Step 1: Identifying Parameters and Constants

Radioactive decay follows first-order kinetics [Per the Integrated Rate Law for first-order reactions].

• Half-life ($t_{1/2}$) = $5730 \text{ years}$
• Remaining concentration fraction ($\frac{[A]_t}{[A]_0}$) = $80\% = 0.80$
• Decay constant ($k$): Calculated via the relation $k = \frac{\ln(2)}{t_{1/2}}$

Step 2: Calculating the Decay Constant ($k$)

Using the definition of half-life for a first-order process:

$k = \frac{0.693}{5730 \text{ years}}$

$k \approx 1.2094 \times 10^{-4} \text{ year}^{-1}$ [Mathematical evaluation of $\frac{0.693}{5730}$]

Step 3: Applying the First-Order Integrated Rate Equation

The age of the sample ($t$) is determined by the formula [Derived from the Integrated Rate Law]:

$t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right)$

Substituting the known values:

$t = \frac{2.303}{1.2094 \times 10^{-4}} \log\left(\frac{1}{0.80}\right)$

Step 4: Algebraic Computation

First, calculate the logarithmic component:

$\log(1.25) \approx 0.0969$ [Logarithmic property: $\log(1) - \log(0.8)$]

Now, solve for $t$:

$t = (1.9042 \times 10^4) \times 0.0969$

$t \approx 1845 \text{ years}$

Visual Representation of Decay

Final Solution: The age of the archaeological artifact is approximately 1845 years.

Step 1: Identifying the Order of Reaction

The rate law for the given reaction is provided as:

$Rate = k[NO]^2$

By definition, the order of a reaction with respect to a specific reactant is the exponent to which its concentration term is raised in the rate law expression [Per the Rate Law definition].

• The concentration term is $[NO]$.
• The exponent (power) to which $[NO]$ is raised is $2$.

Therefore, the overall order of the reaction $n$ is $2$. This identifies the reaction as a second-order reaction.

Step 2: Deriving Dimensions of the Rate Constant ($k$)

The general rate expression for a reaction of order $n$ is given by:

$Rate = k[Concentration]^n$

We rearrange the formula to solve for $k$:

$k = \frac{Rate}{[Concentration]^n}$

Substituting the standard units for the variables involved:

• Units of Rate: $\text{mol L}^{-1} \text{s}^{-1}$
• Units of Concentration: $\text{mol L}^{-1}$
• Order ($n$): $2$

Substituting these into the expression for $k$:

$k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2}$

Step 3: Algebraic Simplification of Units

Performing the division of units:

$k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^2 \text{L}^{-2}}$

$k = (\text{mol L}^{-1})^{1-2} \text{s}^{-1}$

$k = (\text{mol L}^{-1})^{-1} \text{s}^{-1}$

$k = \text{mol}^{-1} \text{L}^{1} \text{s}^{-1}$

The resulting dimensions for the rate constant of a second-order reaction are $\text{L mol}^{-1} \text{s}^{-1}$.

Final Solution: The order of the reaction is 2 (second-order), and the dimensions of the rate constant $k$ are $\text{L mol}^{-1} \text{s}^{-1}$.

Step 1: Establishing the First Order Rate Equation

For a first-order reaction, the rate constant $k$ is given by the integrated rate law:

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

Where $[A]_0 = 100$ and $[A]_t = (100 - x)$. Thus, for completion $x$, $k = \frac{2.303}{t} \log \frac{100}{100-x}$.

Step 2: Relating Rates at 298 K and 308 K

Given the time $t$ for 10% completion at $T_1 = 298\text{ K}$ is equal to the time for 25% completion at $T_2 = 308\text{ K}$:

$t = \frac{2.303}{k_1} \log \frac{100}{90} = \frac{2.303}{k_2} \log \frac{100}{75}$

$\frac{k_2}{k_1} = \frac{\log(1.333)}{\log(1.111)} \approx \frac{0.1249}{0.0457} \approx 2.733$

Step 3: Calculating Activation Energy ($E_a$)

Using the Arrhenius equation in logarithmic form: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$

Substitute the values ($R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$):

$\log(2.733) = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)$

$0.4366 = \frac{E_a}{19.147} \left( \frac{10}{91824} \right)$

$E_a = \frac{0.4366 \times 19.147 \times 91824}{10} \approx 76718\text{ J mol}^{-1} \approx 76.72\text{ kJ mol}^{-1}$

