1. States of Matter & Phase Changes
* Three Physical States: Solid, Liquid, Gas
* Phase Changes:
* Melting: Solid → Liquid
* Freezing: Liquid → Solid
* Vaporization: Liquid → Gas
* Condensation: Gas → Liquid
* Sublimation: Solid → Gas
* Deposition: Gas → Solid
* Stronger intermolecular forces (IMFs) → more energy required for phase changes.
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2. Intermolecular vs. Intramolecular Forces
* Intramolecular: Within a molecule (e.g., covalent or ionic bonds) → affect chemical properties.
* Intermolecular: Between molecules → affect physical properties (e.g., boiling/melting points).
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3. Types of Intermolecular Forces
a.
Dispersion (London) Forces
* Present in all molecules; arise from temporary dipoles.
* Stronger in:
* Larger atoms/molecules (more electrons)
* More polarizable molecules
* Linear molecules (more surface area)
b.
Dipole-Dipole Forces
* Between polar molecules
* Strength increases with greater dipole moment
c.
Hydrogen Bonding
* Special dipole-dipole interaction
* Occurs when H is bonded to N, O, or F and attracted to lone pairs on N, O, or F
* Strongest type of IMF among neutral molecules
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4. Properties of Liquids
a.
Viscosity
* Resistance to flow
* Increases with stronger IMFs and larger molecular size
* Decreases with temperature
b.
Surface Tension
* Energy needed to increase liquid’s surface area
* Related to cohesive forces (same molecules attracting each other)
c.
Capillary Action
* Movement of liquid up a narrow tube due to:
* Adhesive forces (to tube walls)
* Cohesive forces (within the liquid)
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5. Enthalpies of Phase Changes
* Units: J/mol or kJ/mol
* Important Enthalpies:
* ΔH_fus: Enthalpy of fusion (melting)
* ΔH_vap: Enthalpy of vaporization (boiling)
* ΔH_sub: Enthalpy of sublimation
* Phase changes occur at constant temperature; energy goes into breaking/forming IMFs
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6. Vapor Pressure and Boiling Point
* Vapor pressure: Pressure of vapor in dynamic equilibrium with its liquid
* Affected by:
* Temperature (↑T → ↑vapor pressure)
* IMFs (stronger IMFs → ↓vapor pressure)
* Boiling point: T where vapor pressure = external pressure
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7. Clausius-Clapeyron Equation
* Describes how vapor pressure varies with temperature:
\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
* Use this to find enthalpy of vaporization or predict vapor pressure at a different temperature.
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8. Heating and Cooling Curves
* Show temperature vs. heat added
* Use:
* q = mc\Delta T for temperature changes
* q = n\Delta H for phase changes
* Total energy = sum of all q values during heating/cooling
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9. Phase Diagrams
* Graph of pressure vs. temperature
* Shows regions of solid, liquid, gas
* Lines represent equilibrium between phases (e.g., melting point)
* Critical point: beyond this, no distinction between liquid and gas (supercritical fluid)
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10. Types of Solids
a.
Ionic Solids
* Cations + anions, held by electrostatic forces
* High melting points, brittle, non-conductive (unless molten)
b.
Metallic Solids
* Metal atoms in a “sea” of electrons
* Malleable, ductile, conductive
c.
Covalent Network Solids
* Atoms connected by covalent bonds
* Hard, very high melting points, non-conductive (e.g., diamond, SiO₂)
d.
Molecular Solids
* Neutral molecules held by IMFs
* Properties depend on polarity and size
* Generally low melting points, poor conductors
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Example Problem Types
* Identify strongest IMFs in a set of compounds
* Rank substances by boiling/vapor pressure
* Calculate heat required for full phase changes
* Determine state of matter from phase diagram
* Use Clausius-Clapeyron to estimate ΔHvap
Practice Problems
Intermolecular Forces & Boiling Points
1. Which of the following has the highest boiling point? Explain why.
a) CH₄
b) CH₃Cl
c) CH₃OH
d) CH₃CH₃
2. Arrange the following in order of increasing intermolecular force strength:
H₂O, CO₂, CH₃F
3. Which substance exhibits hydrogen bonding?
a) H₂S
b) CH₄
c) NH₃
d) CCl₄
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Vapor Pressure and Clausius-Clapeyron
4. Which liquid has the highest vapor pressure at room temperature?
a) Water
b) Acetone
c) Glycerol
d) Ethylene glycol
5. Use the Clausius-Clapeyron equation to calculate ΔHvap:
Vapor pressure of a substance is 5.00 kPa at 25°C and 15.0 kPa at 45°C.
\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
Use R = 8.314 J/mol·K.
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Heating Curves and Phase Changes
6. How much energy is required to convert 50.0 g of ice at -10°C to steam at 120°C?
Use:
* c_ice = 2.09 J/g·°C
* c_water = 4.18 J/g·°C
* c_steam = 1.84 J/g·°C
* ΔH_fus = 6.01 kJ/mol
* ΔH_vap = 40.67 kJ/mol
* Molar mass of H₂O = 18.02 g/mol
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Phase Diagrams
7. Given a phase diagram, what phase is present at 0.5 atm and 80°C?
(Use water’s phase diagram as a reference.)
