Chapter 16: Energy and Chemical Change Table of Contents Chapter 16: Energy and Chemical Change 16.1: Energy 16.2: Heat in Chemical Reactions and Processes 16.3: Thermochemistry 16.3: Thermochemical Equations 16.4: Calculating Enthalpy Change

universe = system + surroundings Energy and Chemical Change: Basic Concepts Chemical Energy and the Universe universe = system + surroundings The Universe is everything around us the System: the specific part of the universe that you are studying. the Surroundings: everything else

Energy is the ability to do work or produce heat. Energy and Chemical Change: Basic Concepts The Nature of Energy Energy is the ability to do work or produce heat. Potential energy: stored energy. Kinetic energy: the energy of movement.

Chemical potential energy: stored in bonds Energy and Chemical Change: Basic Concepts The Nature of Energy Chemical potential energy: stored in bonds The energy stored in a substance because of its composition : the type of atoms in the substance the number and type of chemical bonds joining the atoms the way atoms are arranged.

Molecular kinetic energy: Measure of molecular “speed” Energy and Chemical Change: Basic Concepts The Nature of Energy Molecular kinetic energy: Measure of molecular “speed” Measured by a thermometer: As temperature increases, the motion of submicroscopic particles increases. So a thermometer is like a molecular speedometer.

Chemical systems contain both kinetic energy and potential energy. Energy and Chemical Change: Basic Concepts The Nature of Energy Chemical systems contain both kinetic energy and potential energy. Law of conservation of energy: energy can be converted from one form to another, but it is neither created nor destroyed. suppose you have money in two accounts at a bank and you transfer funds from one account to the other. Although the amount of money in each account has changed, the total amount of your money in the bank remains the same. the first law of thermodynamics

Chemical potential energy Energy and Chemical Change: Basic Concepts Chemical potential energy Heat (symbol=q), is the flow of kinetic energy due to temperature changes. Exothermic = gives off heat (hot) Forming bonds – releases energy Endothermic = takes in heat (cold) Breaking bonds – requires energy Heat always travels from hot to cold.

1,000 calories = 1 kilocal = 1 Calorie Energy and Chemical Change: Basic Concepts Measuring heat UNITS calorie (cal): metric unit of energy the amount of energy required to raise the 1 gram of water 1° Celsius 1,000 calories = 1 kilocal = 1 Calorie dietary Calories chemistry calories

joule (J): The SI unit of heat and energy Energy and Chemical Change: Basic Concepts Measuring heat UNITS joule (J): The SI unit of heat and energy 4.184 joules = 1 calorie 1,000 J = 1 kJ (Calories are bigger than joules)

Converting Energy Units Steps for converting units: Energy and Chemical Change: Basic Concepts Converting Energy Units Steps for converting units: 1. Find your goal 2. Start with what you know 3. Use conversion factors to change units 4. Check that units cancel

Converting Energy Units Energy and Chemical Change: Basic Concepts Converting Energy Units How much energy in joules is 230 nutritional Calories? Goal = Joules Known = 230 Calories J = 230 Cal 1000 cal 4.184 J = J 962,320 1 Cal 1 cal = 9.6 x 105 J convert Cal to cal convert cal to J

Practice converting energy units Convert 142 Nutritional Calories into calories. An exothermal reaction releases 86.5 kJ. How many kilocalories of energy are released? If an endothermic process absorbs 256 J, how many kilocalories are absorbed? 142,000 cal 20.7 kcal 0.0612 kcal

used to determine the specific heat of unknown metals Energy and Chemical Change: Basic Concepts Measuring Heat Calorimeter: a device used to measure the amount of heat absorbed or released during a chemical or physical process. used to determine the specific heat of unknown metals

Each substance has its own specific heat. Energy and Chemical Change: Basic Concepts Specific Heat The specific heat (c): the amount of heat required to raise the temperature of 1 gram of any substance by 1° C. Each substance has its own specific heat. Why? Because different substances have different compositions. specific heat (c) of water = 4.184 J/(g∙°C)

∆T = |Tfinal – Tinitial| Energy and Chemical Change: Basic Concepts Calculating heat evolved and absorbed Heat absorbed or released by a substance during a change in temperature: specific heat (J/g ·°C) mass (g) Change in temperature (°C) HEAT (J) ∆T = |Tfinal – Tinitial|

Calculating Specific Heat Energy and Chemical Change: Basic Concepts Calculating Specific Heat The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron? q 114 J c = = = m · ΔT (10.0 g) |25.0°C-50.4°C| 114 J = = .449 J/(g·°C) (10.0 g) (25.4°C)

Using Data from Calorimetry Energy and Chemical Change: Basic Concepts Using Data from Calorimetry A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C. What is the specific heat of the metal? Known Unknown mass of metal = 4.68 g metal specific heat c = ? J/(g·°C) heat absorbed, q = 256 J ∆T = 182°C c = .301 J/(g·°C)

Practice specific heat equation If the temperature of 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed? 4.50 g of gold [c = 0.129 J/(g·°C)] absorbed 276 J of heat. If the initial temperature was 25.0°C, what was the final temperature? 4.52 x 103 J 5.00 x 102 °C

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Chapter 16: Energy and Chemical Change Table of Contents Chapter 16: Energy and Chemical Change 16.2-3: Thermochemistry

Practice specific heat equation If the temperature of 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed? 4.50 g of gold [c = 0.129 J/(g·°C)] absorbed 276 J of heat. If the initial temperature was 25.0°C, what was the final temperature? 4.52 x 103 J 5.00 x 102 °C

Practice specific heat equation Endothermic or exothermic? Melting ice Boiling water Freezing water Making water from steam endo endo exo exo

Endo- or exothermic? http://www.youtube.com/watch?v=yTzcoyzPQE0

Chemical Energy and the Universe Energy and Chemical Change: Basic Concepts Chemical Energy and the Universe Thermochemistry: the study of heat changes that accompany chemical reactions and phase changes. Enthalpy (H): the heat content of a system at constant pressure. Enthalpy (heat) of reaction (∆Hrxn) = the change in enthalpy for a reaction. always measured as change.

Thermochemical Equations Exothermic 4Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ 4Fe(s) + 3O2 → 2Fe2O3(s); ∆H = -1625 kJ Endothermic 27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) NH4NO3(s) → NH4+(aq) + NO3-(aq); ∆H = 27 kJ

Standard enthalpy changes have the symbol ∆H°. Energy and Chemical Change: Basic Concepts Enthalpy changes Standard enthalpy changes have the symbol ∆H°. Standard conditions: 1 atm pressure 298 K (25°C) Units of enthalpy: kJ released 1 mol substance

∆Hcomb is exothermic (-) Energy and Chemical Change: Basic Concepts Enthalpy changes Types of enthalpy Enthalpy (heat) of combustion (∆Hcomb) = enthalpy change for the complete burning of one mole of the substance. ∆Hcomb is exothermic (-)

PHASE CHANGES TEMPERATURE boiling evaporating gas BP condensing liquid melting MP freezing solid ENERGY

∆Hvap and ∆Hfus are endothermic (+) Energy and Chemical Change: Basic Concepts Changes of state Molar enthalpy (heat) of vaporization (∆Hvap) = heat required to evaporate one mole of a liquid (liquid → gas) Molar enthalpy (heat) of fusion (∆Hfus) = heat required to melt one mole of a solid (solid → liquid) ∆Hvap and ∆Hfus are endothermic (+)

∆Hcondensation = -∆Hvap ∆Hsolidification = -∆Hfus Energy and Chemical Change: Basic Concepts Changes of state Molar enthalpy (heat) of condensation (∆Hcond) = heat required to condense one mole of a liquid (gas → liquid) Molar enthalpy (heat) of solidification (∆Hsolid) = heat required to solidify one mole of a solid (liquid → solid) ∆Hcondensation = -∆Hvap ∆Hsolidification = -∆Hfus

PHASE CHANGES TEMPERATURE ΔHchange boiling evaporating gas BP condensing liquid melting MP freezing q=c∙m∙ΔT solid ENERGY

Converting Energy Units Energy and Chemical Change: Basic Concepts Converting Energy Units How much heat is released when 82.1 g of methanol is burned? Goal = heat released Known = 82.1 g methanol 82.1 g CH3OH 1 mol CH3OH - 726 kJ 32.05 g CH3OH 1 mol CH3OH Use ENTHALPY to convert mol to kJ convert g to mol MOLAR MASS q = kJ 1860

