Chapter 7: Chemical Reactions - Energy, Rates, and Equilibrium

Energy Fundamentals and Chemical Bonds

• Definition of Energy and Work:

  • Energy represents the capacity to supply heat or do work.

  • Work ($w$) is defined as the process of causing matter to move against an opposing force (e.g., inflating a bicycle tire involves moving air matter in a pump against the opposing atmospheric and tire air pressure).

• Classification of Energy:

  • Potential Energy: The energy an object possesses due to its relative position, composition, or condition; often termed stored energy.

  • Examples: Water at the top of a waterfall or dam (gravitational position); chemical energy stored in molecular bonds (composition); chemical batteries containing potential energy capable of driving electric currents.

  • Kinetic Energy: The energy that an object possesses because of its motion.

  • Examples: Downward-flowing water driving hydroelectric generators to produce electricity; rapidly expanding gaseous products in an internal combustion engine exerting mechanical kinetic energy on engine pistons.

• Solar Energy Origin and Utilization:

  • Over $90\%$ of all energy consumed on Earth originates from the sun.

  • Daily solar input provides approximately $10{,}000$ times the total energy required to satisfy global daily human energy demands.

  • Photosynthetic organisms (plants and photosynthetic bacteria) trap solar radiation and store it as chemical bond energy.

  • Stored solar energy is released via the combustion of biomass (wood, ethanol) or fossil fuels (petroleum, coal), which represent fossilized animal and plant matter.

  • Nutritional metabolism breaks down dietary macronutrients (proteins, carbohydrates, lipids) to sustain cellular processes.

  • Industrial manufacturing expends vast amounts of chemical and thermal energy to refine raw metals (such as iron and aluminum) into structural items.

• Conservation Laws:

  • Law of Conservation of Energy (First Law of Thermodynamics): Energy can neither be created nor destroyed during chemical or physical changes; it can only transition from one form to another.

  • Law of Conservation of Matter: During a chemical transformation, there is no detectable change in total matter quantity.

  • Chemical vs. Nuclear Reactions: In ordinary chemical reactions, mass conversions are immeasurably small, meaning mass and energy are independently conserved. In nuclear reactions, energy releases are larger by factors of approximately $10^6$, making matter-energy transformations measurable. Combining these concepts yields the unified principle: The total quantity of matter and energy in the universe is fixed.

Bond Dissociation Energy and Enthalpy Calculations

• Molecular Stability and Energy Minimum:

  • Atoms assemble into chemical bonds because bonded arrangements attain lower potential energy than isolated, individual atoms.

  • The formation of chemical bonds releases energy (predominantly as heat), behaving analogously to an object rolling down a slide toward minimum energy.

  • Breaking chemical bonds requires an input of energy (an endothermic process).

• Bond Dissociation Energy (BDE):

  • Defined as the characteristic energy required to cleave a specific covalent bond between atoms in a molecule, generating separated neutral fragments.

  • The quantity of energy absorbed to rupture a bond is identical to the quantity liberated when that bond forms.

  • Whereas covalent bond strength is quantified by bond dissociation energy, ionic crystal stability is quantified by lattice energy.

  • Multiple bonds exhibit greater bond dissociation energies than corresponding single bonds (e.g., $\text{C–C} = 85\,\text{kcal/mol}$, $\text{C=C} = 145\,\text{kcal/mol}$, $\text{C}\equiv\text{C} = 200\,\text{kcal/mol}$; note that some texts approximate $\text{C–C}$ at $80\,\text{kcal/mol}$).

• Standard Covalent Bond Dissociation Energies ($\text{kcal/mol}$):

  • $\text{C–H} = 100\,\text{kcal/mol}$

  • $\text{C–O} = 86\,\text{kcal/mol}$

  • $\text{C=O} = 190\,\text{kcal/mol}$

  • $\text{C–N} = 70\,\text{kcal/mol}$

  • $\text{C–C} = 85\,\text{kcal/mol}$

  • $\text{C=C} = 145\,\text{kcal/mol}$

  • $\text{C}\equiv\text{C} = 200\,\text{kcal/mol}$

  • $\text{N–H} = 93\,\text{kcal/mol}$

  • $\text{H–H} = 105\,\text{kcal/mol}$

  • $\text{Br–Br} = 46\,\text{kcal/mol}$

  • $\text{Cl–Cl} = 58\,\text{kcal/mol}$

  • $\text{O–H} = 110\,\text{kcal/mol}$

  • $\text{O=O} = 119\,\text{kcal/mol}$

  • $\text{H–Br} = 87\,\text{kcal/mol}$

  • $\text{H–Cl} = 103\,\text{kcal/mol}$

• Enthalpy Change ($\Delta H$) or Heat of Reaction:

  • The difference between total heat absorbed during reactant bond cleavage and total heat released during product bond formation at constant pressure is the heat of reaction ($\Delta H$).

  • Estimated using average bond dissociation energies:   $\Delta H \approx \sum (\text{BDE of bonds broken}) - \sum (\text{BDE of bonds formed})$

• Worked Example: Synthesis of Water:

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

  • Reactant bonds cleaved:

  • $2\,\text{mol } \text{H–H} = 2\,\text{mol} \times 105\,\text{kcal/mol} = 210\,\text{kcal}$

  • $1\,\text{mol } \text{O=O} = 1\,\text{mol} \times 119\,\text{kcal/mol} = 119\,\text{kcal}$

  • Total energy absorbed = $210\,\text{kcal} + 119\,\text{kcal} = 329\,\text{kcal}$

  • Product bonds synthesized:

  • $4\,\text{mol } \text{O–H} = 4\,\text{mol} \times 110\,\text{kcal/mol} = 440\,\text{kcal}$

  • Total energy liberated = $440\,\text{kcal}$

  • Net enthalpy calculation:   $\Delta H \approx 329\,\text{kcal} - 440\,\text{kcal} = -111\,\text{kcal}$

  • Thermochemical notation:   $2H_2(g) + O_2(g) \rightarrow 2H_2O(g) + 111\,\text{kcal}$

• Worked Example 7.2.1: Formation of Hydrogen Bromide:

  • Reaction: $H_2(g) + Br_2(g) \rightarrow 2HBr(g)$

  • Bonds broken:

  • $1\,\text{mol } \text{H–H} = 105\,\text{kcal}$

  • $1\,\text{mol } \text{Br–Br} = 46\,\text{kcal}$

  • Total reactant input = $151\,\text{kcal}$

  • Bonds formed:

  • $2\,\text{mol } \text{H–Br} = 2 \times 87\,\text{kcal} = 174\,\text{kcal}$

  • Total product output = $174\,\text{kcal}$

  • Calculation:   $\Delta H \approx 151\,\text{kcal} - 174\,\text{kcal} = -23\,\text{kcal}$

  • Classification: Exothermic ($\Delta H < 0$).

• Exercise 7.2.1: Formation of Hydrogen Chloride:

  • Reaction: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

  • Bonds broken:

  • $1\,\text{mol } \text{H–H} = 105\,\text{kcal}$

  • $1\,\text{mol } \text{Cl–Cl} = 58\,\text{kcal}$

  • Total = $163\,\text{kcal}$

  • Bonds formed:

  • $2\,\text{mol } \text{H–Cl} = 2 \times 103\,\text{kcal} = 206\,\text{kcal}$

  • Calculation:   $\Delta H \approx 163\,\text{kcal} - 206\,\text{kcal} = -43\,\text{kcal}$

  • Classification: Exothermic ($\Delta H = -43\,\text{kcal}$).

Exothermic and Endothermic Processes

• Thermodynamic Definitions:

  • System: The specific reaction mixture undergoing chemical transformation.

  • Surroundings: Everything outside the reaction system (solvent, container, atmosphere).

