Thermochemistry

The study of the heat flow of a chemical reaction or

physical change

1st Law of Thermodynamics

Energy cannot be created or destroyed but only converted from one form to another

The energy of the universe is constant

State Function

A state function is independent of the pathway (does not rely on the past or future of the system

A state function depends only on the present state of the system

Ex - Elevation vs distance

EnergyThe capacity to do work or provide heat

Potential Energy (PE) = mhv

Kinetic Energy (KE) = 1/2mv2

The total energy of the system is equal to the sum of the potential and kinetic energy

Can be changed by a flow of heat/work

ΔE = q + w

Chemical Energy

Energy will flow either from the system to its surroundings (exothermic) or from the surroundings to the system (endothermic)

The system will be the reactants and products

ΔE < 0 energy flows out of the systemΔE > 0 energy flows into the system

From the system’s point of view…

q +Heat flows

into the system

q -Heat flows out of the system

w + Work done on the system

w - Work done by the system

Internal Energy

Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system

ΔE = q + wΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ

Work vs Energy flow

Work Done by or to Gases

h x F distance x force Work AF

P

A x P For F/A P since

hA x x P A x F Work hA x volumeinitial - volumefinal V

VP Work :onsubstitutiby out) flows (works VP- w

PV Work

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm

w = -PΔVΔV = 64 L – 46 L = 18 L

w = -(15 atm x 18 L)w = -270 L∙atm

Work is done by the system so w is negative

Internal Energy, Heat & Work

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. Use 1 L∙atm = 101.3 J

ΔE = q + w

q = +1.3 x 108 J

w = -PΔV

ΔV = 5.0 x 105 L

w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm

ΔE = (+1.3 x 108 J) + (-5.1 x 107 J)

ΔE = 8 x 107 J

J 10 x 5.1- atm L 1

J 101.3 x atm L 10 x 0.5 75

w

Enthalpy (H)

H = E + PV

E = internal energy

P = pressure of the system

V = volume of the system

Enthalpy is a state function

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

Enthalpy

How much heat is released when 4.03 g of hydrogen is reacted with excess oxygen?

2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ

kJ 572- H mol 2

kJ 572-

g 2.02

H mol 1 H g 03.4

2

22

Enthalpy #2

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.

qp = ΔH = -890 kJ/mole CH4

44

44 CH mol 0.36

CH g 16.0

CH mole 1 CH g 8.5

kJ 320- CH mol

kJ 890-CH mol 36.0

44

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)

Heat Capacity

Specific heat capacity – the amount of heat needed to raise one gram on a substance by 1oC (J/Co∙g or J/K∙g)

Molar heat capacity – the amount of heat needed to raise one mole of a substance by 1oC (J/Co∙mol or J/K∙mol)

in temp increase

absorbedheat C

Heat Gained or Lost Depends on…

1. The change in temperature during the reaction

2. The amount of substance present

3. The heat capacity of the substance

Tcm q

Calorimetry

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

Calorimetry #2When 1.00 L of 1.00 M Ba(NO3)2 solution at

25.0oC is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC. Assuming that the calorimeter absorbs only a negligible amount of heat, that the specific heat capacity of the solution is 4.18 J/oC∙g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

Solution (no pun intended)

Species present: Ba2+, NO3-, Na+, and

SO42-

Spectator ions: Na+ and NO3-

Net Ionic equation: Ba2+(aq) + SO4

2-(aq) BaSO4(s)

Heat evolved by reaction = heat absorbed by the solution= m x c x ΔT

Since 1.00 L of each solution is used, total volume of the mixture is 2.00 L (2.0 x 103 g)

Temperature increase =28.1oC – 25.0oC = 3.1oC

q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC) 2.6 x 104 J = ΔH

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

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

Solution

N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ

2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ

+ 68 kJ

The second equation has been reversed including ