Thermochemistry

Thermochemistry- measurement & prediction of the effects of heat. Energy of reactants & products is important as is the transfer of energy. Key: Heat = stoichiometric quantity.

Heat (q)– involves transfer of energy between 2 objects with temperature differences.

Unit = calorie (cal) or joule; 1 cal = 4.184 J

Temperature (T) – property that reflects the random motion of particles in a particular substance.

Thermal equilibrium – a condition in which temperature is constant throughout a material and no heat flow occurs from point to point.

Energy (E) – capacity to do work or produce heat. Types: kinetic & potential. Unit = joule (J)

Work is done when a force moves an object through a distance.

1st Law of Thermodynamics: energy cannot be created or destroyed.

Potential: due to position or composition - can be converted to work PE = m x g x h

(m = mass, g = force of gravity, and h = vertical distance)◦ chemical energy is a form of potential energy

Kinetic: due to motion of the object KE = 1/2 mv 2 (m = mass, v = velocity)

Depends only on the present state of the system - not how it arrived there.◦ It is independent of pathway.

KE at the molecular level depends on the mass and velocity of the particle. Velocity depends on temperature so KE depends on temperaturealso.◦ As T ↑, KE ↑

◦ Total energy of matter at the micro level is sum of KE due to random motion and PE due to arrangement of particles.

One of the most important forms of PE at the atomic-molecular level arises from electrostatic interactions.

Coulombic attraction, not gravitational force, determines the potential energy of matter at the atomic level.

Eel ∝ Q1 x Q2

dEel is the electrostatic potential energy

System: the part of the universe that is the focus of a thermodynamic study.

Surroundings: everything in the universe that is not part of the system.

Universe = System + Surroundings

An isolated system exchanges neither energy nor matter with the surroundings.

Reactions can evolve heat – EXOTHERMIC q < 0

heat from system surroundingsEx. Liquid solid (freezing)

Reactions can absorb heat –ENDOTHERMICq > 0

heat from surroundings systemEx. Solid liquid (melting)

The internal energy of a system is the sum of all the KE and PE of all of the components of the system.

ΔE = q + w◦ ΔE = change in system’s internal energy◦ q = heat◦ w = work

Work done by the system = −PΔV (P is the atmospheric pressure and ΔV is the change in volume).

Enthalpy (H) = E + PV

Change in Enthalpy (ΔH) = ΔE + Δ(PV)

At constant pressure qP = ΔE + PΔV, therefore qP = ΔH

ΔH = change in enthalpy: an energy flow as heat (at constant pressure)

key: ΔH > 0, Endothermic

ΔH < 0, Exothermic

Molar heat capacity (cp) is the heat required to raise the temperature of 1 mole of a substance by 1oC at constant pressure.

Specific heat (cs) is the heat required to raise the temperature of 1 gram of a substance by 1oC at constant pressure.

q = mcsΔTHeat capacity (Cp) is the quantity of heat needed to raise the temperature of some specific object by 1oC at constant pressure.

21

Specific heat values for some elements in J/g oC

A 150.0 g sample of lead is heated to the temperature of boiling water (100.0 ºC). (b) A 50.0 g sample of water is added to a thermally insulated beaker, and its temperature is found to be 22.0 ºC. (c) The hot lead is dumped into the cold water, and the temperature of the final lead-water mixture is 28.8 ºC. Calculate CS for lead.

What quantities do we need to solve problem?(1) the quantity of heat supplied(2) the mass of metal involved (150.0 g Pb)(3) the change in temperature (T final - T initial ) for both the water and the metal

Also - realize that the energy lost by the metal is equivalent to the heat gained by the water – this is IMPORTANT!

in equation form: q water = -q metal

(1) the heat gained by the water ...

(2) The heat lost by lead ...

Calorimetry◦ Coffee cup (constant P)◦ Bomb (constant V)

KEY: heat from the system = heat transferred into the surroundings

the system? = the surroundings? =

KEY: heat from the system = heat transferred into the surroundings

the system? = the surroundings? =

To determine the heat capacity of a calorimeter, run a reaction that gives off a known quantity of heat in the calorimeter. (i.e., calibrate your calorimeter)

EX. The heat of combustion of propane is -2200 kJ/mol. When a 0.650 g sample of propane is burned in a bomb calorimeter, the temperature of the water bath rises 4.47 oC. What is the heat capacity (Ccal) of the calorimeter?