• Endothermic Reactions:

  • Chemical processes that absorb thermal energy from their surroundings.

  • Heat is treated as a reactant.

  • Enthalpy change is positive ($\Delta H > 0$).

  • Temperature of surroundings decreases (container feels cold).

  • Product bonds are weaker (higher in potential energy) than reactant bonds.

  • Example: Thermal decomposition of limestone:   $CaCO_3(s) + 177.8\,\text{kJ} \rightarrow CaO(s) + CO_2(g)$   $\Delta H = +177.8\,\text{kJ}$

• Exothermic Reactions:

  • Chemical processes that release thermal energy to their surroundings.

  • Heat is treated as a product.

  • Enthalpy change is negative ($\Delta H < 0$).

  • Temperature of surroundings increases (container feels hot).

  • Product bonds are stronger (lower in potential energy) than reactant bonds.

  • Example: Combustion of methane:   $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) + 890.4\,\text{kJ}$   $\Delta H = -890.4\,\text{kJ}$

  • Alternative representation in kilocalories:   $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) + 213\,\text{kcal}$

• Comparison Summary:

• Classification Practice:

  • $N_2(g) + O_2(g) + 45\,\text{kcal} \rightarrow 2NO(g)$: Endothermic (heat is consumed as a reactant; $\Delta H = +45\,\text{kcal}$).

  • $H_2(g) + F_2(g) \rightarrow 2HF(g) + 130\,\text{kcal}$: Exothermic (heat is evolved as a product; $\Delta H = -130\,\text{kcal}$).

  • $2C(s) + H_2(g) + 5.3\,\text{kcal} \rightarrow C_2H_2(g)$: Endothermic (heat is consumed; $\Delta H = +5.3\,\text{kcal}$).

  • Combustion of paper: Exothermic (oxidative breakdown produces thermal energy and light).

Reaction Spontaneity, Entropy, and Gibbs Free Energy

• Spontaneous vs. Nonspontaneous Processes:

  • Spontaneous Reaction: A reaction that favors the formation of products under the specific conditions of temperature and pressure without continuous external intervention.

  • Example: A roaring bonfire releases thermal energy ($\Delta H < 0$) and yields gaseous products ($CO_2$, $H_2O$ vapor), increasing disorder.

  • Nonspontaneous Reaction: A reaction that does not favor product formation under specified conditions.

  • Example: Synthesis of nitric oxide from atmospheric gases: $N_2(g) + O_2(g) \rightarrow 2NO(g)$ is nonspontaneous at standard conditions because it is highly endothermic; it occurs only under extreme energetic input, such as lightning discharges.

  • Spontaneity vs. Reaction Rate: Spontaneity describes thermodynamic favorability, not velocity. A spontaneous process can occur at an immeasurably slow speed (e.g., room-temperature paper does not ignite spontaneously because it lacks sufficient activation energy to initiate combustion).

• Entropy ($S$) as a Driving Force:

  • Entropy is the thermodynamic property measuring system disorder or randomness, quantified in units of Joules per mole-kelvin ($\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$).

  • Natural systems exhibit a fundamental drive to progress toward maximum entropy.

  • Physical phase progression of entropy: $S_{solid} < S_{liquid} \ll S_{gas}$.

  • Example: Melting ice transfers water molecules from an ordered hydrogen-bonded crystalline lattice into a disordered liquid state, increasing system entropy.

  • Rules for Predicting Sign of $\Delta S$:

  • Phase Changes: Conversions from solid to liquid, liquid to gas, or solid directly to gas result in $\Delta S > 0$.

  • Dissolution: Dissolving a solute in a solvent increases entropy ($\Delta S > 0$) as solute particles disperse throughout the solution.

  • Thermal Variations: Raising temperature increases particle kinetic velocities and vibrational states, leading to $\Delta S > 0$.

  • Molecular Stoichiometry: Reactions yielding a greater count of product gas molecules than reactant gas molecules display an increase in entropy ($\Delta S > 0$).

• Entropy Case Predictions:

  • Condensation of steam ($\text{gas} \rightarrow \text{liquid}$): Entropy decreases ($\Delta S < 0$).

  • Thermal decomposition producing gas and more moles: Entropy increases ($\Delta S > 0$).

  • Gaseous combination where four reactant moles form two product moles ($4\,\text{mol gas} \rightarrow 2\,\text{mol gas}$): Entropy decreases ($\Delta S < 0$).

  • Precipitation of a solid from aqueous solution: Entropy decreases ($\Delta S < 0$).

  • Reactions with equal gas mole counts on both sides: Entropy change is undetermined without precise thermodynamic data.

• Gibbs Free Energy ($G$):

  • Formulated by Josiah Willard Gibbs (1839–1903).

  • Defined as the portion of energy released during a process that is available to do useful thermodynamic work.

  • Gibbs Free Energy Equation:   $\Delta G = \Delta H - T\Delta S$   (where $T$ is absolute temperature in Kelvin, always a positive value).

  • Spontaneity Criteria:

  • $\Delta G < 0$: Spontaneous process.

  • $\Delta G > 0$: Nonspontaneous process.

  • $\Delta G = 0$: System at dynamic chemical equilibrium.

  • Biochemical Terminology:

  • Exergonic: Processes that release free energy ($\Delta G < 0$).

  • Endergonic: Processes that require input of free energy ($\Delta G > 0$).

• Spontaneity Conditions via Enthalpy and Entropy Combinations:

• Temperature Dependence Examples:

  • Ice Melting ($H_2O(s) \rightarrow H_2O(l)$): Endothermic ($\Delta H > 0$) and order-disrupting ($\Delta S > 0$). Below $273\,\text{K}$ ($0^\circ\text{C}$, where ice remains frozen), $\Delta H$ outweighs $T\Delta S$, yielding $\Delta G > 0$ (nonspontaneous). Above $273\,\text{K}$, the positive $T\Delta S$ term outweighs $\Delta H$, making $\Delta G < 0$ (spontaneous).

  • Water Freezing ($H_2O(l) \rightarrow H_2O(s)$): Exothermic ($\Delta H < 0$) with decreased entropy ($\Delta S < 0$). Above $273\,\text{K}$, $-T\Delta S$ dominates, giving $\Delta G > 0$. Below $273\,\text{K}$, the enthalpy term dominates, making $\Delta G < 0$ (spontaneous).

Chemical Kinetics and Collision Theory

• Chemical Kinetics Definition:

  • The quantitative branch of chemistry that investigates the speeds (rates) of chemical reactions and the mechanisms through which transformations take place.

• Reaction Rate Measurement:

  • Rate represents the change in the concentration of a reactant or product per unit time.

  • Primary units: Molarity per second ($\text{M/s}$ or $\text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$).

  • General rate formula for disappearing reactant $A$:   $\text{Rate} = -\frac{\Delta [A]}{\Delta t} = -\frac{[A]_{\text{final}} - [A]_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$   (The negative sign ensures that the calculated reaction rate remains a positive quantity).

• Reaction Rate Calculation Example:

  • An aqueous solution of reactant $A$ decreases from $1.00\,\text{M}$ to $0.70\,\text{M}$ over an elapsed duration of $20.0\,\text{s}$:   $\text{Rate} = -\frac{0.70\,\text{M} - 1.00\,\text{M}}{20.0\,\text{s}} = -\frac{-0.30\,\text{M}}{20.0\,\text{s}} = 0.015\,\text{M/s}$

• Speed Analogy:

  • A sprinter covering a $100\,\text{m}$ dash in $11.0\,\text{s}$ maintains an average rate of:   $\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{100\,\text{m}}{11.0\,\text{s}} = 9.09\,\text{m/s}$

  • By comparison, world-record sprinter Usain Bolt completed $100\,\text{m}$ in $9.58\,\text{s}$, achieving an average velocity of $10.4\,\text{m/s}$ ($23.4\,\text{mph}$).

