Chapter 6 Thermochemistry

Energy Is important macroscopically and microscopically Def: the capacity to do work or produce heat

Law of Conservation of Energy Energy can neither be created nor destroyed, but can be converted from one form to another. The energy of the universe is constant. FIRST LAW OF THERMODYNAMICS

Classifications of Energy Potential Energy: energy due to position or composition –Ex: dam water, attractive/repulsive forces Kinetic Energy: energy due to motion, depends on mass (m) and velocity (v) –KE = (1/2)mv 2 Energy can be converted between these

Energy Transfer Some energy can be lost as heat (ex: frictional heating), represented by q Heat vs. Temperature: TEMPERATURE reflects movement of particles. HEAT deals with transfer of energy between two objects due to a temperature difference. Energy can also be transferred through work (force acting over a distance)

Important Vocabulary Pathway determines how energy changes to heat/work –Includes condition of the surface Total energy transferred will be constant, amounts of heat/work will differ

State Function/Property Property of the system that depends only on its present state (not past or future) Changes in the state properties when switching from one state to another is independent of the particular pathway taken between the two states. Example: energy is a state function, but work and heat are not. –Internal energy, pressure, volume, enthalpy

Heat Represented by “q” Flows from warm to cold

System vs. Surroundings System is what you are focusing on, surroundings is everything else. Energy lost/gained by the system = energy gained/lost by the surroundings

Endothermic System gains heat, surroundings cool Heat INTO (“endo”) system Example: Ice melting “q” is positive Heat is a reactant N 2(g) + O 2(g) + HEAT(kJ) --> 2NO (g)

Exothermic System loses heat to surrounding “Exo” like EXIT Example: fire “q” is negative Heat is a product CH 4 (g) + 2O 2(g) --> CO 2(g) + 2H 2 O (g) + HEAT(kJ)

Potential Energy Energy is stored in chemical bonds as potential energy. When bonds are broken (requires energy) and formed (releases energy), it changes the potential energy.

Units for Energy Two common units: –The “calorie” (old school) –The “joule” (metric) 1 cal = 4.18 J J = kgm J = 1 L atm s2s2

Energy Stoichiometry Energy can be added into stoichiometry equations… C 6 H 12 O 6 + 6O 2 --> 6CO 2 + 6H 2 O kJ You can substitute it in as part of the mole to mole ratio! How much heat is given off when 3.72 moles of oxygen react with glucose? Answer: 1736 kJ

Internal Energy Represented by E Sum of kinetic and potential energy in the system  E = q + w  E = change in system’s internal energy q = heat (usually in J) w = work (usually units are L atm which can be converted to J)

Example Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J. Answer: J

Expansion vs. Compression w = -P  V w = work P = pressure  V = change in volume

Results… If a gas is expanding,  V is positive If a gas is compressed,  V is negative When w is negative, work if flowing out of the system (into surroundings) When w is positive, work is flowing into the system

Example A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system. Answer: 1400 J

Enthalpy Represented by “H” H = E + PV E = internal energy P = pressure of the system V = volume of the system

This means at constant pressure…  H = q Negative  H is exothermic Positive  H is endothermic  H = H products - H reactants

Calorimetry Energy can’t be created nor destroyed. If one object loses heat, another object must gain that heat. If a cool object is placed into a hot one, the hot object gives energy to the cool one until they arrive at the same final temp.

Conservation of Energy The amount of heat lost by the hotter object equals the amount of heat gained by the cooler object: q gained = -q lost (one is system, other surroundings)

Specific Heat Capacity (J/g°C) The amount of energy needed to raise the temp. of 1 gram of an object by 1°C. A high specific heat means the object requires a lot of energy to change temp. (pg. 245) If unit is J/mol°C or J/molK, then it’s called molar heat capacity

Constant Pressure Calorimeter Also have constant volume calorimeters

q = amount of heat gained (negative if lost) m = mass of object (grams) c = specific heat of object  T = change in temperature (T final – T initial ) q = m*c*  T

This equation only works when the temperature is changing. This is used in calorimetry! q = m*c*  T

Example A bar of iron at 21.0°C is heated to 83.5°C. If the iron’s mass is 551 grams, how much heat was added? Answer: 1.5X10 4 J

Calorimetry Example A 36.9 g sample of metal is heated to 100.0ºC, and then added to a calorimeter containing g of water at 23.1ºC. The temperature of the water rises to a maximum of 25.2ºC before cooling back down. What is the specific heat of the metal? Answer: 0.45 J/gºC

Hess’s Law If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Allows you to determine the heat of reaction.

Manipulations If you reverse a reaction, change the sign of  H. C + O 2  CO 2(g) = kJ CO 2(g)  C + O 2 = 393.5kJ

Manipulations If you double/triple/etc. a reaction, you also must double/triple/etc.  H. C + O 2  CO 2(g) = kJ 2C +2O 2  2CO 2(g) = 2*-393.5kJ

Example Work backwards… Given the following, calculate ∆ H for 4NH 3(g) + 3O 2(g) --> 2N 2(g) + 6H 2 O (l) 2N 2 O (g) --> O 2(g) + 2N 2(g)  H = -164kJ 2NH 3(g) + 3N 2 O (g) --> 4N 2(g) + 3H 2 O (l)  H = -1012kJ Answer: kJ

Standard Enthalpy of Formation Represented by  H°f Sometimes calorimeters can not be used to find  H (like if the process is very slow) Shows how much energy is required to make 1 mole of a compound from its elements with all substances in their standard state Table 6.2 in your book

Standard State Definition COMPOUND: Pressure = 1 atm Pure liquid or solid If in solution, concentration = 1M ELEMENT: Pressure = 1 atm Temperature = 25°C State = whatever state it exists in these conditions

Calculating  H 0 reaction = E n p  H f 0 (products)–En r  H f 0 (reactants) E means sum of… n p is moles of product…n r is moles of reactant If dealing with an element, the  H f 0 is zero

Example Calculate the change in enthalpy: CH 4(g) + 2O 2(g)  CO 2(g) +2H 2 O (l) Answer: Products: [-393.5kJ + 2(-285.8kJ)]= kJ Reactants: [-74.86kJ + 2(0.0kJ)] = kJ P-R = kJ-(-74.86kJ) = kJ

Fossil Fuels When decayed plants are burned, the energy stored in them can be used –Petroleum and natural gas (hydrocarbons) –Coal: creates about 23% of U.S. energy

Hydrocarbons Prefixes are the same in organic chemistry Meth- Eth- Prop- But- Pent- Hex- Hept- Oct- Suffixes represent different bonding/compounds

Environmental Impacts Greenhouse effect: CO 2 created from combustion of fossil fuels absorbs infrared radiation given off by the earth and does not allow it to leave the atmosphere H 2 O can also absorb radiation (humidity), but earth’s H 2 O has not changed much…

New Energy Sources Coal gasification: creates a gaseous form of coal (syngas) that is easily transported and reacted with oxygen in a combustion reaction, releasing energy –This material can be used to produce other fuels like methanol

New Energy Sources Coal Slurries: coal ground up and mixed with water and used instead of solid coal in power plants

New Energy Sources Hydrogen has a heat of combustion about 2.5 times that of natural gas –Only biproduct of its combustion rxn is water (no carbon dioxide) –VERY expensive (takes a lot of energy to do), hard to store/transport

New Energy Sources Oil shale (kerogen): expensive and lots of waste rock Ethanol (gasohol): doesn’t work well in low temps because of a lower vapor pressure Methanol: being currently tested Seed oil (sunflower oil): renewable (can be grown and replaced easily)