1}{2}[A]_0}\right) = kt$

$\ln(2) = k \times 23$

Given that $\log_{10}(2) = 0.301$, and knowing that $\ln(2) = \log_{e}(2)$ (where $\log_{e}(2) \approx 0.693$, for conversion from base 10 to e we can use the natural logarithm of 2 directly), we have:

$0.693 = k \times 23$

$k = \frac{0.693}{23} = 0.0301 \, \mathrm{s}^{-1}$

When expressed in the format requested, $k = 3.01 \times 10^{-2} \, \mathrm{s}^{-1}$. Rounding to the nearest integer, $k = 3 \times 10^{-2} \, \mathrm{s}^{-1}$.

$\begin{aligned} & x=8, y=0 \\ & \frac{r_2}{r_1}=\left(\frac{C_{\mathrm{A}_2}}{C_{\mathrm{A}_1}}\right)^2\left(\frac{\mathrm{C}_{\mathrm{B}_2}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{2 \mathrm{C}_{\mathrm{A}_1}}{\mathrm{C}_{\mathrm{A}_1}}\right)^2\left(\frac{2 \mathrm{C}_{\mathrm{B}_1}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \mathrm{r}_2=8 \mathrm{r}_1 \\ & x=8 \end{aligned}$

$\begin{aligned} & \Rightarrow \text { Zero order } \\ & y=0 \end{aligned}$

To determine the overall order of the reaction given by $\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$, we can derive the information based on the given details about the kinetics of reactant A and the graphical behavior of reactant B.

The first piece of information tells us that the time for the concentration of A to reduce to $1/4$ of its initial concentration ($[A]_0/4$) is twice the time it takes to reduce to half of its initial concentration ($[A]_0/2$). This characteristic is a hallmark of first-order reactions. In first-order reactions, the time it takes for the concentration of the reactant to reduce to half of its initial value (known as the half-life, $t_{1/2}$) is constant and does not depend on the initial concentration. Specifically, the fact that the concentration decreases by a factor of 4 (to $1/4$ its initial value) in twice the time it takes to decrease by a factor of 2 (to $1/2$ its initial value) indicates a constant half-life, consistent with first-order kinetics for reactant A.

As for reactant B, the information provided is that a plot of its concentration change over time is a straight line with a negative slope and a positive intercept on the concentration axis. This description matches the behavior of a reactant in a reaction of zero-order kinetics with respect to B, where the rate of reaction is constant and independent of the concentration of B. In zero-order reactions, the concentration of the reactant decreases linearly over time, since the rate of decrease is constant.

Therefore, considering the kinetics of both A and B:

• Reactant A shows first-order kinetics.
• Reactant B shows zero-order kinetics.

The overall order of the reaction is the sum of the individual orders with respect to each reactant, which gives us:

$\text{Overall order} = \text{Order with respect to A} + \text{Order with respect to B} = 1 (A) + 0 (B) = 1.$

Hence, the overall order of the reaction is 1.

$\mathrm{A+B \rightarrow C}$ Reaction .... (1)

$\mathrm{P} \rightarrow \mathrm{Q} \quad$ Reaction .... (2)

$\frac{\left(t_{\frac{1}{2}}\right)_1}{\left(t_{\frac{1}{2}}\right)_2}=\frac{k}{k_1}= \frac{5}{2}$

$\begin{aligned} & \frac{\frac{t_2}{3}}{t_{\frac{4}{5}}^5}=\frac{k_2}{k_1} \frac{\log 3}{\log 5} \\ & =\left(\frac{5}{2}\right) \frac{0.477}{0.699}=1.7 \\ & =17 \times 10^{-1} \\ & =x=17 \end{aligned}$

To determine the time required for a certain level of completion ($x\%$) of a first-order reaction, we can use the formula that relates the time $ t $, the rate constant $ k $, and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ and the concentration at time $t$, $[A]_t$, is given by the integrated rate law for first-order reactions:

$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$

For a given percentage of completion, we use the fact that $[A]_t = [A]_0(1-\frac{x}{100})$, where $x$ is the percentage completion. Thus, the equation becomes:

$\ln\left(\frac{[A]_0}{[A]_0(1-\frac{x}{100})}\right) = kt$

Simplifying, we get:

$\ln\left(\frac{1}{1-\frac{x}{100}}\right) = kt$

Let's calculate the time required for $99.9\%$ and $90\%$:

• For $99.9\%$ completion ($x = 99.9\%$):

$\ln\left(\frac{1}{1-\frac{99.9}{100}}\right) = kt_{99.9}$

$\ln\left(\frac{1}{0.001}\right) = kt_{99.9}$

$\ln(1000) = kt_{99.9}$

• For $90\%$ completion ($x = 90\%$):

$\ln\left(\frac{1}{1-\frac{90}{100}}\right) = kt_{90}$

$\ln\left(\frac{1}{0.1}\right) = kt_{90}$

$\ln(10) = kt_{90}$

Since $k$ is a constant for a given reaction at a constant temperature, we can compare the times $t_{99.9}$ and $t_{90}$ directly by comparing the logarithmic values:

$\frac{t_{99.9}}{t_{90}} = \frac{\ln(1000)}{\ln(10)}$

Using the property of logarithms, we know that $\ln(1000) = 3\ln(10)$ because $1000 = 10^3$, so:

$\frac{t_{99.9}}{t_{90}} = \frac{3\ln(10)}{\ln(10)} = 3$

Therefore, the time required for $99.9\%$ completion of a first-order reaction is $3$ times the time required for completion of $90\%$ of the reaction. The nearest integer to this value is $3$.

$2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})$

As this is single step reaction,

$\mathrm{r}=\mathrm{k}_{\mathrm{f}}[\mathrm{A}]^2[\mathrm{~B}]^2$

$\begin{array}{lllll} & 2 \mathrm{~A}(\mathrm{~g})+ & \mathrm{B}(\mathrm{g}) \rightarrow & \mathrm{C}(\mathrm{g}) \\ \mathrm{t}=0 & 1.5 & 0.7 & 0 \\ \mathrm{t}=\mathrm{t}^{\prime} & 1.5-2 \mathrm{x} \quad & 0.7-\mathrm{x} & \mathrm{x} \end{array}$

$\begin{aligned} & \Rightarrow x=0.5 \mathrm{~atm} \\ & \mathrm{r}_1=\mathrm{k}_{\mathrm{f}}(1.5)^2(0.7) \\ & \mathrm{r}_2=\mathrm{k}_{\mathrm{f}}(0.5)^2(0.2) \\ & \frac{r_1}{r_2}=\frac{9 \times 7}{2}=31.5 \\ & =315 \times 10^{-1} \end{aligned}$

$\begin{aligned} & \text { Rate }=k[A]^x[B]^y \\ & \frac{6 \times 10^{-}}{2.4 \times 10^{-}}=\left(\frac{0.1}{0.4}\right)^x \Rightarrow x=1 \\ & \frac{7.2 \times 10^{-}}{2.88 \times 10^{-}}=\left(\frac{0.2}{0.4}\right)^y \Rightarrow y=2 \end{aligned}$

$\begin{aligned} & A+B \rightarrow C \\ & \frac{-d[A]}{d t}=k[A]^{1 / 2}[B]^{1 / 2} \end{aligned}$

Since, $[A]=[B]$

$\begin{aligned} & \Rightarrow \quad \frac{-d[A]}{d t}=k[A] \\ & \Rightarrow \quad k t=\ln \frac{[A]_0}{[A]} \\ & \Rightarrow \quad t=\frac{1}{4.6 \times 10^{-2}} \times \ln \left(\frac{1}{0.1}\right) \\ & =\frac{2.303}{4.6} \times 100 \approx 50 \end{aligned}$

$\begin{aligned} & k=\frac{k_1 k_2}{k_3} \\ & E_{a_{e f f}}=E_{a_1}+E_{a_2}-E_{a_3} \end{aligned}$

$\begin{aligned} & 400=300+200-E_{\mathrm{a}_3} \\ & 400=500-E_{\mathrm{a}_3} \\ & E_{\mathrm{a}_3}=100 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned}$

The given problem involves calculating the age of a wood sample using the carbon-14 dating method. Carbon-14 ($^{14}C$) is a radioactive isotope of carbon that decays over time, and its ratio compared to carbon-12 ($^{12}C$) can be used to date organic materials. The ratio of $\frac{^{14}C}{^{12}C}$ in the wood is $\frac{1}{8}$th of that in the atmosphere, suggesting the $^{14}C$ has decayed. The half-life of $^{14}C$ is given as 5730 years, which is the time for half the $^{14}C$ to decay.

