Chemical Kinetics

Introduction to Chemical Kinetics

• Chemical kinetics is defined as the scientific study of reaction rates, the factors and conditions under which these rates change, and the specific molecular mechanisms and sequence of elementary events that occur during an overall chemical transformation.

• Chemical reactions span vast time scales:

  • Rapid reactions occur almost instantaneously, such as photosynthesis or fast precipitation reactions.

  • Extremely slow reactions occur over geological or human timescales, such as the thermodynamic conversion of diamond to graphite.

• Primary objectives and rationale for studying reaction rates include:

  • Monitoring and controlling the exact speeds at which chemical processes proceed.

  • Evaluating whether a chemical reaction is kinetically feasible for practical, commercial, or industrial applications.

  • Investigating detailed reaction mechanisms to deduce the exact molecular-level steps, intermediates, and bond-breaking/forming events involved.

Reaction Rates and Measurement

• For a generalized reaction:   $\text{Reaction: } A \rightarrow B$

• Over time, reactant molecules ($A$) decrease in number as they convert into product molecules ($B$):

  • At time $t = 0\,s$, there are $40$ molecules of reactant $A$ and $0$ molecules of product $B$.

  • At time $t = 10\,s$, $30$ molecules of $A$ remain, and $10$ molecules of $B$ have formed.

  • At time $t = 20\,s$, $22$ molecules of $A$ remain, and $18$ molecules of $B$ have formed.

  • At time $t = 30\,s$, $16$ molecules of $A$ remain, and $24$ molecules of $B$ have formed.

  • At time $t = 40\,s$, $11$ molecules of $A$ remain, and $29$ molecules of $B$ have formed.

  • At time $t = 50\,s$, $8$ molecules of $A$ remain, and $32$ molecules of $B$ have formed.

  • At time $t = 60\,s$, $6$ molecules of $A$ remain, and $34$ molecules of $B$ have formed.

• For a general multi-reactant reaction $A + B \rightarrow C + D$, the average rate of reaction can be defined from two equivalent perspectives:

  • Rate of disappearance of reactants (expressed as a negative change in concentration):     $\text{Rate of disappearance} = -\frac{\Delta [A]}{\Delta t} \quad \text{or} \quad -\frac{\Delta [B]}{\Delta t}$

  • Rate of appearance of products (expressed as a positive change in concentration):     $\text{Rate of appearance} = \frac{\Delta [C]}{\Delta t} \quad \text{or} \quad \frac{\Delta [D]}{\Delta t}$

• Quantitative Example: Decomposition of Aqueous Bromine

  • Chemical equation:     $Br_2(aq) + HCOOH(aq) \rightarrow 2Br^-(aq) + 2H^+(aq) + CO_2(g)$

• As the reaction proceeds, the distinct dark orange color of aqueous molecular bromine ($Br_2$) gradually fades as it is consumed.

• Measured experimental data for molecular bromine concentration at $25\,^\circ\text{C}$:

  • $t = 0.0\,s \Rightarrow [Br_2] = 0.0120\,M$

  • $t = 50.0\,s \Rightarrow [Br_2] = 0.0101\,M$

  • $t = 100.0\,s \Rightarrow [Br_2] = 0.00846\,M$

  • $t = 150.0\,s \Rightarrow [Br_2] = 0.00710\,M$

  • $t = 200.0\,s \Rightarrow [Br_2] = 0.00596\,M$

  • $t = 250.0\,s \Rightarrow [Br_2] = 0.00500\,M$

  • $t = 300.0\,s \Rightarrow [Br_2] = 0.00420\,M$

  • $t = 350.0\,s \Rightarrow [Br_2] = 0.00353\,M$

  • $t = 400.0\,s \Rightarrow [Br_2] = 0.00296\,M$

• Mathematical formulation for average rate of bromine disappearance:     $\text{Average Rate} = -\frac{[Br_2]_{\text{final}} - [Br_2]_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$

• Calculation over the initial $50.0\,s$ time interval ($t = 0.0\,s$ to $t = 50.0\,s$):     $\text{Average Rate} = -\frac{0.0101\,M - 0.0120\,M}{50.0\,s - 0.0\,s} = \frac{0.0019\,M}{50.0\,s} = 3.80 \times 10^{-5}\,M/s$

  • Practice Problem: Reaction of Chlorobutane with Water

• Chemical equation:     $C_4H_9Cl + H_2O \rightarrow C_4H_9OH + HCl$

• Experimental time and concentration data:

  • $t = 0\,s \Rightarrow [C_4H_9Cl] = 0.100\,M$

  • $t = 50\,s \Rightarrow [C_4H_9Cl] = 0.0905\,M$

  • $t = 100\,s \Rightarrow [C_4H_9Cl] = 0.0820\,M$

  • $t = 150\,s \Rightarrow [C_4H_9Cl] = 0.0741\,M$

  • $t = 300\,s \Rightarrow [C_4H_9Cl] = 0.0549\,M$

  • $t = 500\,s \Rightarrow [C_4H_9Cl] = 0.0368\,M$

  • $t = 800\,s \Rightarrow [C_4H_9Cl] = 0.0200\,M$

• Average rate calculation over the entire $800\,s$ interval:     $\text{Average Rate} = -\frac{0.0200\,M - 0.100\,M}{800\,s - 0\,s} = \frac{0.0800\,M}{800\,s} = 1.0 \times 10^{-4}\,M/s$

Reaction Stoichiometry and Relative Rates

• When stoichiometric coefficients in a balanced chemical equation are not $1:1$, the rates of disappearance of reactants and appearance of products differ according to their stoichiometric ratios.

• For a general reaction:   $aA + bB \rightarrow cC + dD$

• The general relative rate expression is:   $\text{Relative Rate} = -\frac{1}{a}\frac{\Delta [A]}{\Delta t} = -\frac{1}{b}\frac{\Delta [B]}{\Delta t} = \frac{1}{c}\frac{\Delta [C]}{\Delta t} = \frac{1}{d}\frac{\Delta [D]}{\Delta t}$

• Stoichiometric Example: Decomposition of Phosphine

  • Balanced reaction equation:     $4PH_3(g) \rightarrow P_4(g) + 6H_2(g)$

  • Relative rate relation for this specific reaction:     $-\frac{1}{4}\frac{\Delta [PH_3]}{\Delta t} = \frac{1}{1}\frac{\Delta [P_4]}{\Delta t} = \frac{1}{6}\frac{\Delta [H_2]}{\Delta t}$

  • Given that molecular hydrogen ($H_2$) is formed at a rate of $0.168\,M/s$, the rate of tetraphosphorus ($P_4$) production is:     $\frac{\Delta [P_4]}{\Delta t} = \frac{1}{6}\frac{\Delta [H_2]}{\Delta t} = \frac{1}{6}(0.168\,M/s) = 0.028\,M/s$

Factors Influencing Chemical Reaction Rates

• Five major factors govern the rate at which chemical reactions take place:

• 1. Nature of the Reactants:

  • The intrinsic physical and chemical properties of the reacting substances dictate their reactivity and activation barrier.

  • Comparative Example:

  • White phosphorus ($P_4$) ignites spontaneously in air at low temperatures around $50\,^\circ\text{C}$.

  • Red phosphorus (a polymeric allotrope) is considerably more stable and ignites only at temperatures exceeding $240\,^\circ\text{C}$.

• 2. Concentration of Reactants:

  • Increasing the concentration of dissolved or gaseous reactants increases the density of particles per unit volume, increasing collision frequency.

  • Comparative Example: Oxidation of Steel Wool ($Fe$)

  • Steel wool heated in ambient air (about $20\%\,O_2$) glows red-hot but oxidizes to iron(III) oxide ($Fe_2O_3$) relatively slowly.