Step 4: Calculating Rate Constant at 318 K ($k_3$)

Using the Arrhenius equation: $k = A e^{-E_a / RT}$

Given $A = 4 \times 10^{10}\text{ s}^{-1}$ and $T_3 = 318\text{ K}$:

$\ln k_3 = \ln A - \frac{E_a}{RT_3}$

$\ln k_3 = \ln(4 \times 10^{10}) - \frac{76718}{8.314 \times 318}$

$\ln k_3 = 25.412 - 29.023 = -3.611$

$k_3 = e^{-3.611} \approx 0.027\text{ s}^{-1}$

Final Solution: The activation energy $E_a$ is $76.72\text{ kJ mol}^{-1}$ and the rate constant at 318 K is $0.027\text{ s}^{-1}$.

Step 1: Establishing the Rate Law Expression

The rate law for the reaction $2A + B \rightarrow C + D$ can be expressed as:

$Rate = k[A]^x[B]^y$

Where $k$ is the rate constant, $x$ is the order with respect to $A$, and $y$ is the order with respect to $B$. [Per the Law of Mass Action and Empirical Kinetics]

Step 2: Tabulating the Experimental Data

Step 3: Determining Order with Respect to B ($y$)

Compare Experiments II and III, where $[A]$ is held constant:

$\frac{Rate_{III}}{Rate_{II}} = \frac{k[0.3]^x[0.4]^y}{k[0.3]^x[0.2]^y}$

$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \left(\frac{0.4}{0.2}\right)^y$

$4 = 2^y \implies y = 2$

Step 4: Determining Order with Respect to A ($x$)

Compare Experiments I and IV, where $[B]$ is held constant:

$\frac{Rate_{IV}}{Rate_{I}} = \frac{k[0.4]^x[0.1]^y}{k[0.1]^x[0.1]^y}$

$\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \left(\frac{0.4}{0.1}\right)^x$

$4 = 4^x \implies x = 1$

Step 5: Calculating the Rate Constant ($k$)

Substitute the orders back into the rate law: $Rate = k[A]^1[B]^2$. Using data from Experiment I:

$6.0 \times 10^{-3} = k(0.1)^1(0.1)^2$

$6.0 \times 10^{-3} = k(0.1)(0.01)$

$6.0 \times 10^{-3} = k(0.001)$

$k = \frac{6.0 \times 10^{-3}}{10^{-3}} = 6.0 \ L^2 \cdot mol^{-2} \cdot min^{-1}$

Step 6: Visual Representation of Kinetic Dependency

Final Solution: The rate law is $Rate = k[A][B]^2$ and the rate constant $k = 6.0 \ L^2 \cdot mol^{-2} \cdot min^{-1}$.

Step 1: Theoretical Foundation of Rate Laws

The rate of a chemical reaction, expressed as the change in concentration of reactants over time, is proportional to the product of the molar concentrations of the reactants, each raised to the power of their respective stoichiometric orders [Per the Law of Mass Action and Rate Law definitions].

Step 2: Defining the Variables and Order

Given the reaction involves reactants $A$ and $B$ with specified orders:

• Order with respect to $A$ ($\alpha$) = $1$
• Order with respect to $B$ ($\beta$) = $2$
• The rate of reaction is represented by the differential change in concentration of a reactant over an infinitesimal time interval $dt$.

Step 3: Constructing the Differential Rate Equation

The general rate expression is given by:

$\text{Rate} = -\frac{d[R]}{dt} = k[A]^\alpha [B]^\beta$

Substituting the given orders into the expression:

$\text{Rate} = -\frac{d[A]}{dt} = k[A]^1 [B]^2$

Step 4: Representation of the Differential Rate Law

To explicitly state the differential rate equation for the consumption of reactants or the formation of products, we represent the rate of disappearance of the reactants relative to the rate constant $k$:

$\frac{dx}{dt} = k[A][B]^2$

Where:

• $\frac{dx}{dt}$ represents the rate of reaction.
• $k$ is the rate constant for the specific temperature.
• $[A]$ is the molar concentration of reactant $A$.
• $[B]^2$ is the molar concentration of reactant $B$ squared.

Visual Representation of Reaction Dependency

Final Solution: The differential rate equation for the reaction is $\frac{dx}{dt} = k[A][B]^2$.

Step 1: Identifying the Order of Reaction

The rate law for the given reaction is provided as:

$Rate = k[NO]^{2}$

The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression [Per the definition of Rate Law].

• The concentration term is $[NO]$.
• The exponent associated with $[NO]$ is $2$.

Therefore, the order of the reaction is $n = 2$ [By summation of rate exponents].

Step 2: Deriving the Dimensions of the Rate Constant ($k$)

The rate equation is given by $Rate = k[Concentration]^{n}$. To find the units of the rate constant $k$, we rearrange the equation as follows:

$k = \frac{Rate}{[Concentration]^{n}}$

Substituting the standard units for the variables involved:

• $Rate$ is expressed in units of concentration per unit time: $mol \cdot L^{-1} \cdot s^{-1}$.
• $[Concentration]$ is expressed in molarity: $mol \cdot L^{-1}$.
• The order of the reaction $n = 2$.

Substituting these values into the rearranged equation:

$k = \frac{mol \cdot L^{-1} \cdot s^{-1}}{(mol \cdot L^{-1})^{2}}$

$k = \frac{mol \cdot L^{-1} \cdot s^{-1}}{mol^{2} \cdot L^{-2}}$

$k = mol^{(1-2)} \cdot L^{(-1 - (-2))} \cdot s^{-1}$

$k = mol^{-1} \cdot L^{1} \cdot s^{-1}$

Step 3: Tabular Summary of Findings

Final Solution: The order of the reaction is 2, and the dimensions of the rate constant are $L \cdot mol^{-1} \cdot s^{-1}$.

Given Variables & Initial Setup

For a first-order chemical reaction, the rate constant ($k$) and the concentration of the reactant at any given time ($t$) are governed by the integrated rate law:

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

Where:

• $k = 60 \, \text{s}^{-1}$ [Given rate constant]
• $[R]_0 = \text{Initial concentration}$
• $[R]_t = \frac{1}{16} [R]_0$ [Target concentration]
• $t = \text{Time required (to be determined)}$

Step 1: Substitution of Values into the Rate Equation

Substitute the ratio of concentrations into the integrated rate law equation [Per the First-Order Kinetics derivation]:

$60 = \frac{2.303}{t} \log \frac{[R]_0}{\frac{1}{16}[R]_0}$

$60 = \frac{2.303}{t} \log(16)$

Step 2: Logarithmic Calculation

To evaluate $\log(16)$, we express the value as a power of 2 [By the property of logarithms $\log(a^n) = n \log a$]:

$16 = 2^4$

$\log(16) = \log(2^4) = 4 \times \log(2)$

Given $\log(2) \approx 0.3010$:

$\log(16) = 4 \times 0.3010 = 1.2040$

Step 3: Algebraic Solving for Time ($t$)

Rearrange the equation to isolate $t$:

$t = \frac{2.303 \times 1.2040}{60}$

$t = \frac{2.7728}{60}$

$t = 0.04621 \, \text{s}$

Visual Representation of Concentration Decay

Final Verification

The calculation is based on the standard exponential decay model $N(t) = N_0 e^{-kt}$. Using the conversion $t = \frac{\ln(16)}{k}$, we confirm the consistency of the logarithmic base conversion [Logarithmic Identity: $\log_{10}(x) = \frac{\ln(x)}{2.303}$].

Final Solution: The time taken for the reaction to reach $\frac{1}{16}$ of its initial concentration is approximately $4.62 \times 10^{-2} \, \text{s}$.

Step 1: Given Variables & Theoretical Foundation

To determine the pre-exponential factor ($A$), we utilize the Arrhenius equation, which relates the rate constant ($k$), activation energy ($E_a$), temperature ($T$), and the frequency factor ($A$).