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Solids and Bonding Types
8. Classify each of the following as ionic, molecular, metallic, or covalent network solids:
a) NaCl
b) I₂
c) Diamond
d) Cu
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Flashcards (Term → Definition)
1. Dispersion Force → Weak attraction due to instantaneous dipoles; present in all molecules.
2. Hydrogen Bonding → Strong IMF when H is bonded to N, O, or F and attracted to lone pairs on other electronegative atoms.
3. Viscosity → A liquid’s resistance to flow; increases with stronger IMFs.
4. Surface Tension → Energy required to increase the surface area of a liquid.
5. Vapor Pressure → Pressure exerted by a vapor in dynamic equilibrium with its liquid.
6. Boiling Point → Temperature at which vapor pressure equals atmospheric pressure.
7. ΔH_vap → Enthalpy change required to vaporize a substance at its boiling point.
8. Covalent Network Solid → Solid where atoms are held together by a network of covalent bonds (e.g., diamond, SiO₂).
9. Phase Diagram → Graph showing phase stability as a function of temperature and pressure.
10. Clausius-Clapeyron Equation → Mathematical relationship describing how vapor pressure changes with temperature.
1. Highest Boiling Point
Which has the highest boiling point?
* a) CH₄ → nonpolar, dispersion only
* b) CH₃Cl → polar, dipole-dipole
* c) CH₃OH → polar, hydrogen bonding
* d) CH₃CH₃ → nonpolar, dispersion only
Answer: c) CH₃OH
Reason: Hydrogen bonding is the strongest IMF here → highest boiling point.
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2. Increasing IMF Strength
Rank: H₂O, CO₂, CH₃F
* CO₂: nonpolar → dispersion only
* CH₃F: polar → dipole-dipole
* H₂O: polar → hydrogen bonding
Order: CO₂ < CH₃F < H₂O
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3. Hydrogen Bonding
Which exhibits hydrogen bonding?
* a) H₂S – polar but S is not N, O, or F
* b) CH₄ – nonpolar
* c) NH₃ – yes, H is bonded to N
* d) CCl₄ – nonpolar
Answer: c) NH₃
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4. Highest Vapor Pressure
Which liquid has the highest vapor pressure at room temp?
* Water: strong H-bonding
* Acetone: dipole-dipole (weaker than H-bonding)
* Glycerol: strong H-bonding
* Ethylene glycol: strong H-bonding
Answer: b) Acetone
Reason: Weaker IMFs → higher vapor pressure.
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5. Clausius-Clapeyron Equation
Given:
* P_1 = 5.00 kPa at T_1 = 298.15 K (25°C)
* P_2 = 15.0 kPa at T_2 = 318.15 K (45°C)
* R = 8.314 J/mol·K
\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\ln\left(\frac{15.0}{5.00}\right) = \frac{-\Delta H_{\text{vap}}}{8.314} \left(\frac{1}{318.15} - \frac{1}{298.15}\right)
\ln(3) ≈ 1.0986 \quad \text{and} \quad \left(\frac{1}{318.15} - \frac{1}{298.15}\right) ≈ -2.06 \times 10^{-4}
1.0986 = \frac{-\Delta H_{\text{vap}}}{8.314} \times (-2.06 \times 10^{-4})
\Delta H_{\text{vap}} ≈ \frac{1.0986}{2.06 \times 10^{-4}} \times 8.314 ≈ 44.3 \, \text{kJ/mol}
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6. Total Heat for Ice → Steam
Given:
Mass = 50.0 g
Molar mass H₂O = 18.02 g/mol
Moles = 50.0 / 18.02 ≈ 2.78 mol
Break into stages:
Stage 1: Warm ice from -10°C to 0°C
q = mc\Delta T = (50.0)(2.09)(10) = 1045 \, \text{J}
Stage 2: Melt ice
q = n\Delta H_{\text{fus}} = (2.78)(6.01 \times 10^3) = 16,717.8 \, \text{J}
Stage 3: Heat water from 0°C to 100°C
q = (50.0)(4.18)(100) = 20,900 \, \text{J}
Stage 4: Vaporize water
q = (2.78)(40.67 \times 10^3) = 113,062.6 \, \text{J}
Stage 5: Heat steam from 100°C to 120°C
q = (50.0)(1.84)(20) = 1,840 \, \text{J}
Total heat:
\text{q total} = 1045 + 16,717.8 + 20,900 + 113,062.6 + 1,840 ≈ 153,566 \, \text{J} \approx 153.6 \, \text{kJ}
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7. Phase at 0.5 atm, 80°C
Using a typical water phase diagram:
* Boiling point of water at 1 atm = 100°C
* At lower pressure (0.5 atm), boiling occurs at a lower temperature
At 80°C and 0.5 atm → water is a gas
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8. Solid Classification
* NaCl: Ionic
* I₂: Molecular
* Diamond: Covalent network
* Cu: Metallic
Here is a complete study guide, practice problems, and flashcards for Chapter 11 – Solutions and Colloids based on your PowerPoint.
