Calculating Rates of Reaction Worksheet

Fundamentals of Chemical Kinetics and Reaction Rates

• Definition of Reaction Kinetics:

  • Kinetics is the branch of chemistry concerned with the study of reaction rates, examining how fast a reaction occurs and how chemical entities change over a specific period of time.

  • Reaction rates quantify either the rate at which a reactant is consumed (used up, conventionally represented by a negative rate of change: $-\frac{\Delta[\text{reactant}]}{\Delta t}$) or the rate at which a product is formed (accumulated, represented by a positive rate of change: $+\frac{\Delta[\text{product}]}{\Delta t}$

• Standard Mathematical Representation:

  • Reaction rate is expressed fundamentally as the change in concentration divided by the change in time:     $\text{Reaction Rate} = \frac{\Delta\text{concentration}}{\Delta t} = -\frac{\Delta[\text{reactant}]}{\Delta t} = +\frac{\Delta[\text{product}]}{\Delta t}$

  • Square brackets denote molar concentration in units of moles per liter ($mol/L$ or $M$). For example, $[HCl]$ represents the molar concentration of hydrochloric acid.

  • While mathematical rate expressions for disappearing reactants yield a negative numeric value (e.g., $-0.125\,M/\text{min}$), chemical reaction rates are conventionally expressed as positive numbers with descriptive phrasing, such as: "consumed at a rate of $0.125\,M/\text{min}$".

• Non-Constant Nature of Reaction Rates:

  • Reaction rates are dynamic and change continuously throughout the course of a reaction; rate is not a constant.

  • A reaction proceeds at its fastest rate at the very beginning ($t = 0$) because the concentration of reactant particles is at its maximum, leading to the highest frequency of successful collisions.

  • As reactants are progressively consumed and converted to products, the concentration of reactant particles decreases, reducing collision frequency and causing the reaction rate to slow down over time.

• Limiting Versus Excess Reactants in Reaction Progress:

  • In reactions with unequal stoichiometric quantities, the limiting reactant is completely consumed over time, causing its concentration to decline to zero ($[\text{reactant}] = 0$).

  • The excess reactant is not entirely consumed; its concentration curve slopes downward initially and eventually plateaus at a horizontal line at a non-zero concentration value once the limiting reactant is exhausted.

• Average Rate Versus Instantaneous Rate:

  • Average Rate of Reaction: Measured over a finite time interval ($\Delta t = t_f - t_i$) between two distinct data points on a concentration-time or mass-time curve:     $\text{Average Rate} = \frac{[A]_f - [A]_i}{t_f - t_i}$

  • Instantaneous Rate of Reaction: The rate of reaction occurring at an exact, specific moment in time. It is graphically determined by drawing a tangent line to the concentration-time curve at that precise time point and calculating the slope of the tangent line:     $\text{Instantaneous Rate} = \text{slope of tangent} = \frac{x_2 - x_1}{t_2 - t_1}$     where $(t_1, x_1)$ and $(t_2, x_2)$ represent coordinates selected directly from the linear tangent line.

Experimental Methods for Monitoring Reaction Rates

• Observable Physical Properties:

  • In laboratory investigations, reaction rates are monitored by measuring physical property changes per unit of elapsed time:

  • Gas Volume ($mL$ or $L$): Used when one or more products are insoluble or slightly soluble gases collected in an inverted burette or gas syringe.

  • Color Change: Monitored qualitatively or quantitatively using spectrophotometry or colorimetry when a reactant or product absorbs visible light.

  • Mass Change of Reactant: Direct periodic weighing of a solid reactant as it is consumed.

  • Mass Change of Beaker and Contents: Used when a gas is evolved and allowed to escape an open reaction vessel placed directly onto an analytical balance.

  • Mole Change or Concentration Change ($mol/L$ or $M$): The primary and most common standard metric in chemical kinetics.

• Case Study — Decomposition and Dissolution of Copper in Sulfuric Acid:

  • Consider the redox reaction:     $Cu(s) + 2H_2SO_4(aq) \rightarrow CuSO_4(aq) + 2H_2O(l) + SO_2(g)$

  • Four distinct experimental methods can be used to track the rate of this reaction:

  • Color Change: The reaction mixture transitions from colorless (due to clear $H_2SO_4(aq)$) to distinctly blue as aqueous copper(II) sulfate ($CuSO_4(aq)$, containing hydrated $Cu^{2+}$ ions) is formed.

  • Change in Mass of Solid Copper: Periodically removing, drying, and weighing the unreacted $Cu(s)$ strip to track the decrease in mass over time.

  • Change in Volume of Gas: Collecting and measuring the volume of evolved sulfur dioxide ($SO_2(g)$) in a gas syringe over regular time intervals.

  • Change in Mass of Beaker and Contents: Placing the open reaction beaker on an electronic balance; because $SO_2(g)$ escapes the open vessel, the total mass of the system decreases at a rate proportional to $SO_2$ evolution.

Quantitative Rate Calculations and Reaction Stoichiometry

• Stoichiometric Mole Ratios in Rates:

  • The rates of consumption of reactants and rates of formation of products are strictly related to one another through the stoichiometric coefficients of the balanced chemical equation.

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

  • The generalized rate equality is:     $\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

• Worked Example 1: Rate Comparison in Model Reaction $A \rightarrow B$:

  • Experimental concentration data:

  • At $t = 0.0\,\text{min}$: $[A] = 1.000\,M$, $[B] = 0.000\,M$

  • At $t = 3.0\,\text{min}$: $[A] = 0.400\,M$, $[B] = 0.600\,M$

  • At $t = 6.0\,\text{min}$: $[A] = 0.250\,M$, $[B] = 0.750\,M$

  • Average rate of $A$ consumption over the whole time ($0.0$ to $6.0\,\text{min}$):     $\text{Rate}_A = \frac{[A]_f - [A]_i}{t_f - t_i} = \frac{0.250\,M - 1.000\,M}{6.0\,\text{min} - 0.0\,\text{min}} = \frac{-0.750\,M}{6.0\,\text{min}} = -0.125\,M/\text{min}$     $A$ is consumed at a rate of $0.125\,M/\text{min}$.

  • Average rate of $A$ consumption in the first $3\,\text{min}$ ($0.0$ to $3.0\,\text{min}$):     $\text{Rate}_A = \frac{0.400\,M - 1.000\,M}{3.0\,\text{min} - 0.0\,\text{min}} = \frac{-0.600\,M}{3.0\,\text{min}} = -0.200\,M/\text{min}$     $A$ is consumed at a rate of $0.200\,M/\text{min}$.

  • Average rate of $B$ production over the whole time ($0.0$ to $6.0\,\text{min}$):     $\text{Rate}_B = \frac{[B]_f - [B]_i}{t_f - t_i} = \frac{0.750\,M - 0.000\,M}{6.0\,\text{min} - 0.0\,\text{min}} = \frac{0.750\,M}{6.0\,\text{min}} = 0.125\,M/\text{min}$

  • Observations and Comparative Insights:

  • The consumption rate of $A$ is significantly faster in the first $3\,\text{min}$ ($0.200\,M/\text{min}$) than over the entire $6\,\text{min}$ period ($0.125\,M/\text{min}$) because a higher population of reactant particles at the beginning yields more collisions.

  • Over the entire duration, $[B]$ is produced at the exact same numeric rate ($0.125\,M/\text{min}$) as $[A]$ is consumed ($0.125\,M/\text{min}$) because their stoichiometric ratio in the balanced equation is $1:1$.

• Worked Example 2: Decomposition of Hydrogen Peroxide:

  • Given the reaction:     $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$

  • If the rate of decomposition of hydrogen peroxide ($H_2O_2$) is $4.00\,M/\text{min}$, the rate of formation of $O_2$ is calculated stoichiometrically:     $\text{Rate of } O_2 = 4.00\,\frac{mol\,H_2O_2}{L \cdot \text{min}} \times \frac{1\,mol\,O_2}{2\,mol\,H_2O_2} = 2.00\,M/\text{min} = 2.00\,\frac{mol\,O_2}{L \cdot \text{min}}$

• Worked Example 3: Aluminum Reaction in Excess Acid (Production Rate in $mol/s$):

  • Balanced reaction:     $2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g)$

  • A $10.0\,g$ sample of aluminum reacts completely in excess hydrochloric acid over $300.0\,s$. Rate of hydrogen gas production in $mol/s$:     $\text{Rate of } H_2 = \frac{10.0\,g\,Al}{300.0\,s} \times \frac{1\,mol\,Al}{26.98\,g\,Al} \times \frac{3\,mol\,H_2}{2\,mol\,Al} = 0.00185\,mol/s$