Part I: How much heat does the reaction release?for our sample – heat released:

qrxn = -qcal

The calorimeter calibrated in the previous example was used to determine the heat of combustion of benzoic acid (C7H6O2) in oxygen. A 0.821 g portion of benzoic acid was completely burned in the calorimeter and a temperature rise of 3.690C was observed for the calorimeter. What is the heat of combustion of one mole of benzoic acid?

a chemical reaction = the making and breaking of bondsit all involves ENERGY

bond energies can be used to approximate energies of reaction –we call this enthalpy (ΔH).

Enthalpy is T and P dependent. Energy & enthalpy have SI units:

joule (J) = kg m2/s2

reactants ==(break apart) ==> atoms ==(come together) ==> products

KEY: Breaking bonds costs energy (+) while making bonds releases energy (-)

The energy needed to break 1 mole of covalent bonds in the gas phase is the bond energy of that bond.Breaking bonds consumes energy (+) whereas forming bonds releases energy(-).Bond energies can be used to estimate ΔHrxn.

ΔHorxn = Σ ΔH (bonds breaking) + ΣΔH (bonds forming)

*Note: for C=O in CO2 use 799 kJ/mol

CH4(g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Tally up the energies in the bonds … ΔHo

rxn = Σ ΔH (bonds breaking) + ΣΔH (bonds forming)

broken formed

Recall enthalpy (heat) depends on T & PConvention: tabulate enthalpies for specific conditions

****Thermodynamic STANDARD STATE ****P = 1 bar T = 25oC

(NOTE: 1 bar is 1.01325 atm)

Define: STANDARD STATE = the most stable form of a substance in the physical state in which it exists under standard conditions

hydrogen? ... gasmercury? ... liquid

iron? ... solid

Define: STANDARD ENTHALPY CHANGE OF A REACTION (ΔH0)

ΔH0 = the observed or calculated enthalpy change for a reaction occurring when all reactants and products are in their standard states under standard conditions.

The standard enthalpy change for a reaction involving the formation of one mole of a compound directly from its elements is called the STANDARD MOLAR ENTHALPY OF FORMATION = ΔHf

0

H2 (g) + 1/2 O2 (g) ===> H2O (l) ΔHf0= -285.8 kJ/mol

C (graphite) + 2 H2 (g) ===> CH4 (g) ΔHf0 = -74.81 kJ/mol

KEY: elements in standard states forming 1 mole of product

NOTE: this doesn't mean that it's actually made this way

You can estimate the standard enthalpy change for a reaction if the standard enthalpies of formation of reactants and products are known using: ΔH0 = ΣH0

f (products) - ΣH0f (reactants)

Benzene is a hydrocarbon with the formula C6H6 . Calculate the enthalpy change associated with the combustion of benzene using ΔH0

f values.C6H6 (l) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l)

From appendix A4.3:ΔHo

f [benzene (l)] = + 49.0 kJ/molΔHo

f [CO2 (g)] = -393.5 kJ/molΔHo

f [H2O (l)] = -285.8 kJ/molΔHo

f of any element in its standard state is defined as zero.

ΔH0 = ΣH0f (products) - ΣH0

f (reactants)

Hess's Law: “in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.”

For example:Copper (I) chloride can be formed from the reaction of copper metal and chlorine gas. The enthalpy change for this reaction is -137.3 kJ per mole of CuCl formed. Copper (I) chloride can in turn be reacted with more chlorine gas to form copper (II) chloride. The enthalpy change for this reaction is -82.9 kJ per mole of CuCl2formed.

The sum of these two reactions yields a third equation that describes the process for the formation of copper (II) chloride from copper and chlorine.

It is very difficult to measure the heat of formation of methane (CH4) from graphite and hydrogen gas. However, the following thermochemical equations are known:

CH4 (g) + 2 O2 (g) ===> CO2 (g) + 2 H2O (l) - 889.5 kJ/mol CH4H2 (g) + 1/2 O2 (g) ===> H2O (l) - 285.5 kJ/mol H2OC (graphite) + O2 (g) ===> CO2 (g) - 393.3 kJ/mol C

Calculate the heat of formation of methane from graphite and hydrogen gas.

Lightweight camping stoves typically use white gas which is a mixture of C5 and C6hydrocarbons.

(a) How much heat is released during the combustion of 1.00 kg of pentane (C5 H12) if ΔHcomb = -3535 kJ/mol pentane?

(b) How many grams of pentane must be burned to heat 1.00 kg of water from 20.0 to 90.0oC? Assume that all the heat released during the combustion reaction is used to heat the water.