  • Red-hot steel wool placed in pure oxygen ($100\%\,O_2$) burns intensely and rapidly, forming $Fe_2O_3$ almost instantly.

• 3. Particle Size or Surface Area:

  • For heterogeneous reactions involving solids, increasing the available surface area by finely dividing the solid exposes more reactant sites to collisions.

  • Comparative Example:

  • Iron powder reacts violently and rapidly with dilute hydrochloric acid ($HCl$).

  • An iron nail with identical mass reacts at a significantly slower rate due to restricted surface exposure.

• 4. Temperature:

  • Temperature increases the kinetic energy of reactant particles, raising both collision frequency and the fraction of collisions exceeding activation energy.

  • Empirical Rule of Thumb: For many common chemical reactions near room temperature, a $10\,^\circ\text{C}$ increase in temperature approximately doubles the reaction rate.

• 5. Catalysts:

  • Catalysts are substances that substantially increase reaction rate without undergoing permanent chemical change or being consumed in the overall stoichiometry.

Reaction Orders and the Rate Law Expression

• The Rate Law defines the quantitative relationship between reaction rate and reactant concentrations.

• For the general reaction $aA + bB \rightarrow cC + dD$:   $\text{Rate} = k[A]^x[B]^y$

  • $k$ is the rate constant (a proportionality constant specific to a given reaction at a specific temperature; independent of time and concentration, but varies with temperature $T$).

  • $x$ and $y$ are the reaction orders with respect to reactants $A$ and $B$, respectively (determined exclusively through experiment, not stoichiometric coefficients).

• Meaning of Reaction Orders:

  • First Order ($x = 1$): Rate is directly proportional to concentration $[A]^1$. Doubling $[A]$ doubles the rate.

  • Second Order ($x = 2$): Rate is directly proportional to the square of concentration $[A]^2$. Doubling $[A]$ quadruples the rate ($2^2 = 4$).

  • Zero Order ($x = 0$): Rate is independent of reactant concentration $[A]^0 = 1$. Changing concentration has no effect on rate.

  • Overall Reaction Order: Sum of all individual reactant orders ($x + y + \dots$).

Determination of Rate Laws: Method of Initial Rates

• The method of initial rates involves measuring initial reaction rates across multiple experimental runs where initial concentrations of individual reactants are systematically varied.

• Systematic Procedure:

  • Step 1: Identify two experiments where $[B]$ is held constant while $[A]$ varies to determine the exponent $x$.

  • Step 2: Identify two experiments where $[A]$ is held constant while $[B]$ varies to determine the exponent $y$.

  • Step 3: Substitute concentration and rate values from any experiment into the rate law to compute $k$.

• Detailed Worked Example: Reaction of $F_2$ and $ClO_2$

  • Reaction data at constant temperature:

  • Experiment 1: $[F_2] = 0.10\,M$, $[ClO_2] = 0.010\,M$, $\text{Initial Rate} = 1.2 \times 10^{-3}\,M/s$

  • Experiment 2: $[F_2] = 0.10\,M$, $[ClO_2] = 0.040\,M$, $\text{Initial Rate} = 4.8 \times 10^{-3}\,M/s$

  • Experiment 3: $[F_2] = 0.20\,M$, $[ClO_2] = 0.010\,M$, $\text{Initial Rate} = 2.4 \times 10^{-3}\,M/s$

  • Finding order with respect to $F_2$ (compare Exp 1 and Exp 3 where $[ClO_2]$ is held constant at $0.010\,M$):     $\frac{[F_2]_3}{[F_2]_1} = \frac{0.20\,M}{0.10\,M} = 2$     $\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{2.4 \times 10^{-3}\,M/s}{1.2 \times 10^{-3}\,M/s} = 2$     $2^x = 2 \Rightarrow x = 1 \quad (\text{1st order in } F_2)$