• Worked Example 4: Decomposition of Water (Time Calculation):

  • Balanced reaction:     $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$

  • The rate of production of oxygen is $1.2 \times 10^{-2}\,mol/s$. Time in seconds required to decompose $100.0\,g$ of water:     $t = 100.0\,g\,H_2O \times \frac{1\,mol\,H_2O}{18.02\,g\,H_2O} \times \frac{1\,mol\,O_2}{2\,mol\,H_2O} \times \frac{1\,s}{1.2 \times 10^{-2}\,mol\,O_2} = 231\,s$

• Worked Example 5: Aluminum Reaction Over Extended Duration (Mass Rate in $g/\text{min}$):

  • Reaction:     $2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g)$

  • Experimental mass of $Al(s)$: at $t = 0.0\,\text{min}$, $m = 5.67\,g$; at $t = 90.0\,\text{min}$, $m = 1.10\,g$.

  • Rate of $Al$ consumption:     $\text{Rate}_{Al} = \frac{1.10\,g - 5.67\,g}{90.0\,\text{min} - 0.0\,\text{min}} = \frac{-4.57\,g}{90.0\,\text{min}} = -0.0508\,g/\text{min} \rightarrow 0.0508\,g/\text{min}$

  • Rate of hydrogen gas produced in $g/\text{min}$:     $\text{Rate}_{H_2} = 0.0508\,\frac{g\,Al}{\text{min}} \times \frac{1\,mol\,Al}{26.98\,g\,Al} \times \frac{3\,mol\,H_2}{2\,mol\,Al} \times \frac{2.02\,g\,H_2}{1\,mol\,H_2} = 0.00571\,g/\text{min}$

• Worked Example 6: Combustion of Paraffin Wax Candle ($C_{20}H_{42}$):

  • Balanced combustion reaction:     $2C_{20}H_{42}(s) + 61O_2(g) \rightarrow 40CO_2(g) + 42H_2O(g)$

  • Given a $CO_2$ production rate of $0.98\,g/\text{min}$, the mass of candle consumed in $30.0\,s$ ($0.50\,\text{min}$) is:     $\text{Mass consumed} = \frac{0.98\,g\,CO_2}{1\,\text{min}} \times \frac{1\,mol\,CO_2}{44.01\,g\,CO_2} \times \frac{2\,mol\,C_{20}H_{42}}{40\,mol\,CO_2} \times \frac{282.62\,g\,C_{20}H_{42}}{1\,mol\,C_{20}H_{42}} \times 0.50\,\text{min} = 0.16\,g$

• Worked Example 7: Aluminum and Copper(II) Chloride Reaction Duration:

  • Balanced single replacement equation:     $2Al(s) + 3CuCl_2(aq) \rightarrow 2AlCl_3(aq) + 3Cu(s)$

  • If aluminum is consumed at a rate of $0.46\,g/\text{min}$, the time required to produce $0.89\,g$ of $Cu$ is:     $t = 0.89\,g\,Cu \times \frac{1\,mol\,Cu}{63.55\,g\,Cu} \times \frac{2\,mol\,Al}{3\,mol\,Cu} \times \frac{26.98\,g\,Al}{1\,mol\,Al} \times \frac{1\,\text{min}}{0.46\,g\,Al} = 0.55\,\text{min}$

• Worked Example 8: Combustion of Paraffin Wax ($C_{28}H_{58}$) Over Time Intervals:

  • Mass of candle recorded over time:

  • $0.0\,\text{min}: 25.6\,g$

  • $6.0\,\text{min}: 25.1\,g$

  • $12.0\,\text{min}: 24.5\,g$

  • $18.0\,\text{min}: 23.9\,g$

  • $24.0\,\text{min}: 23.4\,g$

  • $30.0\,\text{min}: 22.8\,g$

  • Average rate of paraffin consumption from $12.0$ to $24.0\,\text{min}$:     $\text{Rate} = \frac{23.4\,g - 24.5\,g}{24.0\,\text{min} - 12.0\,\text{min}} = \frac{-1.1\,g}{12.0\,\text{min}} = -0.0917\,g/\text{min} \rightarrow 0.0917\,g/\text{min}$

