Unit 7: Energy and Rate of Reactions Notes

Energy and Rate of Reactions

Energy Exchange

Temperature vs. Heat

• Temperature reflects the random motion of particles.

• Heat involves the transfer of energy between two objects due to a temperature difference.

• At constant pressure, the change in enthalpy \Delta H is equal to the energy flow as heat.
  \Delta H = H{\text{products}} - H{\text{reactants}}

Systems and Surroundings

• System: The part of the universe on which we focus attention (e.g., reactants and products of a reaction).

• Surroundings: Everything else in the universe outside the system.

• Universe: The system plus the surroundings. \text{Universe} = \text{System} + \text{Surroundings}

Heat Exchange in Chemical Reactions

Endothermic Reaction

• Heat is absorbed by the system from the surroundings.

• \Delta H is positive at constant pressure.

• The reaction feels cold to the touch.

Exothermic Reaction

• Heat is produced and released by the system to the surroundings.

• \Delta H is negative at constant pressure.

• The reaction feels hot to the touch.

• Energy gained by the surroundings is equal to the energy lost by the system.

Phase Changes

• Gas Phase: Can transition to liquid via condensation (exothermic) or from liquid via vaporization (endothermic).

• Liquid Phase: Can transition to solid via freezing (exothermic) or from solid via melting (endothermic).

• Solid Phase: Can transition to gas via sublimation (endothermic) or from gas via deposition (exothermic).

Thermo Calculations

Types of Thermo Calculations

• Two types of energy-dependent physical changes for a substance:

  • Phase Changes

  • Temperature Changes

Phase Changes Thermo

• Heat of Fusion (\Delta H_{\text{fus}}) – energy required to melt a solid (breaking intermolecular forces).

• Heat of Vaporization (\Delta H_{\text{vap}}) – energy required to vaporize a liquid into a gas.

• Energy is typically reported in kilojoules (kJ), joules (J), or calories (cal).

• Units for \Delta H{\text{fus}} and \Delta H{\text{vap}}: kJ/mol, J/mol, cal/mol, kJ/g, J/g, cal/g

Phase Change Practice #1

Problem:

Solution:

Phase Change Practice #2

Problem:

Solution:

Temperature Changes Thermo

• Specific Heat Capacity – the energy required to raise one gram of a substance by one degree Celsius (^{\circ}C). Units are J/^{\circ}C \cdotg or J/K\cdotg

  • Substances with lower specific heat capacity are easier to heat up.

• Formula: Q = mC\Delta T

  • Q = heat/energy gained or lost

  • m = mass of substance (grams)

  • c = Specific Heat of substance

  • \Delta T = Change in Temperature (Final Temp - Initial Temp)

• Specific Heat Capacity of Water = 4.184 \frac{J}{g^{\circ}C}

Temperature Change Practice #3

Problem:

Solution:

Temperature Change Practice #4

Problem:

Solution:

Combined Phase & Temperature Change Equations

• Specific Heat of Solid: q = mC\Delta T

• Specific Heat of Liquid: q = mC\Delta T

• Specific Heat of Gas: q = mC\Delta T

• Heat of Fusion: \Delta H_{\text{fus}}

• Heat of Vaporization: \Delta H_{\text{vap}}

Combined Phase & Temperature Change Practice #5

Problem:

You have a sample of H_2O with a mass of 23.0 g at a temperature of -46.0 ^{\circ}C

Solution:

Measuring Enthalpy

• Calorimetry is the science of measuring heat.

• Constant-Pressure Calorimetry is used to determine the changes in enthalpy for reactions occurring in solution.

  • Uses a "Coffee-Cup Calorimeter"

  • Two nested Styrofoam cups with a cover through which a stirrer and thermometer can be inserted

• Energy released by the reaction = Energy absorbed by the solution.

Transfer of Energy Practice #6

Problem:

Solution:

Transfer of Energy Practice #7

Problem:

Solution:

Enthalpy of Reaction

Breaking Bonds vs Forming Bonds

• Breaking bonds requires an input of energy (endothermic, positive \Delta H).

• Forming bonds releases energy (exothermic, negative \Delta H).

• The overall energy change and whether a reaction is endothermic or exothermic depends on the proportion of energy needed for breaking bonds versus the energy released during the formation of bonds.

• Example: C6H{12}O6 + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)

Calculating Enthalpy of Reaction

Standard Bond Energies Table

• Provides bond energies (kJ/mol) for various bonds such as C-C, C=C, C≡C, C-H, C-O, C=O, O-O, O=O, O-H

Enthalpy of Reaction Calculation Example

• Example: C6H{12}O6 + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)

• Breaking Bonds (requires energy, +): involves breaking 5 C-C, 7 C-O, 7 C-H, 5 O-H bonds and 6 O=O bonds;

  • Energies: (5 \times 347) + (7 \times 358) + (7 \times 413) + (5 \times 467)+ (6 \times 498) = 12455 \text{ kJ/mol}

• Forming Bonds (releases energy, -): involves forming 12 C=O and 12 O-H bonds;

  • Energies: (12 \times 745) + (12 \times 467) = 14544 \text{ kJ/mol}

• \Delta H = \text{Breaking} - \text{Forming} = 12455 - 14544 = -2089 \text{ kJ/mol}

Thermochemical Equations

• Thermochemical equations illustrate the energy change associated with a reaction per coefficient moles of each substance.

