THERMOCHEMISTRY

Study of heat change in chemical reactions

Systems, Surroundings & Internal Energy

Surroundings = rest of the universe (or as much as needed…) the flask. perhaps the flask and this classroom. perhaps the flask and all of the building, etc.

Universe = System + Surroundings

System = the part of the universe under study chemicals in a flask. the coffee in your coffee cup. my textbook.

The First Law of Thermodynamics• Thermodynamics

• Deals with all kinds of energy effects in all kinds of processes

Two types of energy:• Heat (q) • Work (w)

The Law of Conservation of Energy• ΔEsystem = - ΔEsurroundings

The First Law:• ΔE = q + w• The total change in energy is equal to the

sum of the heat and work transferred between the system and the surroundings

The First Law of Thermodynamics

Internal energy = E within the system because of nanoscale position or motion

• Conservation of energy becomes:

ΔE = q + w

Heat • Thermal E transfer caused by a T

difference.• Heat flows from hotter to cooler objects

until they reach thermal equilibrium (have equal T ).

Work

• In an internal combustion engine, a significant fraction of the energy of combustion is converted to useful work– The expansion of the combustion gases

produces a volume and a pressure change

– The system does work on its surroundings

• Propels the car forward• Overcomes friction• Charges battery

Pressure-Volume Work

Note the same sign convention for q and w

ΔE = q + w

Heat transfer outq < 0

Work transfer inw > 0

Heat transfer inq > 0

Work transfer outw < 0

SURROUNDINGS

SYSTEM

1. A liquid cools from 45°C to 30°C, transferring 911 J to the surroundings. No work is done on or by the liquid. What is ΔEliquid?

SAMPLE PROBLEMS

2. A system does 50.2 J of work on its surroundings and there is a simultaneous 90.1 J heat transfer from the surroundings to the system. What is ΔEsystem?

State functionsAlways have the same value whenever the system is in the same state.

Two equal mass samples of water produced by:1. Heating one from 20°C to 50°C.2. Cooling the other from 100°C to 50°C.

have identical final H (and V, P, E…).

State function

s

H E P V T etc.

State Functions and Path Independence

State function changes are path independent.

ΔH = Hfinal – Hinitial is constant.

ENTHALPY of Chemical Reactions

• At constant V, ΔV = 0

• At constant p,

Enthalpy, ΔH

• H = E + PV• ΔH = ΔE + PΔV ; ΔH = qp

– The PV product is important only where gases are involved; it is negligible when only liquids or solids are involved

• ΔH = ΔE + ΔngRT– Δng is the change in the number of

moles of gas as the reaction proceeds

Measurement of Heat Flow: Calorimetry

• A calorimeter is a device used to measure the heat flow of a reaction– The walls of the calorimeter are

insulated to block heat flow between the reaction and the surroundings

– The heat flow for the system is equal in magnitude and opposite in sign from the heat flow of the calorimeter

• qreaction = - qcalorimeter

• qreaction = - Ccal Δt

The Calorimetry Equation

q = C x Δt – Δt = tfinal – tinitial

– C (uppercase) is the heat capacity of the system: it is the quantity of heat needed to raise the temperature of the system by 1 °C

q = m x c x Δt– c (lowercase) is the specific heat: the

quantity of heat needed to raise the temperature of one gram of a substance by 1 °C

• c depends on the identity and phase of the substance

Coffee-cup calorimeterNested styrofoam cups prevent heat transfer with the surroundings.

Constant P. ΔT measured.q = qp = ΔH

Assume the cups do not absorb heat. (Ccal = 0 and qcal = 0)

Measuring Enthalpy Changes

or−qreaction = qbomb + qwater

withqbomb = mcalccalΔT = CcalΔT

Bomb CalorimeterMeasure ΔT of the water. Constant V: qV = ΔE

Conservation of E:qreaction + qbomb + qwater = 0

Measuring Enthalpy Changes

A constant for a calorimeter

Sample Problem

Copper is used in building the integrated circuits, chips and printed circuit boards for computers. When 228 J of heat are absorbed by 125 g of copper at 22.38oC, the temperature increases to 27.12oC. What is the specific heat of copper?

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 100.0 g of water at 58.5°C with 100.0 g of water, already in the calorimeter, at 22.8°C, the final temperature of the water is 39.7°C.Calculate the heat capacity of the calorimeter in J/°C. Use 4.184 J/g - °C as the specific heat of water.

Octane (0.600 g) was burned in a bomb calorimeter containing 751 g of water. T increased from 22.15°C to 29.12°C. Calculate the heat evolved per mole of octane burned. Ccal = 895 J°C-1.

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l) qsys + qsurr = 0

qreaction + qbomb + qwater = 0

qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J

qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C

= +2.190 x 104 JSo −qreaction = +6238 + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ

qreaction = −28.1 kJ

Measuring Enthalpy Changes

Octane (0.600 g) was burned in a bomb calorimeter… Calculate the heat evolved per mole of octane burned

Molar mass of C8H18 = 114.23 g/mol.

nC8H18 = (0.600 g) / (114.23 g/mol)

= 0.00525 mol C8H18

Heat evolved /mol octane = −28.1 kJ 0.00525 mol

= −5.35 x 103 kJ/mol = −5.35 MJ/mol

Measuring Enthalpy Changes

Thermochemical Equations

• a chemical equation with the ΔH for the reaction included

ExampleNH4NO3 (s) NH4

+ (aq) + NO3- (aq) ΔH =

+28.1kJThis means:

+28.1 kJ +28.1 kJ +28.1 kJ

Or or 1 mol NH4NO3(s) 1 mol NH4

+(aq) 1 mol

NO3-(aq)

Conventions for Thermochemical Equations

1. The sign of ΔH indicates whether the reaction is endothermic or exothermic

2. The coefficients of the thermochemical equation represent the number of moles of reactant and product

3. The phases of all reactant and product species must be stated

4. The value of ΔH applies when products and reactants are at the same temperature, usually 25 °C

Rules of Thermochemistry

• ΔH is directly proportional to the amount of reactant or product– If a reaction is divided by 2, so is ΔH – If a reaction is multiplied by 6, so is ΔH

• ΔH changes sign when the reaction is reversed

• ΔH has the same value regardless of the number of steps

Sample Problem

In photosynthesis, the following reaction takes place:

6CO2(g) + 6H2O(l) 6O2(g) + C6H12O6(s) H = + 2801 KJ

a. Calculate H when one mole of carbon dioxide reacts?b. How many kilojoules of energy are liberated when 15.00 g of glucose is burned in oxygen?

METHODS OF DETERMINING Hreaction

1. DIRECT METHOD- using Ho

f of every reactants and products of chemical reaction

2. INDIRECT METHOD- using HESS LAW (for reaction involving several steps)

DIRECT METHOD

ΔH° ={(nproducts)(ΔHf° products)} – {(nreactants)(ΔHf°

reactants)}

Where: ΔHf° = standard molar enthalpy of formation taken from Thermodynamic data

Standard state = most stable form of the pure element at P = 1 bar.

e.g. C standard state = graphite (not diamond)

ΔHf° for any element in its standard state is zero. (take 1 mol of the element and make… 1 mol of element)

ΔHf° (Br2(l) ) = 0 at 298 K

ΔHf° (Br2(g) ) ≠ 0 at 298 K

Standard Molar Enthalpy of Formation

1

but the formation reaction is:H2(g) + ½ O2(g) H2O(l) ΔHf° = −285.83 kJ

H2 combustion:

2 H2(g) + O2(g) 2 H2O(l) ΔH° = −571.66 kJ

f = formation

Formation reactionMake 1 mol of compound from its elements in their standard states.

ΔH° ={(nproducts)(ΔHf° products)} – {(nreactants)

(ΔHf° reactants)}

ΔH° = [1ΔHf°(HCN) + 3 ΔHf°(H2)] −

[1ΔHf°(NH3) +

1ΔHf°(CH4)]

DIRECT METHOD

= [+134 + 3(0)] − [− 46.11 + (−74.85)] = 255 kJ

ExampleCalculate ΔH° for:

CH4(g) + NH3(g) HCN(g) + 3 H2(g)

ΔHf° : -46.11 -74.85 +134 0

INDIRECT METHOD - Hess’s Law

“If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔH° for the 1st reaction must be the sum of the ΔH° values of the other reactions.”Another version:“ΔH° for a reaction is the same whether it takes place in a single step or several steps.”

H is a state function

SAMPLE PROBLEMS