To solve for the age of the wood, we use the formula for radioactive decay:

where $N$ is the remaining amount of $^{14}C$, $N_0$ is the original amount of $^{14}C$, $\lambda$ is the decay constant, and $t$ is the time (age of the wood).

Since the ratio $\frac{N}{N_0} = \frac{1}{8}$, we infer that $N = \frac{N_0}{8}$. Substituting into the decay equation and solving for $t$, first we find the decay constant $\lambda$:

Substitute $\lambda$ and rearrange to solve for $t$:

This calculation shows that the wood sample is 17190 years old.

$\mathrm{r}=\mathrm{k}[\mathrm{A}]$

So, order of reaction $=1$

$\mathrm{t}_{1 / 2}=120 \mathrm{~min}$

For $90 \%$ completion of reaction

$\begin{aligned} & \Rightarrow \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right) \\ & \Rightarrow \frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{2.303}{\mathrm{t}} \log \frac{100}{10} \\ & \therefore \mathrm{t}=399 \mathrm{~min} . \end{aligned}$

Rate of reaction (ROR)

$\begin{aligned} & =-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \text { ROR }=\frac{1}{240} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}$

Rate of formation of $\mathrm{NO}_2=\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}$

$=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3} \text {. }$

$\begin{aligned} & 0.04=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \quad \text{..... (1)}\\ & 0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \quad \text{..... (2)} \end{aligned}$

$\begin{aligned} \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}(2-1)} /(2) \\ \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}} \\ \ln \frac{4}{3} & =600 \mathrm{k} \\ \ln \frac{4}{3} & =600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}} \\ \mathrm{t}_{1 / 2} & =600 \frac{\ln 2}{\ln \frac{4}{3}} \sec \\ \mathrm{t}_{1 / 2} & =600 \times \frac{\log 2}{\log 4-\log 3} \mathrm{sec} .=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min} \\ \mathrm{t}_{1 / 2} & =24.08 \mathrm{~min} \end{aligned}$

Ans. 24

Half life of bromine $-82=36$ hours

$\begin{aligned} & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{0.693}{36}=0.01925 \mathrm{~hr}^{-1} \\ & 1^{\text {st }} \text { order rxn kinetic equation } \\ & \mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{\mathrm{t} \times \mathrm{K}}{2.303}(\mathrm{t}=1 \text { day }=24 \mathrm{~hr}) \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{24 \mathrm{hr} \times 0.01925 \mathrm{hr}^{-1}}{2.303} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=0.2006 \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\mathrm{anti} \log (0.2006) \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=1.587 \\ & \text { If }=1 \\ & \frac{1}{1-\mathrm{x}}=1.587 \Rightarrow 1-\mathrm{x}=0.6301=\text { Fraction remain } \\ & \mathrm{after} \text { one day } \end{aligned}$

$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}=\frac{\mathrm{A}_1 \cdot \mathrm{A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}} \\ & \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}}\end{aligned}$

$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}=40+50-60=30 \mathrm{~kJ} / \mathrm{mole} .$

$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{1 / 2}}=\frac{\frac{2.303}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)}{\frac{2.303}{\mathrm{k}} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^3}{\log 2}=\frac{3}{0.3}=10$

Let, $\mathrm{R}=\mathrm{k}[\mathrm{HI}]^{\mathrm{n}}$

using any two of given data,

$\begin{aligned} & \frac{3 \times 10^{-3}}{7.5 \times 10^{-4}}=\left(\frac{0.01}{0.005}\right)^{\mathrm{n}} \\\\ & \mathrm{n}=2 \end{aligned}$

1. Given Data:

• Ratio of mass of $ \mathrm{Pb}^{206} $ to $ \mathrm{U}^{238} $ = $ \frac{7}{1} $


• Mass of $ \mathrm{Pb}^{206} $ = 7 g


• Mass of $ \mathrm{U}^{238} $ = 1 g


• Half-life of $ \mathrm{U}^{238} $ = $ 4.5 \times 10^9 $ years


• $ \log_e 2 = 0.693 $

1. Calculate Moles:

• Moles of $ \mathrm{Pb}^{206} $:

$ \text{Moles of } \mathrm{Pb}^{206} = \frac{7}{206} = 34 \times 10^{-3} \text{ moles} $

• Moles of $ \mathrm{U}^{238} $:

$ \text{Moles of } \mathrm{U}^{238} = \frac{1}{238} = 4.2 \times 10^{-3} \text{ moles} $

1. Initial Moles of $ \mathrm{U}^{238} $ ( $ N_0 $ ):

$ N_0 = \text{Moles of } \mathrm{U}^{238} \text{ initially} = \text{Moles of } \mathrm{Pb}^{206} + \text{Moles of remaining } \mathrm{U}^{238} $

$ N_0 = 34 \times 10^{-3} + 4.2 \times 10^{-3} = 38.2 \times 10^{-3} \text{ moles} $

1. Calculate the Time:

• Using the decay formula:

$ \lambda t = \ln \frac{N_0}{N_t} $

• Where $ \lambda $ (decay constant):

$ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9 \text{ years}} $

• Calculate $ \ln \frac{N_0}{N_t} $:

$ \ln \frac{N_0}{N_t} = \ln \frac{38.2}{4.2} = \ln 9.09 $

• Simplify $ \ln 9.09 $:

$ \ln 9.09 \approx 2.21 $

1. Plug in the Values:

$ t = \frac{4.5 \times 10^9 \text{ years}}{0.693} \times \ln 9.09 $

$ t = 4.5 \times 10^9 \text{ years} \times \frac{2.21}{0.693} $

$ t \approx 4.5 \times 10^9 \times 3.184 \approx 14.33 \times 10^9 \text{ years} $

1. Convert to $ P \times 10^8 $:

$ t = P \times 10^8 \text{ years} $

$ P \times 10^8 = 14.33 \times 10^9 $

$ P = 143.28 $

Thus, the value of $ P $ is $ \boxed{143.28} $.

The given reaction is:

$2 \mathrm{H}_2(\mathrm{~g}) + 2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g}) + 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

And it follows the mechanism below:

$ \begin{array}{ll} 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) & \text { (fast equilibrum) } \\\\ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (slow reaction) } \\\\ \mathrm{N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_3} \mathrm{~N}_2(\mathrm{~g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (fast reaction) } \end{array} $

To determine the order of the reaction, we need to focus on the slow step, as it is the rate-determining step. The slow step is:

$ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) $

The rate law for this step can be written as:

$ \text{Rate} = k_2 [\mathrm{N}_2\mathrm{O}_2][\mathrm{H}_2] $

However, $[\mathrm{N}_2 \mathrm{O}_2]$ is an intermediate and its concentration can be expressed in terms of the reactants. From the fast equilibrium step:

$ 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) $

The equilibrium constant for this step can be written as:

$\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}$ = $ K = \frac{[\mathrm{N}_2 \mathrm{O}_2]}{[\mathrm{NO}]^2} $

Hence, the concentration of the intermediate $\mathrm{N}_2 \mathrm{O}_2$ can be expressed as:

$ [\mathrm{N}_2 \mathrm{O}_2] = K[\mathrm{NO}]^2 $

Substituting this into the rate law for the slow step gives:

$ \text{Rate} = k_2 K [\mathrm{NO}]^2[\mathrm{H}_2] $

Let's combine $k_2$ and $K$ into a single constant, say $k'$:

$ \text{Rate} = k' [\mathrm{NO}]^2[\mathrm{H}_2] $

This implies that the reaction is second-order in $\mathrm{NO}$ and first-order in $\mathrm{H}_2$. Therefore, the overall order of the reaction is:

$ 2 + 1 = 3 $

Thus, the order of the reaction is 3.

The overall order of a reaction is determined by the slow (rate-determining) step.

Here, the slow step is: $ \text{NOBr}_2 + \text{NO} \rightarrow 2 \text{NOBr} $

This is a second-order reaction: first-order with respect to $\text{NOBr}_2$ and first-order with respect to NO.

However, $\text{NOBr}_2$ is not a reactant in the overall reaction. It's an intermediate. So, we need to express it in terms of the initial reactants.