• Example: 2 H2S (g) +3 O2 (g) \rightarrow 2 H2O(l) + 2 SO2 (g) \quad \Delta H_{rxn} = -1120 \text{ kJ/mol}

Example Problems

Problem 1

Problem 2

Determine amount of energy released when 1.45 grams of SO_2(g) are formed.

Determine the mass of H_2S(g) used when 120 kJ of energy are released.

Hydrocarbon Combustion Example

• The hydrocarbon pentane, C5H{12} (72.15 g/mol) is combusted to produce carbon dioxide and water, as shown in the unbalanced reaction below.
  _C5H{12}(l) + _O2(g) \rightarrow _H2O(l) + _CO_2(g)

• The complete combustion of 5.00 g of pentane releases 243 kJ of heat. On the basis of this information, calculate the value of \Delta H for the complete combustion of one mole of pentane.

Neutralization Example

• A neutralization reaction between NaOH and HCl is carried out in a coffee cup calorimeter. A 10.0 mL sample of a 1.0 M NaOH solution is titrated against a 10.0 mL sample of a 1.0 M HCl solution, as shown in the equation below.
  HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)

• If both solutions started at a temperature of 20.0 ^{\circ}C and ended at 26.0 ^{\circ}C, what is the \Delta H_{rxn} in kJ/molrxn? (Assume the specific heat capacity of the total solution is the same as water, 4.18 J/g ^{\circ}C, and the density is 1.0 g/mL.)

Student Experiment and Calculations

• Reaction: 4 Fe(s) + 3O2(g) \rightarrow 2 Fe2O3(s) \quad \Delta H^{\circ} = 1650 \frac{kJ}{mol{rxn}}

Part (a)

Part (b)

Part (c)

Entropy and Gibbs Free Energy

Spontaneity

• Processes have a natural tendency to occur in one direction under a given set of conditions.

• A spontaneous process is one that occurs naturally under certain conditions (thermodynamically favorable).

• A nonspontaneous process requires a continual input of energy from an external source to occur.

• A process spontaneous in one direction is nonspontaneous in the reverse direction under the same conditions.

• Thermodynamics predicts whether a process will occur but doesn't indicate the time required for the process.

Predicting Spontaneity

• Spontaneous processes release usable energy.

• Nonspontaneous processes require an input of energy.

• An exothermic process (\Delta H < 0) is more likely to be spontaneous but not guaranteed, as both enthalpy and entropy must be considered to calculate usable energy change.

Entropy

• Entropy, S, measures the distribution of matter and energy (J/mol\cdotK).

• Phases of Matter example: S{\text{solid}} < S{\text{liquid}} << S_{\text{gas}}

• Thermal Equilibrium example

Entropy Change, ΔS

• Qualitative predictions for a positive change in entropy (\Delta S):

  • Increase # of independently moving molecules

  • Increase # of gas molecules

  • Increase Temperature

  • Increase Volume (for Gases)

  • Solid or liquid \rightarrow Gas

  • Solid \rightarrow Liquid

  • Solid + Liquid \rightarrow Solution

Gibbs Free Energy, ΔG

• If we can measure entropy and enthalpy changes, we can calculate the change in usable energy, known as Gibbs free energy, \Delta G. \Delta G = \Delta H - T\Delta S

  • \Delta H = enthalpy change of reaction/process (kJ/mol)

  • \Delta S = entropy change of reaction/process (J/mol\cdotK)

  • T = temperature in Kelvin

• Spontaneous processes occur when \Delta G < 0 (releasing usable energy).

Gibbs Free Energy Interpretation

• Usable energy (\Delta G) represents the difference between energy produced (\Delta H) and energy lost to surroundings (T\Delta S).

Coupled Reactions

• In biological systems, spontaneous processes (releasing usable energy) can drive non-spontaneous processes through coupled reactions.

Example Calculation

Problem

Solution

Reaction Spontaneity Table

Spontaneity Table Expanded

Free Energy Check-In #1

Problem

Free Energy Check-In #2

Explanation Question: Reaction Spontaneity at Different Temperatures

Biochemical Relevance of Delta G: Protein Folding

• Biochemical Example: Protein Folding

  • Folded state

  • Unfolded state

Protein Folding and Free Energy

• Dissecting the free energy of protein folding: \Delta G = \Delta H-T\Delta S < 0, \Delta G = -50 \frac{kJ}{mol} Hydrophobic effect -T\Delta S < 0 ~-200 kJ/mol H-bonds \Delta H < 0 ~-500 kJ/mol chain conformational entropy -T\Delta S > 0 50 kJ/mol
  VDW Electrostatic \Delta H ~-50 kJ/mol

Kinetics

Factors Affecting Reaction Rates: Questions to Consider

• How does temperature affect the rate of a chemical reaction?