• Collision Theory Fundamentals:

  • Chemical reactions occur via collisions between reactant particles.

  • For a collision to lead to product formation (an effective collision), two criteria must be satisfied:

  1. Sufficient Kinetic Energy: Particles must collide with kinetic energy equal to or surpassing the activation energy ($E_a$).

  2. Correct Spatial Orientation: Molecular geometries must align so that breaking and forming bonds can occur.

  • Ineffective collisions lack either sufficient kinetic energy or appropriate orientation, causing colliding species to rebound unchanged.

• Activation Energy ($E_a$):

  • The minimum kinetic energy required by colliding reactant molecules to overcome repulsive electron clouds, distort existing bonds, and initiate reaction.

  • Determines reaction speed: High $E_a$ corresponds to slow reactions at ambient temperatures; low $E_a$ corresponds to rapid reactions.

• Activated Complex (Transition State):

  • A transient, highly unstable chemical species existing at the apex of the potential energy barrier.

  • Lifetime is brief, lasting approximately $10^{-13}\,\text{s}$.

  • Carries equal probability of decomposing backward into intact reactants or continuing forward into products.

Factors Influencing Reaction Rates

• Concentration Effects:

  • Increasing the concentration of dissolved or gaseous reactants packs more particles into a given volume.

  • Increases the total collision frequency per unit time, directly accelerating the observed reaction rate.

• Temperature Effects:

  • Increasing reaction temperature increases the average kinetic energy of the molecules.

  • Molecules travel faster, moderately raising collision frequency.

  • Primary Factor: A significantly greater fraction of colliding molecules possesses kinetic energy exceeding the activation energy threshold ($E_k \ge E_a$), increasing the proportion of effective collisions.

• Catalytic Effects:

  • A catalyst is a chemical substance that enhances the speed of a reaction without being consumed or permanently altered in the process.

  • Returns to its initial composition and can participate repeatedly in catalytic cycles.

  • Appears written over the reaction yield arrow rather than as a reactant or product stoichiometry term.

  • Mechanism: Catalysts provide an alternative reaction pathway featuring a lower activation energy ($E'_a < E_a$).

  • Does not alter the initial or final chemical states; therefore, the overall enthalpy change ($\Delta H$) is unchanged.

• Demonstrated Catalytic Systems:

  • Decomposition of Hydrogen Peroxide:   $2H_2O_2(aq) \xrightarrow{MnO_2} 2H_2O(l) + O_2(g)$

  • Uncatalyzed hydrogen peroxide ($3\%$ household solution) degrades slowly over several years.

  • The addition of a trace amount of solid manganese(IV) oxide ($MnO_2$) drives rapid decomposition, reaching completion within minutes.

  • Biological Catalysts: Enzymes are specialized protein catalysts that accelerate biochemical metabolic processes under physiological conditions.

Reversible Reactions and Dynamic Equilibrium

• Reversibility of Chemical Reactions:

  • A reaction that proceeds in both forward and reverse directions simultaneously is termed a reversible reaction.

  • Example: Hydrogen and iodine reacting in a closed vessel:   $\text{Forward:} \quad H_2(g) + I_2(g) \rightarrow 2HI(g)$   $\text{Reverse:} \quad 2HI(g) \rightarrow H_2(g) + I_2(g)$

• Concept of Dynamic Equilibrium:

  • As product concentrations accumulate, the rate of the reverse reaction accelerates until it equals the rate of the forward reaction.

  • At chemical equilibrium, the forward and reverse reaction velocities are equal:   $\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}$

  • Macroscopic properties (concentrations, color, total pressure) remain constant.

  • Equilibrium is dynamic: reactions do not cease; both transformations occur continuously, canceling any net macroscopic change.