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Study Guide: Chapter 11 – Solutions and Colloids
1. Types of Mixtures
* Solution: Homogeneous mixture (e.g., saltwater, air)
* Colloid: Intermediate particle size; scatters light (Tyndall effect); doesn’t settle (e.g., milk, fog)
* Suspension: Particles eventually settle (e.g., muddy water)
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2. Dissolution and Solubility
* Solute: Substance dissolved (lesser amount)
* Solvent: Substance doing the dissolving (greater amount)
* Like dissolves like: Polar solvents dissolve polar solutes; nonpolar dissolves nonpolar
Types of Solute-Solvent Interactions
* Ion-dipole: Ionic solute + polar solvent (e.g., NaCl in water)
* Hydrogen bonding: Polar solutes (e.g., alcohols in water)
* Dispersion forces: Nonpolar solutes and solvents
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3. Thermodynamics of Solution Formation
* ΔH_soln = ΔH_solute + ΔH_solvent + ΔH_mix
* Can be endothermic or exothermic
* Entropy (ΔS) increases when solutions form → favors dissolving
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4. Electrolytes
* Electrolyte: Produces ions in solution → conducts electricity
* Strong: Fully dissociates (e.g., NaCl)
* Weak: Partially dissociates (e.g., acetic acid)
* Nonelectrolyte: Does not produce ions (e.g., sugar, alcohol)
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5. Solubility
* Saturated: Max solute at equilibrium
* Unsaturated: Less than max solute
* Supersaturated: More than equilibrium; unstable
Factors Affecting Solubility
* Temperature:
* Solids: ↑T → ↑solubility
* Gases: ↑T → ↓solubility
* Pressure (Henry’s Law):
S_{\text{gas}} = k_H \times P_{\text{gas}}
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6. Concentration Units
* Molarity (M): mol/L
* Molality (m): mol/kg solvent
* Mole fraction (X): mol component / total mol
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7. Colligative Properties
Depend only on the number of particles, not type.
a.
Vapor Pressure Lowering (Raoult’s Law)
P_{\text{solvent}} = X_{\text{solvent}} \times P^0_{\text{solvent}}
b.
Boiling Point Elevation
\Delta T_b = i \cdot K_b \cdot m
c.
Freezing Point Depression
\Delta T_f = i \cdot K_f \cdot m
d.
Osmotic Pressure
\Pi = i \cdot M \cdot R \cdot T
van’t Hoff factor (i) = number of particles the solute forms in solution
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8. Colloids
* Mixture with dispersed particles (~1–1000 nm)
* Uniform, stable, Tyndall effect
* Examples: fog, milk, whipped cream
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Practice Problems
1. Predict Solubility
Which will dissolve in water?
a) CCl₄
b) CH₃OH
c) I₂
d) Benzene
Answer: b) CH₃OH (polar with hydrogen bonding)
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2. Henry’s Law
At 25°C, the solubility of O₂ is 2.86 × 10⁻⁴ mol/L at 1 atm. What is solubility at 3 atm?
S = k_H \cdot P \Rightarrow 2.86 \times 10^{-4} \cdot 3 = 8.58 \times 10^{-4} \, \text{mol/L}
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3. Molarity vs Molality
Find molality of 8.0 g NaCl (MW = 58.44 g/mol) in 250 mL water (density = 0.962 g/mL)
n = \frac{8.0}{58.44} \approx 0.137 \, \text{mol} \quad m = \frac{0.137}{0.250 \cdot 0.962} \approx 0.57 \, m
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4. Freezing Point Depression
What is the freezing point of 0.20 m solution in benzene? (Kf = 5.12°C/m, Tf = 5.5°C)
\Delta T_f = 5.12 \cdot 0.20 = 1.024 \quad \Rightarrow \text{Tf solution} = 5.5 - 1.024 = 4.48°C
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5. Osmotic Pressure
Find osmotic pressure of 0.15 M glucose at 298 K.
\Pi = MRT = 0.15 \cdot 0.0821 \cdot 298 = 3.67 \, \text{atm}
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Flashcards (Term → Definition)
1. Solution → Homogeneous mixture of solute + solvent
2. Colloid → Uniform dispersion with light scattering (Tyndall effect)
3. Electrolyte → Produces ions; conducts electricity
4. Nonelectrolyte → Does not produce ions in solution
5. Henry’s Law → Gas solubility ∝ partial pressure
6. Molality → Moles solute / kg solvent
7. Boiling Point Elevation → Solution boils at higher temperature than pure solvent
8. Freezing Point Depression → Solution freezes at lower temperature
9. Osmotic Pressure → Pressure to stop osmosis
10. Raoult’s Law → Vapor pressure of solution < pure solvent
Here’s a complete study guide, practice problems, and flashcards for Chapter 12 – Thermodynamics based on your PowerPoint.
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Study Guide: Chapter 12 – Thermodynamics
1. What Is Thermodynamics?
* Thermodynamics: Study of energy transformations in chemical/physical processes.
* Focuses on heat (q), work (w), and energy changes (ΔE, ΔH, ΔG).
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2. Spontaneity
* Spontaneous Process: Occurs without outside energy once started.
* Nonspontaneous Process: Requires continuous energy input.
* Reversible: Can return to original state without net energy change.
* Irreversible: Cannot return without energy loss.
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3. Entropy (S)
* Entropy = measure of disorder or randomness.
* ΔS > 0: Increase in disorder (favored for spontaneity)
* ΔS = S_final – S_initial
* Microstates (W): Number of possible arrangements.