  • Finding order with respect to $ClO_2$ (compare Exp 1 and Exp 2 where $[F_2]$ is held constant at $0.10\,M$):     $\frac{[ClO_2]_2}{[ClO_2]_1} = \frac{0.040\,M}{0.010\,M} = 4$     $\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{4.8 \times 10^{-3}\,M/s}{1.2 \times 10^{-3}\,M/s} = 4$     $4^y = 4 \Rightarrow y = 1 \quad (\text{1st order in } ClO_2)$

  • Rate Law and Overall Order:     $\text{Rate} = k[F_2]^1[ClO_2]^1$     $\text{Overall Order} = 1 + 1 = 2 \quad (\text{2nd order overall})$

  • Calculating Rate Constant $k$ using Experiment 1:     $k = \frac{\text{Rate}}{[F_2][ClO_2]} = \frac{1.2 \times 10^{-3}\,M/s}{(0.10\,M)(0.010\,M)} = 1.2\,M^{-1}s^{-1}$

• Detailed Worked Example: Reaction $A + B \rightarrow C$

  • Experimental Data Table:

  • Exp 1: $[A] = 0.10\,M$, $[B] = 0.015\,M$, $\text{Initial Rate} = 2.1 \times 10^{-4}\,M/s$

  • Exp 2: $[A] = 0.20\,M$, $[B] = 0.015\,M$, $\text{Initial Rate} = 4.2 \times 10^{-4}\,M/s$

  • Exp 3: $[A] = 0.10\,M$, $[B] = 0.030\,M$, $\text{Initial Rate} = 8.4 \times 10^{-4}\,M/s$

  • Analysis:

  • Comparing Exp 1 and 2: Doubling $[A]$ while holding $[B]$ constant doubles the initial rate ($\frac{4.2 \times 10^{-4}}{2.1 \times 10^{-4}} = 2 = 2^1$), indicating 1st order in $A$.

  • Comparing Exp 1 and 3: Doubling $[B]$ while holding $[A]$ constant quadruples the initial rate ($\frac{8.4 \times 10^{-4}}{2.1 \times 10^{-4}} = 4 = 2^2$), indicating 2nd order in $B$.

  • Rate Law:     $\text{Rate} = k[A][B]^2$     $\text{Overall Order} = 1 + 2 = 3 \quad (\text{3rd order overall})$

  • Calculation of Rate Constant $k$:     $k = \frac{\text{Rate}}{[A][B]^2} = \frac{2.1 \times 10^{-4}\,M/s}{(0.10\,M)(0.015\,M)^2} = \frac{2.1 \times 10^{-4}}{0.10 \times 0.000225} = 9.3\,M^{-2}s^{-1}$

• Practice Problem: Reaction $NH_4^+ + NO_2^- \rightarrow N_2 + 2H_2O$

  • Experimental Data:

  • Run 1: $[NO_2^-] = 0.0100\,M$, $[NH_4^+] = 0.200\,M$, $\text{Rate} = 5.4 \times 10^{-7}\,M/s$

  • Run 2: $[NO_2^-] = 0.0200\,M$, $[NH_4^+] = 0.200\,M$, $\text{Rate} = 10.8 \times 10^{-7}\,M/s$

  • Run 3: $[NO_2^-] = 0.200\,M$, $[NH_4^+] = 0.0202\,M$, $\text{Rate} = 10.8 \times 10^{-7}\,M/s$

  • Run 4: $[NO_2^-] = 0.200\,M$, $[NH_4^+] = 0.0404\,M$, $\text{Rate} = 21.6 \times 10^{-7}\,M/s$

  • Deduction:

  • Runs 1 & 2: $[NO_2^-]$ doubles $\Rightarrow$ rate doubles ($10.8/5.4 = 2$) $\Rightarrow$ 1st order in $NO_2^-$.