• Rate constant ($k$) = $2.418 \times 10^{-5} \, s^{-1}$
• Activation energy ($E_a$) = $179.9 \, kJ/mol = 179,900 \, J/mol$ [Conversion: $1 \, kJ = 1000 \, J$]
• Temperature ($T$) = $546 \, K$
• Gas constant ($R$) = $8.314 \, J \cdot K^{-1} \cdot mol^{-1}$

[Per the Arrhenius Equation: $k = A \cdot e^{-E_a / RT}$]

Step 2: Linearizing the Arrhenius Equation

To solve for $A$, we apply the natural logarithm ($\ln$) to both sides of the equation:

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

Rearranging to isolate the pre-exponential factor ($A$):

$\ln(A) = \ln(k) + \frac{E_a}{RT}$

Step 3: Numerical Calculation

First, calculate the exponent component $\frac{E_a}{RT}$:

$\frac{E_a}{RT} = \frac{179900 \, J/mol}{(8.314 \, J/K \cdot mol) \times 546 \, K}$

$\frac{E_a}{RT} = \frac{179900}{4539.444} \approx 39.630$

Next, calculate $\ln(k)$:

$\ln(2.418 \times 10^{-5}) = \ln(2.418) + \ln(10^{-5})$

$\ln(2.418 \times 10^{-5}) = 0.8829 - 11.5129 = -10.630$

Now, calculate $\ln(A)$:

$\ln(A) = -10.630 + 39.630 = 29.00$

Step 4: Exponential Conversion

Taking the antilog (exponential) of both sides:

$A = e^{29.00}$

$A \approx 3.92 \times 10^{12} \, s^{-1}$

Visual Representation of Arrhenius Relationship

Final Solution: The pre-exponential factor ($A$) for the decomposition is $3.92 \times 10^{12} \, s^{-1}$.

Step 1: Determining the Order of Reaction and Rate Constant

Based on the provided experimental data for the decomposition of $N_{2}O_{5}$ at $318\text{ K}$, the reaction follows first-order kinetics. This is confirmed by the linear relationship between $\ln[N_{2}O_{5}]$ and time, or by checking the consistency of the rate constant $k$ using the first-order integrated rate law: $k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}}$ [Per the First-Order Integrated Rate Law].

Using the data points at $t = 0$ ($[A]_{0} = 1.63 \times 10^{-2} \text{ M}$) and $t = 184 \text{ s}$ ($[A]_{t} = 1.36 \times 10^{-2} \text{ M}$):

Step 2: Deriving the Half-Life Period

For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and is defined as the time taken for the concentration of the reactant to reduce to half of its initial value. The relationship is derived from the integrated rate law where $[A]_{t} = \frac{[A]_{0}}{2}$ at $t = t_{1/2}$:

Step 3: Numerical Calculation

Substituting the calculated value of $k$ into the half-life formula:

Visual Representation of Decomposition

Final Solution: The half-life period ($t_{1/2}$) for the decomposition of $N_{2}O_{5}$ at 318 K is approximately 704.27 seconds.

Given Variables & Initial Setup

To determine the Activation Energy ($E_a$), we utilize the Arrhenius equation in its integrated logarithmic form, which relates the rate constants ($k$) at two different temperatures ($T_1$ and $T_2$).

• Initial Temperature ($T_1$): $293\text{ K}$
• Final Temperature ($T_2$): $313\text{ K}$
• Rate Constant Ratio: $\frac{k_2}{k_1} = 4$ [Given: Rate quadruples]
• Universal Gas Constant ($R$): $8.314\text{ J K}^{-1}\text{ mol}^{-1}$

Step 1: The Arrhenius Equation Formulation

The Arrhenius equation for two temperatures is defined as [Per the Arrhenius Principle]:

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

Step 2: Substitution of Numerical Values

Substitute the given variables into the equation:

$\log(4) = \frac{E_a}{2.303 \times 8.314} \left( \frac{313 - 293}{293 \times 313} \right)$

Knowing $\log(4) \approx 0.6021$:

$0.6021 = \frac{E_a}{19.147} \left( \frac{20}{91709} \right)$

Step 3: Algebraic Isolation of Activation Energy ($E_a$)

Rearranging the equation to solve for $E_a$:

$E_a = \frac{0.6021 \times 19.147 \times 91709}{20}$

$E_a = \frac{1056721.2}{20}$

$E_a = 52836.06\text{ J mol}^{-1}$

Step 4: Final Conversion to Standard Units

Converting Joules to Kilojoules for standard thermodynamic reporting [By standard SI convention]:

$E_a = 52.84\text{ kJ mol}^{-1}$

Final Solution: The energy of activation for the reaction is $52.84\text{ kJ mol}^{-1}$.