  • Represented using a double arrow ($\rightleftharpoons$):   $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

  • Equilibrium expressions can be inverted if the reaction is considered in reverse:   $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$

• Formulation Examples:

  • Decomposition of calcium carbonate:   $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

  • Reversible synthesis of water from elements:   $2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)$

Equilibrium Constants and Mathematical Formulations

• The Law of Mass Action:

  • Discovered in the mid-1860s by Norwegian scientists Cato Maximilian Guldberg and Peter Waage.

  • States that for any reversible chemical reaction at equilibrium, a characteristic ratio of product concentrations to reactant concentrations remains constant at a fixed temperature.

• Concentration Equilibrium Constant ($K_{eq}$):

  • For the general balanced reversible reaction:   $aA + bB \rightleftharpoons cC + dD$

  • The equilibrium constant expression is formulated as:   $K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

  • Square brackets represent equilibrium concentrations in molarity ($\text{M}$).

  • Stoichiometric coefficients from the balanced chemical equation serve as exponents.

  • By convention, units of molarity are omitted from $K_{eq}$.

  • $K_{eq}$ is constant for a given reaction at a fixed temperature; changes in concentration shift relative amounts without altering the numerical value of $K_{eq}$.

• Simple Theoretical Demonstration:

  • For the system $A \rightleftharpoons B$ with $K_{eq} = 2.0$:   $K_{eq} = \frac{[B]}{[A]} = 2.0$

  • If $[B] = 4.0\,\text{M}$, then $[A] = 2.0\,\text{M}$ (since $4.0 / 2.0 = 2.0$).

  • If $[B] = 6.0\,\text{M}$, then $[A] = 3.0\,\text{M}$ (since $6.0 / 3.0 = 2.0$).

• Worked Quantitative Examples ($K_{eq}$):

  • Example 1: Reaction: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

  • Given: $K_{eq} = 0.40$, $[HI] = 0.75\,\text{M}$, $[H_2] = 0.20\,\text{M}$. Calculate $[I_2]$.     $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$     $0.40 = \frac{(0.75)^2}{(0.20)[I_2]} = \frac{0.5625}{0.20[I_2]}$     $[I_2] = \frac{0.5625}{0.40 \times 0.20} = 7.0\,\text{M}$

  • Exercise 1: Reaction: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

  • Given: $K_{eq} = 0.40$, $[HI] = 0.060\,\text{M}$, $[I_2] = 0.90\,\text{M}$. Calculate $[H_2]$.     $0.40 = \frac{(0.060)^2}{[H_2](0.90)} = \frac{0.0036}{0.90[H_2]}$     $[H_2] = \frac{0.0036}{0.40 \times 0.90} = 0.010\,\text{M}$

  • Example 2: Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • Given: $K_{eq} = 13.7$, $[N_2] = 1.88\,\text{M}$, $[NH_3] = 6.62\,\text{M}$. Calculate $[H_2]$.     $K_{eq} = \frac{[NH_3]^2}{[N_2][H_2]^3}$     $13.7 = \frac{(6.62)^2}{(1.88)[H_2]^3} = \frac{43.8244}{1.88[H_2]^3}$     $[H_2]^3 = \frac{43.8244}{13.7 \times 1.88} = \frac{43.8244}{25.756} \approx 1.7015$     $[H_2] = \sqrt[3]{1.7015} = 1.19\,\text{M}$

  • Exercise 2: Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • Given: $K_{eq} = 13.7$, $[N_2] = 0.055\,\text{M}$, $[H_2] = 1.62\,\text{M}$. Calculate $[NH_3]$.     $13.7 = \frac{[NH_3]^2}{(0.055)(1.62)^3} = \frac{[NH_3]^2}{(0.055)(4.2515)} = \frac{[NH_3]^2}{0.23383}$     $[NH_3]^2 = 13.7 \times 0.23383 \approx 3.2035$     $[NH_3] = \sqrt{3.2035} = 1.79\,\text{M}$

• Pressure Equilibrium Constant ($K_P$):

  • For gas-phase equilibria: $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$