  • Rate of $CO_2$ production in $mol/\text{min}$ over this same interval:     $2C_{28}H_{58}(s) + 85O_2(g) \rightarrow 56CO_2(g) + 58H_2O(g)$     $\text{Rate of } CO_2 = \frac{0.0917\,g\,C_{28}H_{58}}{1\,\text{min}} \times \frac{1\,mol\,C_{28}H_{58}}{394.86\,g\,C_{28}H_{58}} \times \frac{56\,mol\,CO_2}{2\,mol\,C_{28}H_{58}} = 0.0065\,mol/\text{min}$

• Worked Example 9: Oxidation of Copper by Nitric Acid:

  • Balanced reaction:     $3Cu(s) + 8HNO_3(aq) \rightarrow 3Cu(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)$

  • Mass of beaker and open contents measured over time:

  • $0.0\,\text{min}: 200.00\,g$

  • $1.0\,\text{min}: 197.50\,g$

  • $2.0\,\text{min}: 195.45\,g$

  • $3.0\,\text{min}: 193.55\,g$

  • $4.0\,\text{min}: 191.70\,g$

  • $5.0\,\text{min}: 189.90\,g$

  • $6.0\,\text{min}: 188.15\,g$

  • $7.0\,\text{min}: 186.45\,g$

  • $8.0\,\text{min}: 184.80\,g$

  • Reaction rate for the time period $2.0$ to $6.0\,\text{min}$:     $\text{Rate} = \frac{\Delta\text{mass}}{\Delta t} = \frac{188.15\,g - 195.45\,g}{6.0\,\text{min} - 2.0\,\text{min}} = \frac{-7.30\,g}{4.0\,\text{min}} = -1.825\,g/\text{min}$     Mass is lost (consumed/escaped) at an average rate of $1.825\,g/\text{min}$.

  • Mass of copper consumed in the first $5.0\,\text{min}$:

  • The only substance escaping the open reaction vessel is nitric oxide gas ($NO(g)$). Therefore, total system mass loss equals the mass of $NO(g)$ produced.

  • Mass of $NO$ evolved in first $5.0\,\text{min}$:       $\Delta m_{NO} = 200.00\,g - 189.90\,g = 10.10\,g\,NO$

  • Corresponding rate of $NO$ production:       $\text{Rate}_{NO} = \frac{10.10\,g\,NO}{5.0\,\text{min}} = 2.02\,g/\text{min}$

  • Mass of copper consumed:       $m_{Cu} = 10.10\,g\,NO \times \frac{1\,mol\,NO}{30.01\,g\,NO} \times \frac{3\,mol\,Cu}{2\,mol\,NO} \times \frac{63.55\,g\,Cu}{1\,mol\,Cu} = 32.1\,g\,Cu$

• Worked Example 10: Deducing Stoichiometric Coefficients from Instantaneous Rates:

  • In an unknown reaction involving species $C$, $X$, and $W$, instantaneous rates are measured at $t = 60\,s$ from tangents on a Concentration vs. Time graph:

  • For reactant $W$: Tangent passes through $(0, 0.04)$ and $(150, 0.34)$       $\text{Rate}_W = \frac{0.34\,M - 0.04\,M}{150\,s - 0\,s} = \frac{0.30\,M}{150\,s} = +0.0020\,M/s$

  • For reactant $X$: Tangent passes through $(0, 0.42)$ and $(195, 0)$       $\text{Rate}_X = \frac{0\,M - 0.42\,M}{195\,s - 0\,s} = \frac{-0.42\,M}{195\,s} = -0.0021\,M/s \approx -0.002\,M/s$

  • For reactant $C$: Tangent passes through $(0, 0.48)$ and $(120, 0.22)$       $\text{Rate}_C = \frac{0.22\,M - 0.48\,M}{120\,s - 0\,s} = \frac{-0.26\,M}{120\,s} = -0.0022\,M/s \approx -0.002\,M/s$

  • Balanced Equation:     $C + X \rightarrow W$

  • Stoichiometric Deduction: Because the instantaneous consumption rates of $C$ and $X$ ($0.002\,M/s$) match the instantaneous production rate of $W$ ($0.002\,M/s$), the stoichiometric coefficients for all participating species are identical and equal to 1.

Collision Theory and Molecular Dynamics

• The Three Postulates of Collision Theory:

  • Collision Theory provides a mechanical framework explaining why different reactions occur at different rates. For any reaction to occur:

  1. Collision Requirement: Reactant particles (atoms, ions, or molecules) must physically collide.

  2. Orientation/Geometry Requirement: Reactant particles must collide with the correct spatial orientation to align the reactive centers.