S = k \ln W \quad \text{(Boltzmann equation)}
* S(gas) > S(liquid) > S(solid)
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4. Second Law of Thermodynamics
* ΔS_universe = ΔS_system + ΔS_surroundings
* A process is spontaneous if:
\Delta S_{\text{universe}} > 0
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5. Third Law of Thermodynamics
* Entropy of a perfect crystal at absolute zero = 0 J/mol·K
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6. Predicting Entropy Trends
* ↑ Temperature → ↑ entropy
* ↑ Molecular size or complexity → ↑ entropy
* ↑ Phase (solid → gas) → ↑ entropy
* Dissolving solids/liquids → ↑ entropy
* Dissolving gases → ↓ entropy
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7. Entropy in Chemical Reactions
\Delta S^\circ_{\text{rxn}} = \sum m S^\circ_{\text{products}} - \sum n S^\circ_{\text{reactants}}
* If gas moles increase, ΔS° > 0
* If gas moles decrease, ΔS° < 0
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8. Entropy of Surroundings
\Delta S_{\text{surr}} \propto -\frac{q_{\text{sys}}}{T}
* Exothermic (ΔH < 0) → ΔS_surr > 0
* Endothermic (ΔH > 0) → ΔS_surr < 0
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9. Free Energy (Gibbs Energy)
* G = H - T·S
* ΔG < 0: spontaneous
* ΔG > 0: nonspontaneous
* ΔG = 0: equilibrium
\Delta G = \Delta H - T \Delta S
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10. Temperature and Spontaneity
ΔH
ΔS
ΔG Behavior
–
+
Always spontaneous
+
–
Never spontaneous
–
–
Spontaneous at low T
+
+
Spontaneous at high T
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11. Calculating ΔG°
\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ
Or from formation values:
\Delta G^\circ_{\text{rxn}} = \sum m \Delta G_f^\circ(\text{products}) - \sum n \Delta G_f^\circ(\text{reactants})
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Practice Problems
1. Predict the sign of ΔS
For the reaction:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Answer: ΔS > 0 (produces more gas particles)
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2. Entropy and Microstates
If a process increases W (microstates) from 4 to 16, what happens to entropy?
\Delta S = k \ln\left(\frac{W_{\text{final}}}{W_{\text{initial}}}\right) = k \ln(4)
Answer: Entropy increases
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3. ΔG Calculation
Given: ΔH = -120 kJ, ΔS = -200 J/K. Find ΔG at 298 K.
\Delta G = -120 - (298)(-0.200) = -120 + 59.6 = -60.4 \, \text{kJ}
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4. Predict spontaneity
ΔH = +70 kJ, ΔS = +180 J/K
Will the reaction be spontaneous at 400 K?
\Delta G = 70 - (400)(0.180) = 70 - 72 = -2 \, \text{kJ}
Answer: Yes, spontaneous at high T
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5. Calculate ΔS_rxn
For:
2CO(g) + O₂(g) → 2CO₂(g)
Given:
S°(CO₂) = 213.7 J/mol·K
S°(CO) = 197.9 J/mol·K
S°(O₂) = 205.0 J/mol·K
\Delta S^\circ = [2(213.7)] - [2(197.9) + 205.0] = 427.4 - 600.8 = -173.4 \, \text{J/K}
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6. Find T at which ΔG = 0
Given:
ΔH = -141.6 kJ, ΔS = -187.9 J/K
0 = \Delta H - T \Delta S \Rightarrow T = \frac{\Delta H}{\Delta S} = \frac{-141600}{-187.9} \approx 754 \, \text{K}
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Flashcards (Term → Definition)
1. Spontaneous Process → Occurs without added energy
2. Entropy (S) → Measure of molecular disorder
3. Second Law of Thermodynamics → Universe’s entropy increases
4. Third Law of Thermodynamics → Perfect crystal has S = 0 at 0 K
5. Free Energy (G) → Energy available to do work
6. ΔG < 0 → Spontaneous process
7. ΔG = ΔH - TΔS → Gibbs free energy equation
8. ΔS_univ > 0 → Required for spontaneity
9. Microstate → Specific arrangement of molecules
10. Boltzmann Constant (k) → 1.38 × 10⁻²³ J/K
📘 Chapter 13: Chemical Equilibrium – Study Guide
1.
Understanding Chemical Equilibrium
* Definition: A state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction.
* Dynamic Nature: Even at equilibrium, reactions continue to occur in both directions, but concentrations of reactants and products remain constant.
* Equilibrium Position:
* If the concentration of products is greater than reactants, equilibrium lies to the right.
* If the concentration of reactants is greater than products, equilibrium lies to the left.
2.
The Equilibrium Constant (K)
* Expression: For a general reaction: aA + bB ⇌ cC + dD, the equilibrium constant (K) is given by:
K = [C]^c [D]^d / [A]^a [B]^b
* Types:
* Kc: Based on concentrations (mol/L).
* Kp: Based on partial pressures (atm).
* Interconversion: Kp = Kc(RT)^Δn, where Δn = moles of gaseous products - moles of gaseous reactants.
3.
Reaction Quotient (Q)
* Purpose: Determines the direction in which a reaction will proceed to reach equilibrium.
* Comparison with K:
* If Q < K: Reaction proceeds forward (toward products).
* If Q > K: Reaction proceeds in reverse (toward reactants).
* If Q = K: System is at equilibrium.
4.
Le Châtelier’s Principle
* Concept: If a system at equilibrium is disturbed, it will adjust to minimize the disturbance and re-establish equilibrium.
* Factors Affecting Equilibrium:
* Concentration: Adding or removing reactants/products shifts equilibrium to oppose the change.
