ThermalChemistry

V.B.3a. Explain the law of conservation

of energy in chemical reactions

b. Describe the concept of heat and explain the difference between heat energy and temperature

c. Explain physical and chemical changes and endothermic or exothermic energy changes

Energy

Def: the ability to do work

Classified as either:

Potential Energy

Kinetic Energy

1st Law of Thermodynamics

• The energy of the universe is constant

• Energy can neither be created or destroyed

• Can be converted from one form to another

Energy

• Energy is a state function

• Property of the system that changes independently of the pathway

Temperature & Heat

Temperature is a measure of the motion of the particles of a substance

Heat is the flow of energy due to a temperature difference

System & Surroundings

gsSurroundinSystem

gsSurroundinSystemenergy

energy

Internal EnergyAlways from the viewpoint of the

system

ΔE = q + w

q = heat

w = work

Special Conditions

• If the container is rigid and cannot expand, then no work is done by or on the system

• All energy must be in the form of heat (q) and therefore

ΔE = q

V.B.3.

d. Solve heat capacity and heat transfer problems involving specific heat, heat of fusion, and vaporization

e. Calculate the heat of reaction for a given chemical reaction when given calorimetric data

Measuring Energy

calorie (c) – the amount of energy needed to raise the temperature of 1g of water by 1oC

Joule (J) – 4.184 J = 1 calorie

(SI unit)

Heating Depends On..

1. The amount of substance being heated (in grams)

2. The temperature change

3. Specific heat capacity – the amount of energy needed to raise the temperature of one gram of a substance by 1oC

Calculating Energy Change

q = m c ΔT

q = heat

m = mass in grams

c = specific heat capacity

ΔT = change in temperature

V.B.4.

a. Define enthalpy and explain how changes in enthalpy determine whether a reaction is endothermic or exothermic

b. Compute ΔHrxn from ΔHfo and

explain why the ΔHfo values for

elements are zero

ENTHALPY

What is Enthalpy?

Consider a process at constant pressure where the only work is PV work (w = -PΔV)

ΔE = qp + w

ΔE = qp – PΔV

qp = ΔE + PΔV

H = E + PV or ΔH = ΔE + Δ(PV)ΔH = ΔE + PΔV (pressure is

constant)ΔH = qp

In Other Words…

the terms heat of reaction and change in enthalpy are the same so

ΔH = Hproducts - Hreactants

• Endothermic – absorbs heat during reaction (feels cold)

• Exothermic – gives off heat during reaction (feels hot)

Calculate the Standard Enthalpy Change for the combustion of Methane

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

1. CH4(g) C(s) + 2H2(g)

2H2(g) + C(s) CH4(g) -75 kJ

• O2(g) 0 kJ

• C(s) + O2(g) CO2(g) -394 kJ

• H2(g) + ½ O2(g) H2O(l) -286 kJ

Example

• When 1 mole of CH4 is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which 5.8 g of CH4 is burned at constant pressure.

Calorimetry

Calorimeter – used to determine the heat energy change during a reaction

Carried out under constant pressure measures enthalpy (ΔH)

Carried out under constant volume measures energy (ΔE)

Example

A 110. g sample of copper (specific heat capacity = 0.20 J/Co∙g) is heated to 82.4oC and then placed in a container of water at 22.3oC. The final temperature of the water and the copper is 24.9oC. What was the mass of the water in the original container, assuming complete transfer of heat from the copper to the water?

Heat lost by copper =

-(heat lost by copper) = (heat gained by water)

J 1265 C)24.9 - C(82.4 Cg

J 0.20Cu g 110. oo

o

waterg 120

CgC2.6J 4.184

J 1265

Ts

q water of mass

o

o

Heating Curve

Heat of Fusion

ΔHfus = enthalpy change that occurs in melting a solid at its melting point

Example: What quantity of heat is needed to melt 1.0 kg of ice at its melting point?

ΔHfus =6.0 kJ/mol

kJ 333 mol 1

kJ 6.0

g 18.0

mol 1

kg 1

1000gkg 00.1

Heat of Vaporization

ΔHvap = the energy needed to vaporize one mole of a liquid at a pressure of 1 atm

Example: What quantity of heat is required to vaporize 130. g of water?

kJ 317 mol 1

kJ 43.9

g 18.0

mol 1 g .130

Example

Substance X has the following properties:

ΔHvap = 20. kJ/mol

ΔHfus = 5.0 kJ/molBoiling point = 75oCMelting point = -15oCSpecific heatSolid = 3.0 J/goCLiquid = 2.5 J/goCGas = 1.0 J/goC

Calculate the energy required to convert 250. g of substance X from a solid at -50oC to a gas at 100oC. Assume that X has a molar mass of 75.00 g/mol.

5 Step Process

1. Heating solid

2. Melting solid

3. Heating liquid

4. Boiling liquid

5. Heating gas

Solution1. q = m x c x ΔT

= 250.g x (3.0 J/goC) x 35oC = 26 kJ

2. mol x ΔHfus = 3.33 mol x 5.0 kJ/mol = 17 kJ

3. q = m x c x ΔT= 250.g x (2.5 J/goC) x 90oC = 56 kJ

mol x ΔHvap = 3.33 mol x 20. kJ/mol =67 kJ

q = m x c x ΔT= 250.g x (1.0 J/goC) x 25oC = 6.2 kJ

172 kJ

Heat of Formation

• ΔHfo

• The change in enthalpy of the formation of one mole of a compound from it elements in their standard states

• ΔHf° 25oC at 1 atm and 1 M

ΔHfo = 0

By definition, the standard heat of formation for elements in their standard states equals zero.

Example: Which of the following will have standard heats of formation equal to zero?

H2(g), Hg(s), CO2(g), H2O(l), Br2(l)

Example

Write the balanced molecular equation representing the ΔHf° for ethanol.

Answer:

2C(s) + 3H2(g) + ½ O2 C2H5OH

Hess’s Law

Based on…

1. State function

2. Enthalpy change is same for a reaction whether the reaction takes place in one or many steps

How to use Hess’s Law

• Manipulate equations to reach the desired reaction

• If the reaction given is reversed, so is ΔH

• If multiplying the equation to balance the coefficients also multiply ΔH by the same number

Example

• Calculate the enthapy for the following reaction:

N2(g) + 2O2(g) 2NO2(g) ΔH° = ??? kJ

N2(g) + O2(g) 2NO(g) ΔH° = +180 kJ

2NO2(g) 2NO(g) + O2(g) ΔH° = +112 kJ

For any reaction…

ΔH°reaction = Σ nΔH°f(products) - Σ mΔH°f(reactants)

Calculate the Standard Enthalpy Change for the Combustion of Methane

1. 2H2(g) + C(s) CH4(g) -75 kJ

2. O2(g) 0 kJ

o C(s) + O2(g) CO2(g) -394 kJ

o H2(g) + ½ O2(g) H2O(l) -286 kJ