  3. Energy Requirement: Reactant particles must collide with kinetic energy equal to or greater than a specific threshold known as the activation energy ($E \ge E_a$).

• Effective Versus Ineffective Collisions:

  • An effective collision fulfills both the geometric and energetic criteria simultaneously. Bonds within the reactants are broken or distorted, forming an activated complex that resolves into product molecules.

  • An ineffective collision fails on at least one requirement:

  • Incorrect Orientation: Even if colliding with high kinetic energy, particles that hit with misaligned atoms cannot form the activated complex and rebound elastically with their chemical identities intact.

  • Insufficient Energy ($E < E_a$): Particles colliding with proper orientation but possessing kinetic energy below $E_a$ are repelled by electrostatic repulsion of their electron clouds, resulting in a perfectly elastic bounce without reaction.

• Mechanistic Sequence of Carbon Monoxide and Nitrogen Dioxide Collision:

  • Consider the bimolecular collision:     $CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g)$

  • Case 1 — Effective Collision: The carbon atom of $CO$ collides directly with an oxygen atom of $NO_2$ with kinetic energy exceeding $E_a$. A short-lived transition state (the activated complex, $O-C\cdots O-N-O$) is formed, enabling the oxygen atom to transfer, successfully producing $CO_2$ and $NO$.

  • Case 2 — Ineffective Collision via Incorrect Orientation (Oxygen-to-Oxygen contact): The oxygen atom of $CO$ collides with an oxygen atom of $NO_2$. Despite colliding, the atoms rebound unchanged due to incorrect spatial geometry.

  • Case 3 — Ineffective Collision via Incorrect Orientation (Carbon-to-Nitrogen contact): The carbon atom of $CO$ collides with the central nitrogen atom of $NO_2$. The collision fails to present the correct alignment for an oxygen transfer, resulting in a rebound.

  • Case 4 — Ineffective Collision via Low Kinetic Energy: The carbon atom of $CO$ aligns with an oxygen atom of $NO_2$, but the relative velocity is too low. Electrostatic repulsion halts approach before chemical interaction occurs; the particles rebound elastically.

Maxwell-Boltzmann Distribution and Temperature Dependence

• Core Principles of Kinetic Molecular Theory (KMT):

  • Molecules are in constant, random motion and travel at varying velocities.

  • Intermolecular collisions in an ideal system are perfectly elastic; total kinetic energy is conserved without net dissipation.

  • The average kinetic energy ($KE_{\text{avg}}$) of a molecular population is directly proportional to the absolute temperature in Kelvin ($KE_{\text{avg}} \propto T$).

• Characteristics of the Maxwell-Boltzmann Distribution Curve:

  • In any gas or liquid sample, individual particles possess a broad distribution of kinetic energies due to random, continuous collisions.

  • Total Area Under Curve: Represents the total number of particles in the system and remains constant regardless of temperature changes.

  • Origin: The curve starts at the origin $(0,0)$ because zero particles have zero kinetic energy.

  • Asymmetry: The distribution is positively skewed, possessing a long high-energy tail extending to the right that approaches but never touches the horizontal axis.

• Effect of Temperature on the Distribution Curve:

  • When temperature increases (e.g., from $300\,K$ to $310\,K$):

  • The average kinetic energy increases, shifting the peak of the curve to the right.

  • The curve broadens and the height of the maximum peak decreases, preserving the total area under the curve.

  • A significantly larger proportion of molecules now possess velocities and kinetic energies equal to or exceeding the activation energy ($E_a$).

• Correlating Maxwell-Boltzmann Curves to Activation Energy:

  • A vertical boundary line is placed on the distribution curve at the value of the activation energy ($E_a$):

  • Sub-Threshold Area (White/Unshaded Area, $E < E_a$): Represents the vast fraction of particles that do not possess sufficient energy to undergo an effective collision upon impact.

  • Super-Threshold Area (Shaded Tail, $E \ge E_a$): Represents the proportion of particles capable of overcoming the energetic barrier to react upon impact.

  • A modest temperature increase produces an exponential expansion in the shaded area under the tail beyond $E_a$, providing a direct explanation for the dramatic rate increase observed with small rises in temperature.