* Pressure: Increasing pressure favors the side with fewer gas molecules.
* Temperature:
* For exothermic reactions: Increasing temperature shifts equilibrium to the left.
* For endothermic reactions: Increasing temperature shifts equilibrium to the right.
* Catalysts: Speed up the rate at which equilibrium is achieved but do not affect the position of equilibrium.
5.
Calculating Equilibrium Concentrations
* ICE Tables:
1. I: Initial concentrations.
2. C: Change in concentrations.
3. E: Equilibrium concentrations.
* Steps:
1. Write the balanced chemical equation.
2. Set up the ICE table.
3. Express changes in terms of a variable (usually x).
4. Substitute equilibrium concentrations into the K expression.
5. Solve for x and determine all equilibrium concentrations.
6.
Heterogeneous Equilibria
* Definition: Equilibria involving reactants and products in different phases (solid, liquid, gas).
* Key Point: Concentrations of pure solids and liquids are constant and thus not included in the equilibrium expression.
7.
Applications of Equilibrium Concepts
* Industrial Processes: Optimizing conditions to favor product formation (e.g., Haber process for ammonia synthesis).
* Environmental Chemistry: Understanding pollutant formation and removal.
* Biological Systems: Maintaining homeostasis through equilibrium processes.
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Additional Resources:
* For practice problems and further reading, consider exploring study guides and flashcards available on educational platforms.
Here’s your enhanced study guide for Chapter 13 – Chemical Equilibrium, including flashcards and practice problems with answers:
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Chemical Equilibrium Study Guide with Flashcards & Practice Problems
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Flashcards
Term: Chemical Equilibrium
Definition: A dynamic state where the rate of the forward reaction equals the rate of the reverse reaction.
Term: Equilibrium Constant (K)
Definition: Ratio of product concentrations to reactant concentrations, each raised to the power of its coefficient at equilibrium.
Term: Reaction Quotient (Q)
Definition: Same as the equilibrium expression, but using current concentrations/pressures to determine reaction direction.
Term: Le Châtelier’s Principle
Definition: A system at equilibrium shifts to counteract imposed changes (concentration, pressure, temperature).
Term: Kp vs. Kc
Definition: Kp uses partial pressures; Kc uses concentrations. Related by Kp = Kc(RT)^Δn.
Term: ICE Table
Definition: A method to organize initial, change, and equilibrium values when solving equilibrium problems.
Term: Δn in Equilibrium
Definition: Change in moles of gas: (moles of gaseous products) – (moles of gaseous reactants).
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Practice Problems
1. Writing the Equilibrium Expression
Question:
Write the equilibrium expression for the reaction:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Answer:
K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
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2. Using K to Determine Direction
Question:
At a certain temperature, K = 4.0. For the reaction
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \],
the concentrations are:
[\text{H}_2] = 0.10\;M,\ [\text{I}_2] = 0.10\;M,\ [\text{HI}] = 0.60\;M
Calculate Q and determine direction of shift.
Answer:
Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.60)^2}{0.10 \times 0.10} = \frac{0.36}{0.01} = 36
Since Q > K, the system shifts left (toward reactants).
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3. ICE Table & Solving for Equilibrium
Question:
For the reaction:
\[ \text{A} \rightleftharpoons \text{B} \],
initial [A] = 1.0 M, [B] = 0 M, and K = 0.25. Find the equilibrium concentrations.
Answer:
ICE Table:
A
B
Initial
1.0
0
Change
-x
+x
Equil.
1.0−x
x
K = \frac{x}{1.0 - x} = 0.25 \Rightarrow x = 0.25(1.0 - x) \Rightarrow x = 0.25 - 0.25x \Rightarrow 1.25x = 0.25 \Rightarrow x = 0.20
[A] = 0.80 M, [B] = 0.20 M
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4. Effect of Pressure Change (Le Châtelier)
Question:
For the equilibrium:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \],
what happens if pressure is increased?
Answer:
Pressure increase favors the side with fewer gas moles.
Left: 1 + 3 = 4 mols
Right: 2 mols
Shift → Right (toward NH₃ production)
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5. Kp vs Kc Conversion
Question:
Given:
\[ \text{N}_2O_4(g) \rightleftharpoons 2\text{NO}_2(g) \]
Kc = 0.25 at 298 K. Find Kp. (R = 0.0821 L·atm/mol·K)
Answer:
Δn = 2 – 1 = 1
Kp = Kc(RT)^{\Delta n} = 0.25(0.0821 \times 298)^1 = 0.25(24.5) ≈ 6.13
📘 Chapter 14: Acid-Base Equilibria – Study Guide
1.
Acid-Base Theories
* Arrhenius Definition:
* Acid: Increases [H⁺] in aqueous solution.
* Base: Increases [OH⁻] in aqueous solution.
* Brønsted-Lowry Definition:
* Acid: Proton (H⁺) donor.
* Base: Proton (H⁺) acceptor.
* Lewis Definition:
* Acid: Electron pair acceptor.
* Base: Electron pair donor.
2.
Conjugate Acid-Base Pairs
* Every acid has a conjugate base, and every base has a conjugate acid.
* Example:
* Acid: HCl → Conjugate Base: Cl⁻
* Base: NH₃ → Conjugate Acid: NH₄⁺
3.