Step 1: Theoretical Foundation

For a reaction following first-order kinetics, the rate constant ($k$) is related to the half-life ($t_{1/2}$) of the reaction. The derivation proceeds from the integrated rate equation:

$[R] = [R]_0 e^{-kt}$

At the half-life ($t = t_{1/2}$), the concentration of the reactant is half of its initial value, i.e., $[R] = \frac{[R]_0}{2}$. Substituting this into the integrated rate law results in the standard expression for half-life [Per the derivation of First-Order Integrated Rate Law]:

$t_{1/2} = \frac{\ln(2)}{k}$

Using the numerical approximation $\ln(2) \approx 0.693$, the formula simplifies to:

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

Step 2: Given Parameters

The rate constant ($k$) for the first-order reaction is provided as follows:

• $k = 2 \text{ min}^{-1}$

Step 3: Numerical Substitution and Calculation

Substitute the value of $k$ into the half-life expression:

$t_{1/2} = \frac{0.693}{2 \text{ min}^{-1}}$

Performing the arithmetic division:

$t_{1/2} = 0.3465 \text{ min}$

Visual Representation of First-Order Decay

Step 4: Unit Conversion (Optional)

To express the result in seconds, multiply by 60 seconds per minute [Unit factor method]:

$t_{1/2} = 0.3465 \text{ min} \times 60 \text{ s/min} = 20.79 \text{ s}$

Final Solution: The half-life of the reaction is 0.3465 min (or 20.79 s).

Initial Theoretical Framework

For a chemical reaction involving a reactant $A$, the rate law expression for a reaction that is second-order with respect to that reactant is defined as:

$Rate = k[A]^2$

Where:

• $Rate_1$ = Initial rate of the reaction
• $k$ = Rate constant for the reaction at a constant temperature
• $[A]_1$ = Initial concentration of the reactant

Step 1: Defining the Initial and Final States

Let the initial concentration of the reactant be $[A]_1 = [A]$.

The initial rate of the reaction is represented by:

$Rate_1 = k[A]^2$ [Per the definition of Rate Law]

According to the problem statement, the final concentration is reduced to half of the initial concentration:

$[A]_2 = \frac{1}{2}[A]$

Step 2: Deriving the New Rate of Reaction

Substitute the new concentration $[A]_2$ into the second-order rate equation to find the new rate, $Rate_2$:

$Rate_2 = k([A]_2)^2$

$Rate_2 = k\left(\frac{1}{2}[A]\right)^2$ [By algebraic substitution]

$Rate_2 = k\left(\frac{1}{4}[A]^2\right)$ [Applying the exponent power rule $(ab)^n = a^n b^n$]

$Rate_2 = \frac{1}{4} \cdot k[A]^2$

Step 3: Comparative Analysis

Since $Rate_1 = k[A]^2$, we can substitute $Rate_1$ into the expression for $Rate_2$:

$Rate_2 = \frac{1}{4} \cdot Rate_1$

This demonstrates that the rate of the reaction decreases by a factor of 4 when the concentration of the reactant is reduced by a factor of 2 in a second-order reaction.

Final Solution: The rate of the reaction will be reduced to one-fourth ($\frac{1}{4}$) of its initial rate.

Given Variables & Initial Setup

The decomposition reaction is represented as: $(CH_3)_2CHN=NCH(CH_3)_2(g) \rightarrow C_6H_{14}(g) + N_2(g)$

Given data at $T = 543 \text{ K}$:

Let the initial pressure of azoisopropane be $P_0 = 35.0 \text{ atm}$. Let $x$ be the decrease in pressure of azoisopropane at time $t$.

Step 1: Deriving the Pressure Expression

The total pressure $P_t$ at any time $t$ is the sum of the partial pressures of all gaseous components [Per Dalton’s Law of Partial Pressures].

Total pressure $P_t = (P_0 - x) + x + x = P_0 + x$.

Therefore, $x = P_t - P_0$.

The partial pressure of the reactant at time $t$ is $P_{reactant} = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t$.

Step 2: Determining Rate Constant ($k$)

First-order integrated rate law: $k = \frac{2.303}{t} \log \frac{P_0}{P_{reactant}} = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$ [By Integrated Rate Law for First-Order Reactions].