Potential Energy Diagrams and Reaction Energetics

• Anatomy of a Potential Energy Profile:

  • Potential Energy (PE): Plotted along the vertical axis (typically in $kJ$ or $kJ/mol$), reflecting the stored chemical potential energy of chemical species.

  • Reaction Progress / Coordinate: Plotted along the horizontal axis, tracking the geometric transformation from separated reactants to products over time.

  • Activation Energy ($E_a$): The energy required to transition reactants to the activated complex. It is always calculated as the potential energy difference between the activated complex peak and the starting reactants:     $E_{a(\text{forward})} = PE_{\text{activated complex}} - PE_{\text{reactants}}$     $E_{a(\text{reverse})} = PE_{\text{activated complex}} - PE_{\text{products}}$

  • Enthalpy Change of Reaction (Heat of Reaction, $\Delta H$): The net change in chemical potential energy between products and reactants:     $\Delta H = PE_{\text{products}} - PE_{\text{reactants}}$

• Exothermic Versus Endothermic Criteria:

  • Exothermic Reactions: The potential energy of the products is lower than the potential energy of the reactants ($PE_{\text{products}} < PE_{\text{reactants}}$). The enthalpy change is negative ($\Delta H < 0$), releasing heat to the surroundings.

  • Endothermic Reactions: The potential energy of the products is higher than the potential energy of the reactants ($PE_{\text{products}} > PE_{\text{reactants}}$). The enthalpy change is positive ($\Delta H > 0$), absorbing heat from the surroundings.

• Microscopic Interconversion Between Potential and Kinetic Energy:

  • By the law of conservation of energy, the total energy of an isolated molecular collision is constant ($E_{\text{total}} = KE + PE$).

  • During Particle Approach: As electron clouds of approaching reactant molecules overlap, mutual electrostatic repulsion increases. Particles decelerate, causing kinetic energy to convert into potential energy ($KE \downarrow$, $PE \uparrow$).

  • At the Transition Peak (Activated Complex): Potential energy reaches its absolute maximum, corresponding to the minimum kinetic energy of the system ($PE = \text{maximum}$, $KE = \text{minimum}$, particles moving at their slowest instantaneous speed).

  • During Product Separation: As newly formed products push apart, electrostatic repulsion drives them outward, accelerating the fragments. Potential energy drops and converts back into kinetic energy ($PE \downarrow$, $KE \uparrow$, particles moving faster).

  • Bonds and Stability: Species with the lowest potential energy possess the highest kinetic stability and strongest bonds. Species with the highest potential energy (the activated complex) have the weakest, partially formed bonds.

• Comprehensive Quantitative Analysis — Endothermic Profile ($X_2 + Y_2 \rightarrow 2XY$):

  • Given potential energy benchmarks:

  • Reactants ($X_2 + Y_2$): $50\,kJ$

  • Products ($2XY$): $100\,kJ$

  • Activated Complex ($X_2Y_2$): $130\,kJ$

  • Energetic Calculations:

  • Overall reaction: Endothermic

  • Activation energy for forward reaction: $E_{a(\text{fwd})} = 130\,kJ - 50\,kJ = 80\,kJ$

  • Activation energy for reverse reaction: $E_{a(\text{rev})} = 130\,kJ - 100\,kJ = 30\,kJ$

  • Enthalpy change of forward reaction: $\Delta H_{\text{fwd}} = 100\,kJ - 50\,kJ = +50\,kJ$

  • Enthalpy change of reverse reaction: $\Delta H_{\text{rev}} = 50\,kJ - 100\,kJ = -50\,kJ$ (Exothermic)

  • Species forming the activated complex: $X_2Y_2$

  • Species with the highest potential energy: $X_2Y_2$ ($130\,kJ$

  • Species with the highest kinetic energy (lowest PE): $X_2 + Y_2$ ($50\,kJ$

  • Species with the lowest kinetic energy: $X_2Y_2$

  • Species with the weakest bonds: $X_2Y_2$

  • Species with the strongest bonds: $X_2 + Y_2$

  • Enthalpy change for decomposition of activated complex to reactants ($X_2Y_2 \rightarrow X_2 + Y_2$): $\Delta H = 50\,kJ - 130\,kJ = -80\,kJ$

  • Comparison of forward and reverse rates: The reverse reaction is substantially faster than the forward reaction because