Autoionization of Water
* Water can act as both an acid and a base (amphoteric).
* Reaction: H₂O ⇌ H⁺ + OH⁻
* Ion-product constant for water (Kw):
* At 25°C, Kw = 1.0 × 10⁻¹⁴
4.
pH and pOH
* pH = -log[H⁺]
* pOH = -log[OH⁻]
* pH + pOH = 14
5.
Strong vs. Weak Acids and Bases
* Strong Acids: Completely dissociate in water.
* Examples: HCl, HNO₃, H₂SO₄
* Weak Acids: Partially dissociate in water.
* Examples: CH₃COOH, HF
* Strong Bases: Completely dissociate in water.
* Examples: NaOH, KOH
* Weak Bases: Partially dissociate in water.
* Examples: NH₃
6.
Acid and Base Dissociation Constants
* Ka: Acid dissociation constant.
* Ka = [H⁺][A⁻]/[HA]
* Kb: Base dissociation constant.
* Kb = [BH⁺][OH⁻]/[B]
* Relationship: Ka × Kb = Kw
7.
Calculating pH of Weak Acid/Base Solutions
* Use ICE tables to determine equilibrium concentrations.
* Apply the Ka or Kb expression to solve for [H⁺] or [OH⁻].
* Calculate pH or pOH accordingly.
8.
Percent Ionization
* % Ionization = ([H⁺] at equilibrium / [HA] initial) × 100%
* Indicates the strength of a weak acid.
9.
Polyprotic Acids
* Acids that can donate more than one proton (e.g., H₂SO₄).
* Each dissociation step has its own Ka value.
10.
Acid-Base Properties of Salts
* Salts derived from:
* Strong acid + strong base → neutral solution.
* Strong acid + weak base → acidic solution.
* Weak acid + strong base → basic solution.
11.
Buffer Solutions
* Consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
* Resist changes in pH upon addition of small amounts of acid or base.
* Henderson-Hasselbalch Equation:
* pH = pKa + log([A⁻]/[HA])
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🧠 Flashcards
Q: What is a Brønsted-Lowry acid?
A: A proton (H⁺) donor.
Q: Define conjugate base.
A: The species formed when an acid donates a proton.
Q: What is the value of Kw at 25°C?
A: 1.0 × 10⁻¹⁴
Q: How do you calculate pH from [H⁺]?
A: pH = -log[H⁺]
Q: What does a high Ka value indicate?
A: A stronger acid.
Q: What is the relationship between Ka and Kb?
A: Ka × Kb = Kw
Q: What is a buffer solution?
A: A solution that resists changes in pH upon addition of small amounts of acid or base.
Q: How is percent ionization calculated?
A: ([H⁺] at equilibrium / [HA] initial) × 100%
Q: What is a polyprotic acid?
A: An acid that can donate more than one proton.
Q: How does the pH of a salt solution depend on its parent acid and base?
A: It depends on the strengths of the parent acid and base; strong acid + weak base yields acidic solution, etc.
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📝 Practice Problems
Problem 1: Calculating pH of a Weak Acid
Question:
Calculate the pH of a 0.10 M acetic acid (CH₃COOH) solution. Ka = 1.8 × 10⁻⁵
Solution:
CH₃COOH ⇌ H⁺ + CH₃COO⁻
Initial: [CH₃COOH] = 0.10 M, [H⁺] = [CH₃COO⁻] = 0
Change: [CH₃COOH] = 0.10 - x, [H⁺] = [CH₃COO⁻] = x
Ka = x² / (0.10 - x) ≈ x² / 0.10
1.8 × 10⁻⁵ = x² / 0.10
x² = 1.8 × 10⁻⁶
x = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³
pH = -log(1.34 × 10⁻³) ≈ 2.87
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Problem 2: Determining Ka from pH
Question:
A 0.050 M solution of a weak acid has a pH of 3.00. Calculate Ka.
Solution:
[H⁺] = 10^(-3.00) = 1.0 × 10⁻³ M
Assuming [H⁺] = [A⁻] = x, [HA] = 0.050 - x ≈ 0.050
Ka = x² / [HA] = (1.0 × 10⁻³)² / 0.050 = 2.0 × 10⁻⁵
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Problem 3: Buffer pH Calculation
Question:
Calculate the pH of a buffer solution containing 0.30 M acetic acid and 0.30 M sodium acetate. Ka = 1.8 × 10⁻⁵
Solution:
pKa = -log(1.8 × 10⁻⁵) ≈ 4.74
pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.30/0.30) = 4.74
📘 Chapter 15: Equilibria of Other Reaction Classes – Study Guide
1.
Solubility Equilibria
* Solubility Product Constant (Ksp): Represents the equilibrium between a solid and its ions in a saturated solution.
For a generic salt AB:
\[ AB_{(s)} \rightleftharpoons A^+{(aq)} + B^-{(aq)} \]
K_{sp} = [A^+][B^-]
* Common Ion Effect: The addition of an ion common to the equilibrium shifts the position of equilibrium, affecting solubility.
2.
Predicting Precipitation
* Reaction Quotient (Q): Used to predict whether a precipitate will form.
* If Q < Ksp: No precipitate forms.
* If Q = Ksp: The solution is saturated.
* If Q > Ksp: Precipitation occurs.
3.