Calculation for $t = 360 \text{ s}$ ($P_t = 54.0 \text{ atm}$):

$k_1 = \frac{2.303}{360} \log \left( \frac{35.0}{2(35.0) - 54.0} \right) = \frac{2.303}{360} \log \left( \frac{35.0}{16.0} \right)$

$k_1 = \frac{2.303}{360} \log(2.1875) \approx \frac{2.303 \times 0.3399}{360} \approx 2.17 \times 10^{-3} \text{ s}^{-1}$

Calculation for $t = 720 \text{ s}$ ($P_t = 63.0 \text{ atm}$):

$k_2 = \frac{2.303}{720} \log \left( \frac{35.0}{2(35.0) - 63.0} \right) = \frac{2.303}{720} \log \left( \frac{35.0}{7.0} \right)$

$k_2 = \frac{2.303}{720} \log(5) \approx \frac{2.303 \times 0.6989}{720} \approx 2.23 \times 10^{-3} \text{ s}^{-1}$

Step 3: Calculating Average Rate Constant

$k_{avg} = \frac{k_1 + k_2}{2} = \frac{(2.17 + 2.23) \times 10^{-3}}{2} = 2.20 \times 10^{-3} \text{ s}^{-1}$

Final Solution: The rate constant for the decomposition of azoisopropane is $2.20 \times 10^{-3} \text{ s}^{-1}$.

Step 1: Defining the Rate Law

For a reaction that is second order with respect to a reactant $A$, the rate of the reaction ($r_1$) is expressed by the rate law equation:

$r_1 = k[A]^2$

[Per the definition of Rate Law, where $k$ is the rate constant and $[A]$ is the molar concentration of the reactant].

Step 2: Defining the New Concentration

The problem states that the concentration of the reactant is reduced to half of its initial value. Let the initial concentration be $[A]_1$ and the new concentration be $[A]_2$.

$[A]_2 = \frac{1}{2}[A]_1$

Step 3: Calculating the New Rate of Reaction

The new rate of reaction ($r_2$) can be calculated by substituting the new concentration into the rate law equation:

$r_2 = k[A]_2^2$

$r_2 = k \left( \frac{1}{2}[A]_1 \right)^2$

[By the Principle of Substitution in algebraic expressions].

Step 4: Algebraic Simplification

Expanding the squared term:

$r_2 = k \left( \frac{1}{4} \right) [A]_1^2$

$r_2 = \frac{1}{4} (k[A]_1^2)$

Since $r_1 = k[A]_1^2$, we can substitute $r_1$ into the equation:

$r_2 = \frac{1}{4} r_1$

Visual Representation of Kinetic Dependency

Analytical Conclusion

By comparing the initial rate ($r_1$) and the new rate ($r_2$), it is evident that the rate of the reaction decreases by a factor of 4.

Final Solution: The rate of reaction becomes one-fourth ($\frac{1}{4}$) of its original value when the concentration of the reactant is reduced to half.

1. Effect of Temperature on the Rate Constant

The rate constant ($k$) of a chemical reaction is highly sensitive to changes in temperature. For most chemical reactions, an increase in temperature leads to an exponential increase in the rate constant, typically resulting in a doubling or tripling of the reaction rate for every $10^\circ\text{C}$ rise in temperature.

Qualitative Justification: According to the Maxwell-Boltzmann distribution of molecular energies, an increase in temperature shifts the distribution curve to the right and increases the area under the curve representing molecules with energy equal to or greater than the threshold activation energy ($E_a$). Consequently, the frequency of effective collisions increases, directly elevating the rate constant [$Per\ Arrhenius\ collision\ theory$].

2. Quantitative Representation: The Arrhenius Equation

The quantitative relationship between the rate constant ($k$) and temperature ($T$) is defined by the Arrhenius equation:

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

Where:

• $k$: Rate constant of the reaction.
• $A$: Arrhenius factor or frequency factor (pre-exponential factor).
• $E_a$: Activation energy ($\text{J mol}^{-1}$).
• $R$: Universal gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
• $T$: Temperature in Kelvin.