  • Formulated using equilibrium partial pressures in atmospheres ($\text{atm}$):   $K_P = \frac{P_C^c P_D^d}{P_A^a P_B^b}$

  • Example: Reaction: $2NO_2(g) \rightleftharpoons N_2O_4(g)$

  • Given: $P_{NO_2} = 0.664\,\text{atm}$, $P_{N_2O_4} = 1.09\,\text{atm}$.     $K_P = \frac{P_{N_2O_4}}{(P_{NO_2})^2} = \frac{1.09}{(0.664)^2} = \frac{1.09}{0.4409} = 2.47$

  • Exercise: Reaction: $H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)$

  • Given: $P_{H_2} = 0.44\,\text{atm}$, $P_{Cl_2} = 0.22\,\text{atm}$, $P_{HCl} = 2.98\,\text{atm}$.     $K_P = \frac{(P_{HCl})^2}{P_{H_2} P_{Cl_2}} = \frac{(2.98)^2}{(0.44)(0.22)} = \frac{8.8804}{0.0968} = 91.7$

• Mathematical Relationship Between $K_P$ and $K_{eq}$:

  • Related by the ideal gas expression:   $K_P = K_{eq}(RT)^{\Delta n}$

  • Universal Gas Constant: $R = 0.08205\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.

  • Absolute Temperature: $T$ in Kelvin ($\text{K} = ^\circ\text{C} + 273$).

  • Change in gaseous moles: $\Delta n = n_{\text{gas, products}} - n_{\text{gas, reactants}}$.

  • If $\Delta n = 0$, then $(RT)^0 = 1$, meaning $K_P = K_{eq}$.

  • Worked Conversion Problem 1:

  • Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • Given: $K_{eq} = 4.2 \times 10^{-2}$ at $25^\circ\text{C}$.

  • Temperature: $T = 25 + 273 = 298\,\text{K}$.

  • Mole change: $\Delta n = 2 - (1 + 3) = 2 - 4 = -2$.     $K_P = (4.2 \times 10^{-2}) \times [0.08205 \times 298]^{-2} = (4.2 \times 10^{-2}) \times [24.451]^{-2}$     $K_P = \frac{4.2 \times 10^{-2}}{597.85} = 7.0 \times 10^{-5}$

  • Worked Conversion Problem 2:

  • Given: $K_{eq} = 98.3$ at $25^\circ\text{C}$ with $\Delta n = +1$.     $K_P = 98.3 \times (0.08205 \times 298)^1 = 98.3 \times 24.451 = 2.40 \times 10^3$

• Heterogeneous Equilibria Rules:

  • A heterogeneous equilibrium involves chemical species in multiple phases (solids, liquids, gases, solutions).

  • Pure solids ($(s)$) and pure liquids ($(l)$) do not change concentration during reaction; therefore, they are excluded from $K_{eq}$ and $K_P$ expressions.

  • Only gaseous species ($(g)$) and aqueous solutes ($(aq)$) appear in equilibrium constant expressions.

  • Example: Decomposition of calcium carbonate:   $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

  • Solid $CaCO_3$ and solid $CaO$ are excluded, giving:   $K_{eq} = [CO_2] \quad \text{and} \quad K_P = P_{CO_2}$

Le Chatelier’s Principle and Equilibrium Shifts

• Definition of Le Chatelier’s Principle:

  • If an external stress (change in concentration, pressure, volume, or temperature) is applied to a chemical system at equilibrium, the equilibrium position shifts in the direction that minimizes or counteracts that stress.

• Effects of Concentration Alterations:

  • Addition of a Component: Shifts the equilibrium away from the added species (e.g., adding reactant shifts equilibrium toward products).

  • Removal of a Component: Shifts the equilibrium toward the removed species to replace it (e.g., removing product shifts equilibrium toward products).

  • Note: Altering concentrations shifts species ratios without changing the numerical value of $K_{eq}$.

  • Example: System: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • Adding $H_2$: Shifts toward products.