Complex Ion Equilibria
* Complex Ions: Formed when metal ions bind with ligands (molecules or ions that donate electron pairs).
\[ M^{n+} + xL \rightleftharpoons ML_x^{n+} \]
* Formation Constant (Kf): Equilibrium constant for the formation of a complex ion.
K_f = \frac{[ML_x^{n+}]}{[M^{n+}][L]^x}
* Effect on Solubility: Formation of complex ions can increase the solubility of certain salts.
4.
Amphoterism
* Amphoteric Substances: Compounds that can act as both acids and bases.
* Example: Aluminum hydroxide, Al(OH)₃, can react with both acids and bases.
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🧠 Flashcards
Q: What is the solubility product constant (Ksp)?
A: The equilibrium constant for the dissolution of a sparingly soluble compound.
Q: How does the common ion effect influence solubility?
A: It decreases the solubility of a salt by shifting the equilibrium position.
Q: What is a complex ion?
A: A species formed from a central metal ion bonded to one or more ligands.
Q: Define the formation constant (Kf).
A: The equilibrium constant for the formation of a complex ion from a metal ion and ligands.
Q: What does it mean if Q > Ksp?
A: The solution is supersaturated, and precipitation is likely to occur.
Q: What is an amphoteric substance?
A: A substance that can act as both an acid and a base.
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📝 Practice Problems
Problem 1: Calculating Ksp
Question:
Calculate the Ksp for barium sulfate (BaSO₄) if its solubility in water is 1.1 × 10⁻⁵ mol/L.
Solution:
BaSO₄ ⇌ Ba²⁺ + SO₄²⁻
[Ba²⁺] = [SO₄²⁻] = 1.1 × 10⁻⁵ M
Ksp = [Ba²⁺][SO₄²⁻] = (1.1 × 10⁻⁵)² = 1.21 × 10⁻¹⁰
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Problem 2: Predicting Precipitation
Question:
Will a precipitate form when 50.0 mL of 0.010 M AgNO₃ is mixed with 50.0 mL of 0.010 M NaCl? (Ksp for AgCl = 1.8 × 10⁻¹⁰)
Solution:
After mixing, concentrations halve:
[Ag⁺] = [Cl⁻] = 0.0050 M
Q = [Ag⁺][Cl⁻] = (0.0050)(0.0050) = 2.5 × 10⁻⁵
Since Q > Ksp, a precipitate will form.
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Problem 3: Complex Ion Formation
Question:
Calculate the concentration of free Ag⁺ in a solution where [Ag(NH₃)₂]⁺ = 0.010 M and [NH₃] = 0.10 M. (Kf for Ag(NH₃)₂⁺ = 1.6 × 10⁷)
Solution:
Kf = [Ag(NH₃)₂⁺] / ([Ag⁺][NH₃]²)
Rearranged: [Ag⁺] = [Ag(NH₃)₂⁺] / (Kf × [NH₃]²)
[Ag⁺] = 0.010 / (1.6 × 10⁷ × (0.10)²) = 6.25 × 10⁻⁶ M
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Problem 4: Amphoterism
Question:
Explain how Al(OH)₃ acts as an amphoteric substance.
Solution:
Al(OH)₃ can react with acids:
Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O
And with bases:
Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻
Thus, it can act as both an acid and a base.
📘 Chapter 16: Electrochemistry – Study Guide
1.
Fundamentals of Electrochemistry
* Redox Reactions: Involve the transfer of electrons; oxidation is the loss of electrons, and reduction is the gain of electrons.
* Oxidizing Agent: Species that gains electrons (is reduced).
* Reducing Agent: Species that loses electrons (is oxidized).
2.
Electrochemical Cells
* Galvanic (Voltaic) Cells: Spontaneous redox reactions that generate electrical energy.
* Electrolytic Cells: Non-spontaneous reactions driven by external electrical energy.
* Anode: Electrode where oxidation occurs; electrons are released.
* Cathode: Electrode where reduction occurs; electrons are accepted.
* Salt Bridge: Maintains electrical neutrality by allowing ion flow between half-cells.
3.
Standard Reduction Potentials (E°)
* Each half-reaction has a standard reduction potential measured in volts (V).
* Standard Cell Potential (E°cell): Calculated as:
E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}
* A positive E°cell indicates a spontaneous reaction.
4.
Nernst Equation
* Used to calculate cell potential under non-standard conditions:
E = E° - \frac{0.0592}{n} \log Q
* E: cell potential under non-standard conditions
* n: number of electrons transferred
* Q: reaction quotient
5.
Electrolysis
* Process of driving non-spontaneous chemical reactions using electrical energy.
* Faraday’s Laws: Relate the amount of substance altered at an electrode to the quantity of electricity used.
* \text{Mass} = \frac{Q \cdot M}{n \cdot F}
* Q: total charge (Coulombs)
* M: molar mass
* n: number of electrons transferred
* F: Faraday’s constant (96,485 C/mol)
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🧠 Flashcards
Q: What is the role of the salt bridge in a galvanic cell?
A: It allows the flow of ions to maintain electrical neutrality between the half-cells.
Q: In which electrode does oxidation occur?
A: Oxidation occurs at the anode.
Q: How is the standard cell potential (E°cell) calculated?
A: E°cell = E°cathode - E°anode.
Q: What does a positive E°cell indicate about a reaction?
A: The reaction is spontaneous under standard conditions.
Q: What is the purpose of the Nernst equation?