  • Runs 3 & 4: $[NH_4^+]$ doubles $\Rightarrow$ rate doubles ($21.6/10.8 = 2$) $\Rightarrow$ 1st order in $NH_4^+$.

  • Rate Law:     $\text{Rate} = k[NH_4^+][NO_2^-]$

• Practice Problem: Decomposition of Ethylene Oxide

  • Reaction:     $C_2H_4O(g) \rightarrow CH_4(g) + CO(g) \quad \text{at } 688\,K$

  • Kinetic Data:

  • Exp 1: $[C_2H_4O] = 0.00272\,M$, $\text{Initial Rate} = 5.57 \times 10^{-7}\,M/s$

  • Exp 2: $[C_2H_4O] = 0.00544\,M$, $\text{Initial Rate} = 1.11 \times 10^{-6}\,M/s$

  • Solution:

  • Concentration ratio: $\frac{0.00544}{0.00272} = 2.00$

  • Rate ratio: $\frac{1.11 \times 10^{-6}}{5.57 \times 10^{-7}} = 1.99 \approx 2.00$

  • Rate law is first order: $\text{Rate} = k[C_2H_4O]$

  • Rate constant $k$:       $k = \frac{5.57 \times 10^{-7}\,M/s}{0.00272\,M} = 2.05 \times 10^{-4}\,s^{-1}$

• Practice Problem: Decomposition of Azomethane

  • Reaction:     $CH_3NNCH_3(g) \rightarrow C_2H_6(g) + N_2(g)$

  • Kinetic Data:

  • Exp 1: $[CH_3NNCH_3] = 1.13 \times 10^{-2}\,M$, $\text{Initial Rate} = 2.8 \times 10^{-6}\,M/s$

  • Exp 2: $[CH_3NNCH_3] = 2.26 \times 10^{-2}\,M$, $\text{Initial Rate} = 5.6 \times 10^{-6}\,M/s$

  • Solution:

  • Concentration doubles $\Rightarrow$ Initial rate doubles ($5.6/2.8 = 2.0$).

  • Rate law: $\text{Rate} = k[CH_3NNCH_3]$

  • Rate constant $k$:       $k = \frac{2.8 \times 10^{-6}\,M/s}{1.13 \times 10^{-2}\,M} = 2.48 \times 10^{-4}\,s^{-1}$

Integrated Rate Laws

• Integrated rate laws express reactant concentration directly as a function of elapsed time.

First-Order Integrated Rate Law

• Differential rate expression:   $\text{Rate} = -\frac{\Delta [A]}{\Delta t} = k[A]^1$

• Integrated rate equation (linear form $y = mx + b$):   $\ln[A]_t = -kt + \ln[A]_0$

• Alternative quotient form:   $\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$

• Graphical behavior:

  • A plot of $\ln[A]_t$ on the vertical y-axis versus time $t$ on the horizontal x-axis produces a straight line.

  • Slope of the line $= -k$

  • y-intercept $= \ln[A]_0$

• Worked Example 1:

  • Reaction $2A \rightarrow B$ is first order in $A$, with $k = 7.5 \times 10^{-3}\,s^{-1}$ at $110\,^\circ\text{C}$.

  • Determine time required for $[A]$ to decrease from $1.25\,M$ to $0.71\,M$:     $\ln\left(\frac{0.71\,M}{1.25\,M}\right) = -(7.5 \times 10^{-3}\,s^{-1}) \cdot t$     $\ln(0.568) = -0.5656 = -7.5 \times 10^{-3} \cdot t$     $t = \frac{-0.5656}{-7.5 \times 10^{-3}\,s^{-1}} = 75.4\,s \approx 75\,s$

• Worked Example 2:

  • Decomposition of $N_2O_5$ is first order with $k = 4.8 \times 10^{-4}\,s^{-1}$. Initial $[N_2O_5]_0 = 1.65 \times 10^{-2}\,M$.

  • Find concentration remaining after $825\,s$:     $\ln[N_2O_5]_t = -(4.8 \times 10^{-4}\,s^{-1})(825\,s) + \ln(1.65 \times 10^{-2})$     $\ln[N_2O_5]_t = -0.396 + (-4.1044) = -4.5004$     $[N_2O_5]_t = e^{-4.5004} = 1.11 \times 10^{-2}\,M$

• Worked Example 3:

  • Reaction $2A \rightarrow B$ with $k = 7.5 \times 10^{-3}\,s^{-1}$ and $[A]_0 = 2.25\,M$.

  • Find $[A]$ after $2.0\,\text{minutes}$ ($120\,s$):     $\ln[A]_t = -(7.5 \times 10^{-3}\,s^{-1})(120\,s) + \ln(2.25)$     $\ln[A]_t = -0.900 + 0.8109 = -0.0891$     $[A]_t = e^{-0.0891} = 0.915\,M$

• Worked Example 4: Decomposition of Ethyl Chloride

  • Reaction:     $CH_3CH_2Cl(g) \rightarrow C_2H_4(g) + HCl(g) \quad (\text{First order})$

  • Initial concentration $[CH_3CH_2Cl]_0 = 0.00100\,M$. At $t = 155\,s$ at $500\,^\circ\text{C}$, concentration is reduced to $0.00067\,M$.

  • Step 1: Calculate $k$:     $\ln\left(\frac{0.00067}{0.00100}\right) = -k(155\,s)$     $-0.40048 = -155 \cdot k \Rightarrow k = 2.58 \times 10^{-3}\,s^{-1}$

  • Step 2: Calculate concentration remaining after a total time of $256\,s$:     $\ln[CH_3CH_2Cl]_{256} = -(2.58 \times 10^{-3}\,s^{-1})(256\,s) + \ln(0.00100)$     $\ln[CH_3CH_2Cl]_{256} = -0.66145 + (-6.90775) = -7.5692$     $[CH_3CH_2Cl]_{256} = e^{-7.5692} = 5.16 \times 10^{-4}\,M \quad (0.000516\,M)$

Second-Order Integrated Rate Law

• Rate law for a second-order reaction in reactant $A$:   $\text{Rate} = k[A]^2 \quad \text{or} \quad \text{Rate} = k[A]^1[B]^1$

• Integrated rate equation (linear form $y = mx + b$):   $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$

• Graphical behavior:

  • A plot of reciprocal concentration $\frac{1}{[A]_t}$ versus time $t$ yields a straight line.

  • Slope of the line $= k$ (positive slope)

  • y-intercept $= \frac{1}{[A]_0}$

Zero-Order Integrated Rate Law

• Rate law for a zero-order reaction:   $\text{Rate} = k[A]^0 = k$

• Integrated rate equation (linear form $y = mx + b$):   $[A]_t = -kt + [A]_0$

• Graphical behavior:

  • A plot of concentration $[A]_t$ versus time $t$ yields a straight line.

  • Slope of the line $= -k$

  • y-intercept $= [A]_0$

• Zero-order reactions are relatively rare and occur primarily when rate is limited by surface area or catalyst availability.

Collision Theory and Temperature Dependence

• Collision Theory postulates that gas or solution particles must physically collide in order to react chemically.

• Increasing collision frequency inherently increases the potential rate of reaction.

• However, only a fraction of total collisions are effective collisions that lead to chemical transformation.

• Requirements for an Effective Collision:

  • 1. Molecular Orientation: Reacting molecules must collide in a precise relative spatial orientation that permits old bonds to break and new bonds to form.

- Orientation A (Effective): $NO$ collides with $Cl_2$ such that Nitrogen directly contacts Chlorine, forming $NOCl$ and releasing a Chlorine atom.
- Orientation B (Ineffective): $NO$ collides with Oxygen facing $Cl_2$, resulting in a simple elastic bounce without reaction.

• 2. Activation Energy ($E_a$): Colliding particles must possess a minimum combined kinetic energy equal to or greater than $E_a$ to overcome electrostatic repulsion and strain existing bonds.

• Activated Complex (Transition State):

  • An unstable, short-lived arrangement of atoms formed at the point of maximum potential energy along the reaction pathway.

  • At the transition state, partial chemical bonds are simultaneously breaking and forming.

The Arrhenius Equation

• The Arrhenius equation quantitatively expresses the relationship between rate constant $k$, absolute temperature $T$, and activation energy $E_a$:   $k = A e^{-E_a/RT}$

  • $k$ = Rate constant

  • $A$ = Arrhenius frequency factor (pre-exponential factor related to collision frequency and molecular orientation probability)

  • $E_a$ = Activation energy in Joules per mole ($\text{J/mol}$)

  • $R$ = Universal gas constant ($8.3145\,\text{J/(K}\cdot\text{mol)}$)

  • $T$ = Absolute temperature in Kelvin ($\text{K}$)

  • $e$ = Base of natural logarithms ($\approx 2.71828$)

• Logarithmic / Straight-Line Form ($y = mx + b$):   $\ln k = -\frac{E_a}{RT} + \ln A$   $\ln k = \left(-\frac{E_a}{R}\right)\left(\frac{1}{T}\right) + \ln A$

  • A plot of $\ln k$ versus $\frac{1}{T}$ produces a straight line.

  • Slope of the line $= -\frac{E_a}{R}$

  • Activation energy is calculated directly from the plot slope: $E_a = -\text{slope} \times R$.

• Two-Point Form of the Arrhenius Equation:

  • Comparing rate constants $k_1$ and $k_2$ at two distinct temperatures $T_1$ and $T_2$:     $\ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

  • Rearranged explicitly for Activation Energy ($E_a$):     $E_a = R \left[ \frac{\ln\left(\frac{k_1}{k_2}\right)}{\frac{1}{T_2} - \frac{1}{T_1}} \right]$

• Worked Problem 1: Arrhenius Activation Energy Calculation

  • Measured first-order rate constants across temperatures:

  • $T_1 = 400\,K \Rightarrow k_1 = 2.9 \times 10^{-3}\,s^{-1}$

  • $T_2 = 450\,K \Rightarrow k_2 = 6.1 \times 10^{-2}\,s^{-1}$

  • $T_3 = 500\,K \Rightarrow k_3 = 7.0 \times 10^{-1}\,s^{-1}$

  • Using $T_1 = 400\,K$ and $T_2 = 450\,K$ to compute $E_a$:     $\ln\left(\frac{2.9 \times 10^{-3}}{6.1 \times 10^{-2}}\right) = \ln(0.04754) = -3.0461$     $\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{450\,K} - \frac{1}{400\,K} = 0.002222 - 0.002500 = -0.0002778\,K^{-1}$     $E_a = (8.3145\,\text{J/(K}\cdot\text{mol)}) \cdot \left[ \frac{-3.0461}{-0.0002778\,K^{-1}} \right] = 8.3145 \cdot 10965 = 91168\,\text{J/mol} = 91\,\text{kJ/mol}$

• Worked Problem 2: Hydrolysis of an Organic Chlorine Compound

  • Experimental rate constants at four temperatures:

  • $T = 300\,K \Rightarrow k = 1.0 \times 10^{-5}\,s^{-1}$

  • $T = 320\,K \Rightarrow k = 5.0 \times 10^{-5}\,s^{-1}$

  • $T = 340\,K \Rightarrow k = 2.0 \times 10^{-4}\,s^{-1}$

  • $T = 355\,K \Rightarrow k = 5.0 \times 10^{-4}\,s^{-1}$

  • Step 1: Calculate Activation Energy ($E_a$) using $300\,K$ and $320\,K$:     $\ln\left(\frac{1.0 \times 10^{-5}}{5.0 \times 10^{-5}}\right) = \ln(0.20) = -1.6094$     $\frac{1}{320} - \frac{1}{300} = 0.003125 - 0.003333 = -0.0002083\,K^{-1}$     $E_a = (8.314\,\text{J/(K}\cdot\text{mol)}) \cdot \left[ \frac{-1.6094}{-0.0002083\,K^{-1}} \right] = 64230\,\text{J/mol} = 64.2\,\text{kJ/mol}$