• How do surface area and concentration affect reaction rates?

• How do catalysts affect the rate of a chemical reaction?

Potential Energy Diagrams

Chemical Kinetics: Introduction

• Kinetics: the study of the rate at which a chemical process occurs.
  Besides the speed of reactions, kinetics sheds light on the reaction mechanism (how the reaction occurs).

Factors That Affect Reaction Rates

Physical state of the reactants.

• Molecules must come in contact to react.
  The more homogeneous the mixture, the faster the molecules react.

Concentration of reactants.

• As the concentration of reactants increases, the likelihood that reactant molecules will collide increases.

Temperature

• At higher temperatures, reactant molecules have more kinetic energy and move faster.

Presence of a catalyst.

• Catalysts speed up reactions by changing the mechanism.

• Catalysts are not consumed during the reaction.

The Collision Model

• In a chemical reaction, bonds are broken, and new bonds are formed.
  Molecules can only react if they collide with each other.

Collision Model - Orientation and Energy

• Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

Activation Energy

• Activation Energy, E_a: the minimum amount of energy required for a reaction to occur. Molecules must possess sufficient energy to get over the activation-energy barrier.

Reaction Coordinate Diagrams

Reaction Coordinate Diagrams: Key Features

• The diagram shows the energy of the reactants and products (\Delta E).
  The high point on the diagram is the transition state.
  The species present at the transition state is called the activated complex.
  The energy gap between the reactants and the activated complex is the activation-energy barrier.

Maxwell–Boltzmann Distributions

• Temperature: a measure of the average kinetic energy of the molecules in a sample.

Key Characteristic

Temperature and Distribution

• As the temperature increases, the curve flattens and broadens.

Activation Energy and Temperature

• If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier.

Relating Energy Profiles to Activation Energies and Speeds of Reaction Practice #1

Determining the Fastest Reaction Given Multiple Reactions and Various Energy Profiles

Reaction Rates

• Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

Practice #2: Calculating an Average Rate of Reaction

• From the data in the figure below, calculate the average rate at which A disappears over the time interval from 20 s to 40 s.

Practice #2b: Calculating an Average Rate of Reaction

• Use the data in the same figure to calculate the average rate of appearance of B over the time interval from 0 s to 40 s.

Example Reaction: Butyl Chloride with Water

• In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.
  C4H9Cl(aq) + H2O(l) \longrightarrow C4H_9OH(aq) + HCl(aq)

Calculating the Average Rate

• The average rate of the reaction over each interval is the change in concentration divided by the change in time:
  C4H9Cl(aq) + H2O(l) \longrightarrow C4H9OH(aq) + HCl(aq) \text{Average rate} = \frac{-\Delta[C4H_9Cl]}{\Delta t}
  Note that average rate decreases as the reaction proceeds. With fewer collisions between reactant molecules, the reaction goes forward.

Instantaneous Rate from a Plot

• A plot of [C4H9Cl] versus time for this reaction yields a curve. The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
  C4H9Cl(aq) + H2O(l) \longrightarrow C4H_9OH(aq) + HCl(aq)

The Best Indicator of Rate

• All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.
  C4H9Cl(aq) + H2O(l) \longrightarrow C4H_9OH(aq) + HCl(aq)

Reaction Rates and Stoichiometry

• In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.
  Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
  C4H9Cl(aq) + H2O(l) \longrightarrow C4H9OH(aq) + HCl(aq) \text{Rate} = \frac{-\Delta[C4H9Cl]}{\Delta t} = \frac{\Delta[C4H_9OH]}{\Delta t}

Practice #3: Calculating an Instantaneous Rate of Reaction

• Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate).

General Rate Expression

• To generalize, then, for the reaction:
  aA + bB \longrightarrow cC + dD
  For such a case
  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}

Practice #4: Relating Rates at Which Products Appear and Reactants Disappear

Question A

Question B

Practice #5

Concentration and Rate

• One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

Experiment: NH4 +(aq) + NO2 − (aq) \longrightarrow N2(g) + 2 H2O(l)

If we compare Experiments 1 and 2, we see that when [NH4 ^+] doubles, the initial rate doubles. Likewise, when we compare Experiments 5 and 6, we see that when [NO2 ^−] doubles, the initial rate doubles.

• This means:
  Rate \propto [NH4 ^+] Rate \propto [NO2 ^−]
  Rate \propto [NH4 ^+] [NO2 ^−]
  Which, when written as an equation, becomes
  Rate = k [NH4 ^+] [NO2 ^−]

Rate Laws

• A rate law shows the relationship between the reaction rate and the concentrations of reactants.
  The exponents tell the order of the reaction with respect to each reactant.
  Since the rate law is Rate = k[NH4 ^+] [NO2 ^−]
  the reaction is First-order in [NH4 ^+] and First-order in [NO2 ^−]
  The overall reaction order can be found by adding the exponents on the reactants in the rate law.
  This reaction is second-order overall.

Practice Rate Law to the Effect of Concentration on Rate (Practice #6)

Practice #6 example

Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in