  • Adding $NH_3$: Shifts toward reactants.

  • Removing $NH_3$: Shifts toward products.

  • Example: System: $CO(g) + Br_2(g) \rightleftharpoons COBr_2(g)$

  • Removing $Br_2$: Shifts toward reactants.

  • Adding $COBr_2$: Shifts toward reactants.

• Effects of Pressure and Volume Alterations:

  • Pressure changes affect gaseous equilibria; liquids and solids are practically incompressible.

  • Increasing Pressure (Decreasing Volume): Shifts equilibrium toward the side with fewer moles of gas.

  • Decreasing Pressure (Increasing Volume): Shifts equilibrium toward the side with more moles of gas.

  • Equal Moles of Gas ($\Delta n_{gas} = 0$): Pressure adjustments have no effect on the equilibrium position.

  • Example: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • Reactant gas moles = $4$; Product gas moles = $2$.

  • Increasing pressure shifts equilibrium toward products (fewer gas moles).

  • Decreasing pressure shifts equilibrium toward reactants (greater gas moles).

• Effects of Temperature Alterations:

  • Temperature changes alter the numerical value of $K_{eq}$.

  • Endothermic Reactions ($\text{Heat} + \text{Reactants} \rightleftharpoons \text{Products}$):

  • Increasing temperature acts as adding a reactant, shifting the equilibrium toward products and increasing $K_{eq}$.

  • Decreasing temperature acts as removing a reactant, shifting the equilibrium toward reactants and decreasing $K_{eq}$.

  • Exothermic Reactions ($\text{Reactants} \rightleftharpoons \text{Products} + \text{Heat}$):

  • Increasing temperature acts as adding a product, shifting the equilibrium toward reactants and decreasing $K_{eq}$.

  • Decreasing temperature acts as removing a product, shifting the equilibrium toward products and increasing $K_{eq}$.

• Catalyst Invariance at Equilibrium:

  • Catalysts accelerate both the forward and reverse reaction rates equally by lowering the activation energy barrier for both pathways.

  • A catalyst speeds up the arrival at equilibrium but does not alter the equilibrium position, product yields, or the numerical value of $K_{eq}$.

• Equilibria in Nature: Hydrangea Flower Coloration:

• The flower petal coloration of Hydrangea macrophylla varies based on soil chemistry.

• The color differences are governed by the availability of free aluminum ions ($Al^{3+}$) in the soil rather than direct petal interaction with $H^+$ or $OH^-$.

• Acidic Soil: High hydronium concentration shifts aluminum precipitation equilibria toward dissolution, increasing soluble $Al^{3+}$. Roots absorb $Al^{3+}$, which complexes with plant anthocyanin pigments to produce blue flowers.

• Basic/Alkaline Soil: High $OH^-$ precipitates aluminum into insoluble hydroxides, preventing uptake. Anthocyanin pigments remain uncomplexed, producing red/pink flowers.

  • Equilibrium Shift Review Problems:

• For $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ (endothermic, $\Delta H > 0$):

  • Decreasing temperature: Shifts toward reactants.

  • Increasing pressure ($2\,\text{mol gas} \rightleftharpoons 2\,\text{mol gas}$): No effect.

  • Removing $HI$: Shifts toward reactants.

• For $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{heat}$ (exothermic):

  • Removing $SO_3$: Shifts toward products.

  • Adding $O_2$: Shifts toward products.

  • Decreasing temperature: Shifts toward products.

• Maximizing Ammonia in the Haber-Bosch Process ($N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + \text{heat}$):

  • Increase total pressure.

  • Lower operating temperature.

  • Continuously condense and remove product $NH_3$.

• Maximizing Calcium Carbonate Synthesis ($CaO(s) + CO_2(g) \rightleftharpoons CaCO_3(s) + \text{heat}$):

  • Increase partial pressure or concentration of $CO_2$.

  • Decrease operating temperature.

  • Increase system total pressure.