A: To calculate the cell potential under non-standard conditions.
Q: What is Faraday’s constant?
A: 96,485 C/mol; the charge of one mole of electrons.
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📝 Practice Problems
Problem 1: Calculating Standard Cell Potential
Question:
Given the following standard reduction potentials:
* Zn²⁺ + 2e⁻ → Zn(s) E° = -0.76 V
* Cu²⁺ + 2e⁻ → Cu(s) E° = +0.34 V
Calculate the standard cell potential for a galvanic cell using these half-reactions.
Solution:
E°cell = E°cathode - E°anode = 0.34 V - (-0.76 V) = 1.10 V
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Problem 2: Using the Nernst Equation
Question:
For the cell in Problem 1, calculate the cell potential when [Zn²⁺] = 0.010 M and [Cu²⁺] = 1.0 M at 25°C.
Solution:
E = E° - (0.0592/n) * log(Q)
n = 2
Q = [Zn²⁺]/[Cu²⁺] = 0.010 / 1.0 = 0.010
E = 1.10 V - (0.0592/2) * log(0.010)
E ≈ 1.10 V - 0.0296 * (-2) = 1.10 V + 0.0592 V = 1.1592 V
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Problem 3: Electrolysis Calculation
Question:
How many grams of copper are deposited when a current of 2.00 A is passed through a Cu²⁺ solution for 3.00 hours? (Molar mass of Cu = 63.55 g/mol)
Solution:
Q = I × t = 2.00 A × 3.00 h × 3600 s/h = 21,600 C
n = Q / (n × F) = 21,600 C / (2 × 96,485 C/mol) ≈ 0.112 mol
Mass = n × M = 0.112 mol × 63.55 g/mol ≈ 7.12 g
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Problem 4: Identifying Oxidizing and Reducing Agents
Question:
In the reaction: Fe²⁺ + Ce⁴⁺ → Fe³⁺ + Ce³⁺, identify the oxidizing and reducing agents.
Solution:
Fe²⁺ → Fe³⁺ + e⁻ (oxidation)
Ce⁴⁺ + e⁻ → Ce³⁺ (reduction)
Fe²⁺ is the reducing agent; Ce⁴⁺ is the oxidizing agent.
📘 Chapter 17: Kinetics Study Guide
1.
Reaction Rate
* Definition: The change in concentration of a reactant or product per unit time.
* Formula: Rate = Δ[Concentration] / ΔTime
* Units: Typically mol/L·s
2.
Factors Affecting Reaction Rate
* Temperature: Increasing temperature generally increases reaction rate by providing more energy to reactants.
* Concentration of Reactants: Higher concentrations lead to more frequent collisions, increasing the rate.
* Surface Area: Greater surface area allows more collisions, speeding up the reaction.
* Catalysts: Substances that increase reaction rate without being consumed by lowering activation energy.
3.
Collision Theory
* Key Points:
* Particles must collide to react.
* Collisions must have sufficient energy (activation energy).
* Collisions must occur in the correct orientation.
4.
Reaction Mechanisms
* Definition: The step-by-step sequence of elementary reactions by which the overall chemical change occurs.
* Intermediates: Species produced in one step and consumed in another; not present in the overall reaction.
* Rate-Determining Step: The slowest step in a reaction mechanism that limits the overall rate.
5.
Rate Laws
* General Form: Rate = k[A]^m[B]^n
* k = rate constant
* m, n = reaction orders with respect to reactants A and B
* Determining Rate Law:
* Use experimental data to find how changes in concentration affect the rate.
* Method of initial rates is commonly used.
6.
Integrated Rate Laws
* Zero-Order Reactions: Rate = k; concentration decreases linearly over time.
* First-Order Reactions: Rate = k[A]; concentration decreases exponentially.
* Second-Order Reactions: Rate = k[A]^2; concentration decreases more rapidly.
7.
Activation Energy (Ea)
* Definition: The minimum energy required for a reaction to occur.
* Arrhenius Equation: k = Ae^(-Ea/RT)
* A = frequency factor
* R = gas constant
* T = temperature in Kelvin
8.
Catalysts
* Homogeneous Catalysts: Same phase as reactants; participate in the same phase reactions.
* Heterogeneous Catalysts: Different phase than reactants; often provide a surface for reactions.
* Effect on Activation Energy: Lower the activation energy, increasing reaction rate.
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🃏 Flashcards
1. What is the definition of reaction rate?
* The change in concentration of a reactant or product per unit time.
2. Name two factors that affect reaction rate.
* Temperature and concentration of reactants.
3. What does collision theory state?
* Particles must collide with sufficient energy and in the correct orientation to react.
4. What is a reaction mechanism?
* The step-by-step sequence of elementary reactions by which the overall chemical change occurs.
5. Define activation energy.
* The minimum energy required for a reaction to occur.
6. What is the purpose of a catalyst?
* To increase the reaction rate by lowering the activation energy without being consumed in the reaction.
7. What is the rate law for a reaction?
* An equation that relates the reaction rate to the concentrations of reactants.
8. What is the Arrhenius equation used for?
* To calculate the rate constant (k) and understand the effect of temperature on reaction rate.
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For additional practice, you can explore these Quizlet flashcard sets:
* Chemistry: Chapter 17 - Kinetics Flashcards
* Reaction Kinetics Study Guide Flashcards