3. Mathematical Derivation for Temperature Variation

To evaluate the effect of changing temperature from $T_1$ to $T_2$, we take the natural logarithm ($\ln$) of the Arrhenius equation:

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

For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:

$\ln k_1 = \ln A - \frac{E_a}{RT_1}$

$\ln k_2 = \ln A - \frac{E_a}{RT_2}$

Subtracting the first equation from the second yields the linear form used for calculation [$By\ algebraic\ subtraction$]:

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

Converting to base-10 logarithm ($\log$):

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

Final Solution: The effect of temperature on the rate constant is quantitative expressed via the Arrhenius equation $k = A e^{-E_a/RT}$. This shows that $k$ increases exponentially with temperature because a higher fraction of reactant molecules possess sufficient kinetic energy to overcome the activation energy barrier.

Step 1: Data Tabulation and Normalization

The provided experimental data for the decomposition reaction $2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$ at $318\,K$ is organized into the following table for graphical representation. The concentration is represented as $[A] = [N_{2}O_{5}] \times 10^{2} \, mol\,L^{-1}$.

Step 2: Graphical Construction Strategy

To plot the concentration of the reactant $[N_{2}O_{5}]$ against time ($t$):

• X-axis: Time ($t$) in seconds, ranging from $0$ to $3200\,s$.
• Y-axis: Concentration $[N_{2}O_{5}]$ in $10^{-2} \, mol\,L^{-1}$, ranging from $0$ to $1.8$.
• Interpretation: Since the concentration decreases over time, the plot will exhibit a downward curvature, characteristic of a first-order kinetic process [Per the Integrated Rate Law for First-Order Reactions, $[A]_t = [A]_0 e^{-kt}$].

Step 3: Visual Representation

Analytical Summary

The plot demonstrates that the concentration of $N_{2}O_{5}$ decays exponentially with time. As the gradient of the curve ($\frac{d[N_{2}O_{5}]}{dt}$) becomes less negative over time, it visually confirms the dependence of the rate of reaction on the remaining concentration of the reactant, which is consistent with the kinetics of a first-order reaction where $\text{Rate} = k[N_{2}O_{5}]$.

Final Solution: The plot of $[N_{2}O_{5}]$ versus $t$ results in an exponential decay curve, indicating that the decomposition rate is directly proportional to the reactant concentration at any given instant.

Step 1: Given Parameters & Constants

The disintegration of radioactive isotopes follows first-order kinetics. We identify the following variables:

• Initial mass ($N_0$) = $1 \, \mu\text{g}$
• Half-life ($t_{1/2}$) = $28.1 \, \text{years}$
• Decay constant ($\lambda$) = $\frac{0.693}{t_{1/2}}$ [By the Integrated Rate Law for First-Order Reactions]

Calculating the decay constant ($\lambda$):

$\lambda = \frac{0.693}{28.1 \, \text{years}} \approx 0.02466 \, \text{year}^{-1}$

Step 2: Deriving the First-Order Equation

For radioactive decay, the remaining amount $N_t$ after time $t$ is expressed as:

$N_t = N_0 \cdot e^{-\lambda t}$

Alternatively, using the half-life formula: $N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$

Step 3: Calculation for $t = 10$ years

Using the exponential form:

$N_{10} = 1 \, \mu\text{g} \cdot e^{-(0.02466 \, \text{year}^{-1})(10 \, \text{years})}$

$N_{10} = 1 \cdot e^{-0.2466}$

$N_{10} \approx 1 \cdot 0.7812 = 0.7812 \, \mu\text{g}$

[Justification: Per first-order kinetics, the quantity decreases exponentially over time.]

Step 4: Calculation for $t = 60$ years

Using the half-life power form for higher precision:

$N_{60} = 1 \, \mu\text{g} \cdot \left(\frac{1}{2}\right)^{\frac{60}{28.1}}$

$N_{60} = 1 \cdot (0.5)^{2.1352}$

$N_{60} \approx 1 \cdot 0.2282 = 0.2282 \, \mu\text{g}$

Analytical Summary Table

Graphical Representation of Decay

Final Solution: After 10 years, $0.7812 \, \mu\text{g}$ of $^{90}\text{Sr}$ remains; after 60 years, $0.2282 \, \mu\text{g}$ remains.

1. Effect of Temperature on Rate Constant ($k$)

The rate constant of a chemical reaction is highly sensitive to changes in temperature. For most chemical reactions, an increase in temperature leads to an increase in the rate constant, which subsequently increases the reaction rate. Specifically, for many reactions, a rise in temperature of $10\text{ K}$ causes the rate constant to approximately double [Per the empirical rule of Van't Hoff].