From the first (fast) step, we get: $ \text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2 $

In the steady state approximation, the rate of formation of $\text{NOBr}_2$ equals the rate of its consumption. We can write this as:

$ k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2] + k_2[\text{NOBr}_2][\text{NO}] $

Since the second reaction (slow step) is much slower than the first one, $k_2[\text{NOBr}_2][\text{NO}]$ term is negligible in comparison to $k_{-1}[\text{NOBr}_2]$.

So, we can simplify to:

$ k_1[\text{NO}][\text{Br}_2] \approx k_{-1}[\text{NOBr}_2] $

Solving for $[\text{NOBr}_2]$, we get: $ [\text{NOBr}_2] \approx \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] $

Substitute $[\text{NOBr}_2]$ into the rate equation for the slow step:

$ \text{Rate} = k_2[\text{NO}][\text{NOBr}_2] $

$ \text{Rate} = k_2[\text{NO}] \left( \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] \right) $

So the overall reaction order is 3 (2 with respect to NO and 1 with respect to $\text{Br}_2$).

A. The successive half lives of zero order reactions decreases with time. This statement is correct. The half-life of a zero-order reaction is given by the formula $t_{1/2} = [A]_0/2k$, where $[A]_0$ is the initial concentration and k is the rate constant. As the concentration decreases, the half-life also decreases.

B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction. This statement is correct. For example, in a zero-order reaction, the rate of reaction does not depend on the concentration of reactants.

C. Order and molecularity of a chemical reaction can be a fractional number. This statement is incorrect. The order of a reaction can be zero, fractional or negative. However, molecularity, which refers to the number of reactant molecules taking part in an elementary reaction, can only be a whole number.

D. The rate constant units of zero and second order reaction are $\mathrm{mol} ~\mathrm{L}^{-1} \mathrm{~s}^{-1}$ and $\mathrm{mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}$ respectively. This statement is correct. The units of the rate constant for a zero-order reaction are M/s or mol·L⁻¹·s⁻¹ . The units of the rate constant for a second-order reaction are 1/(M·s) or L·mol⁻¹·s⁻¹ .

In the given problem, a molecule is undergoing two separate, simultaneous first-order reactions. Each of these reactions has its own rate constant and half-life. The half-life ($T$) of a first-order reaction is related to its rate constant ($k$) by the equation:

$T = \frac{\ln(2)}{k}$

When two or more first-order reactions are occurring independently and simultaneously (also known as parallel or competing reactions), the overall rate constant for the disappearance of the reactant is the sum of the rate constants for the individual reactions:

$k_{\text{total}} = k_1 + k_2$

Conversely, the overall half-life for the disappearance of the reactant is determined by the reciprocal of the sum of the reciprocals of the individual half-lives:

$\frac{1}{T_{\text{total}}} = \frac{1}{T_1} + \frac{1}{T_2}$

This equation reflects the fact that the reactant is disappearing more quickly than it would due to any single reaction, because it is being consumed by two reactions at once.

Substituting the given half-lives of $12$ minutes and $3$ minutes into this equation gives:

$\begin{align} \frac{1}{T_{\text{total}}} &= \frac{1}{12\text{ min}} + \frac{1}{3\text{ min}} \\\\ &= \frac{5}{12\text{ min}} \end{align}$

Solving for $T_{\text{total}}$ then gives:

$T_{\text{total}} = \frac{12}{5}\text{ min} = 2.4\text{ min}$

This is the time taken for $50\%$ of the reactant to be consumed when both reactions are occurring. Rounding to the nearest integer gives an answer of $2$ minutes.

Clearly, if $\mathrm{E}_a=0, \mathrm{~K}$ is temperature independent if $\mathrm{E}_a>0, \mathrm{~K}$ increase with increase in temperature if $\mathrm{E}_a<0, \mathrm{~K}$ decrease with increase in temperature

• Rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.


• Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.


• The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
  
  a. A high activation energy usually implies a slow reaction.
  
  b. $k=\mathrm{P} \times \mathrm{Z} \times \mathrm{e}^{-\mathrm{E}_a / \mathrm{RT}}$
  
  c. The pre-exponential factor $(\mathrm{A}=\mathrm{P} \times \mathrm{Z})$ is independent of the activation energy and the energy of molecules.