  • Step 2: Calculate Arrhenius constant $A$ using data at $300\,K$:     $\ln(1.0 \times 10^{-5}) = -\frac{64230\,\text{J/mol}}{(8.314\,\text{J/(K}\cdot\text{mol)})(300\,K)} + \ln A$     $-11.513 = -25.751 + \ln A$     $\ln A = 14.238 \Rightarrow A = e^{14.238} = 1.52 \times 10^6\,s^{-1}$

Catalysis

• A catalyst is a substance that increases the rate of a chemical reaction without undergoing permanent chemical transformation or being consumed in the overall process.

• Mechanism of Action:

  • Provides an alternative reaction pathway involving a different series of elementary steps with a significantly lower activation energy ($E_a' < E_a$).

• Lowering activation energy exponentially increases the fraction of molecular collisions that possess sufficient energy to react.

• Ozone Decomposition Catalysis Example:

  • Uncatalyzed reaction: $O_3 + O \rightarrow 2O_2$ (High $E_a \approx 20\,\text{kJ}$)

  • Chlorine-catalyzed reaction proceeds via a two-step mechanism:     $\text{Step 1: } Cl + O_3 \rightarrow ClO + O_2$     $\text{Step 2: } ClO + O \rightarrow Cl + O_2$     $\text{Overall Reaction: } O_3 + O \rightarrow 2O_2$

  • $Cl$ acts as a catalyst; $ClO$ acts as a reaction intermediate.

Categories of Catalysts

• 1. Homogeneous Catalysts:

  • Reactants and catalyst exist entirely in a single phase (most commonly gas-phase mixtures or aqueous solutions).

• 2. Heterogeneous Catalysts:

  • Reactants and catalyst exist in different physical phases (typically a solid catalyst interacting with liquid or gaseous reactants).

  • Industrial Example: Hydrogenation of Ethylene on a Solid Nickel Surface     $C_2H_4(g) + H_2(g) \xrightarrow{Ni(s)} C_2H_6(g)$

• Four sequential steps on the nickel catalyst surface:

  • (a) Adsorption of Hydrogen: $H_2$ molecules adsorb on active sites of the $Ni$ surface, breaking $H-H$ bonds to form surface-bound hydrogen atoms.

  • (b) Adsorption of Ethylene: $C_2H_4$ adsorbs on the $Ni$ surface, breaking its carbon-carbon $\pi$ bond.

  • (c) Reaction / Migration: Hydrogen atoms migrate across the surface, sequentially forming new $C-H$ bonds with adsorbed ethylene fragments.

  • (d) Desorption: The fully saturated ethane ($C_2H_6$) product desorbs from the nickel surface into the gas phase.

  • 3. Biological Catalysts (Enzymes):

• Enzymes are large, specialized protein molecules that catalyze metabolic reactions in biological systems.

• Structure & Specificity:

  • Contain one or more specialized three-dimensional regions called active sites where substrate molecules bind.

  • High specificity is described by the Lock-and-Key Model:

- Specific substrate ($S$) binds tightly into the complementary active site of the enzyme ($E$), forming an **Enzyme-Substrate Complex** ($ES$).
- Catalytic conversion occurs, yielding products ($P$) which desorb from the active site, leaving the enzyme unchanged to catalyze subsequent cycles:

      $E + S \rightleftharpoons ES \rightarrow E + P$