2. Quantitative Representation: The Arrhenius Equation

The quantitative relationship between the rate constant ($k$) and temperature ($T$) is defined by the Arrhenius Equation:

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

Where:

• $k$ = Rate constant
• $A$ = Arrhenius factor (Frequency factor or Pre-exponential factor)
• $E_a$ = Activation energy ($J \cdot mol^{-1}$)
• $R$ = Gas constant ($8.314\text{ J} \cdot K^{-1} \cdot mol^{-1}$)
• $T$ = Temperature in Kelvin

3. Linearized Form and Graphical Interpretation

To determine $E_a$ and $A$ experimentally, we take the natural logarithm of the Arrhenius equation:

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

This follows the linear equation form $y = mx + c$, where:

• $y = \ln k$
• $x = \frac{1}{T}$
• Slope ($m$) $= -\frac{E_a}{R}$
• Intercept ($c$) $= \ln A$

4. Calculation of Activation Energy (Two-Temperature Form)

For two different temperatures $T_1$ and $T_2$, with rate constants $k_1$ and $k_2$ respectively, the equation is rearranged to eliminate $A$:

$\ln k_1 = \ln A - \frac{E_a}{RT_1}$

$\ln k_2 = \ln A - \frac{E_a}{RT_2}$

Subtracting the two equations yields the final operational form:

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

Final Solution: The effect of temperature on the rate constant is exponential, as dictated by the Arrhenius equation $k = Ae^{-E_a/RT}$. This indicates that as temperature increases, the fraction of molecules possessing kinetic energy greater than or equal to $E_a$ increases, thereby increasing the frequency of successful collisions and the magnitude of the rate constant.

Step 1: Establishing Theoretical Foundation

The decomposition of sucrose follows first-order kinetics. The relationship between the rate constant ($k$) and the half-life ($t_{1/2}$) is defined by the integrated rate law [Per the First-Order Kinetics derivation]:

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

Given $t_{1/2} = 3.00$ hours, we calculate $k$:

$k = \frac{0.693}{3.00 \text{ h}} = 0.231 \text{ h}^{-1}$

Step 2: Applying the Integrated Rate Law

For a first-order reaction, the concentration of the reactant $[A]$ at time $t$ relative to the initial concentration $[A]_0$ is given by:

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

Alternatively, expressed in exponential form to find the fraction remaining ($\frac{[A]}{[A]_0}$):

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

Step 3: Quantitative Calculation

Substitute the known variables into the equation where $k = 0.231 \text{ h}^{-1}$ and $t = 8.00 \text{ h}$:

$\frac{[A]}{[A]_0} = e^{-(0.231 \text{ h}^{-1} \times 8.00 \text{ h})}$

$\frac{[A]}{[A]_0} = e^{-1.848}$

Using the value of the exponential function $e^{-1.848} \approx 0.15755$:

$\frac{[A]}{[A]_0} \approx 0.1576$

Visual Representation of Decay

Analytical Summary

Final Solution: The fraction of the sucrose sample remaining after 8 hours is approximately 0.1576 (or 15.76%).

Step 1: Data Tabulation and Variable Assignment

To analyze the decomposition of $N_{2}O_{5}$, we represent the concentration $[N_{2}O_{5}]$ as the dependent variable ($y$-axis) and time $t$ as the independent variable ($x$-axis) based on the provided experimental set:

Step 2: Graphical Representation Protocol

Plotting $[N_{2}O_{5}]$ vs $t$ yields a downward-sloping curve, characteristic of a first-order kinetic process [Per Rate Law: $Rate = k[N_{2}O_{5}]$]. As time progresses, the concentration decreases non-linearly, indicating that the reaction rate slows down as reactants are consumed.

Step 3: Analytical Interpretation

The resulting graph demonstrates that the concentration of $N_{2}O_{5}$ decays exponentially with respect to time. By applying the First-Order Integrated Rate Equation: $\ln[N_{2}O_{5}]_t = -kt + \ln[N_{2}O_{5}]_0$ [By Fundamental Law of Chemical Kinetics], the slope of the tangent at any point $t$ provides the instantaneous rate of the reaction at that specific time interval.

Final Solution: The plot of $[N_{2}O_{5}]$ versus $t$ results in an exponential decay curve, confirming that the concentration of the reactant decreases as the reaction proceeds towards equilibrium.