Practice finding heat ∆H = enthalpy = (+) endothermic = (-) exothermic WHY? It tells us if it’s endo- or exothermic q = heat = always (+) WHY? Because it tells us HOW MUCH energy there is, NOT if it’s endo- or exothermic

Practice finding heat 2.58 kJ 377 kJ 232 g Calculate the heat required to melt 25.7 g of solid methanol. How much heat is evolved when 275 g of ammonia gas condenses to a liquid? What mass of methane must be burned in order to release 12,880 kJ of heat? 2.58 kJ 377 kJ 232 g

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WebAssign #14 q=c∙m∙ΔT ΔHchange 693 kJ How much heat is required to change 213 g of ice from -46.5ᵒC to 0.0ᵒC, melt the ice, warm the water from 0.0ᵒC to 100ᵒC, boil the water, and heat the steam to 173.0ᵒC? 100ᵒ to 173ᵒC q=c∙m∙ΔT boil ΔHchange melting 0.0ᵒC to 100ᵒC -46.5ᵒC to 0.0ᵒC 693 kJ

Hess’ Law (adding equations) Comparing products to reactants Energy and Chemical Change: Basic Concepts There are two ways that we will calculate the standard enthalpy of reaction: Hess’ Law (adding equations) Comparing products to reactants - both use enthalpies of formation The standard enthalpy (heat) of formation (∆H°f) of a compound is the change in enthalpy for the formation of one mole of a compound

Calculating Enthalpy Change Energy and Chemical Change: Basic Concepts Calculating Enthalpy Change Hess’s law: If you can add two equations, you can add their enthalpies Used to find the enthalpy of the overall reaction. Think of it like you’re doing a puzzle

Some have 3 equations – follow the same principles Applying Hess’ Law Use the following chemical equations to determine ΔH for this equation: 2H2O2(l) → 2H2O(l) +O2(g) 2H2(g) +O2(g) → 2H2O(l) ΔH = -572 kJ H2(g) +O2(g) → H2O2(l) ΔH = -188 kJ Identify products and reactants Match coefficients to overall equation Add equations (cancel spectators) Add enthalpies ΔH = -196 kJ Some have 3 equations – follow the same principles

Determine ΔH for each reaction 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) ∆H = ? 2CO(g) + O2(g) → 2CO2(g) ∆H = -556.0 kJ N2(g) + O2(g) → 2NO(g) ∆H = 180.6 kJ 4Al(s) + 3MnO2(s) → 2Al2O3(s) + 3Mn(s) ∆H = ? 4Al(s) + 3O2(g) → 2Al2O3(s) ∆H = -3352 kJ Mn(s) + O2(g) → MnO2(s) ∆H = -521kJ -746.6 kJ -1789 kJ

Uses the standard enthalpies of formation for each individual molecule Energy and Chemical Change: Basic Concepts A simpler way: Uses the standard enthalpies of formation for each individual molecule f reactants f products The standard heat of formation of an element in its standard state is zero. Ex) Au, O2; ΔH°f = 0 kJ

Standard Enthalpies of formation Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l) 1. Find ∆H°f for known elements ∆H°f (CH4(g)) = -75 kJ ∆H°f (O2(g)) = 0.0 kJ ∆H°f (CO2(g)) = -394 kJ ∆H°f (H2O(g)) = -286 kJ

Standard Enthalpies of formation Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l) 2. Use formula ∆H°rxn = ∆H°f(prod) - ∆H°f(react) ∆H°rxn = (∆H°f(CO2) + ∆H°f(H2O)) – (∆H°f(CH4) + ∆H°f(O2)) ∆H°rxn = [(-394 kJ + 2(-286 kJ)] – [(-75 kJ + 2(0.0 kJ)] = [-966 kJ] – [-75 kJ] = -966 kJ + 75 kJ = -891 kJ The combustion of 1 mol CH4 releases 891 kJ

Practice ∆H°rxn ∆H°rxn = 178.3 kJ ∆H°rxn = 66.36 kJ Use standard enthalpies of formation to calculate ∆H°rxn for each of the following reactions. CaCO3(s) → CaO(s) + CO2(g) N2(g) + 2O2(g) → 2NO2(g) 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l) ∆H°rxn = 178.3 kJ ∆H°rxn = 66.36 kJ ∆H°rxn = -1660.3 kJ

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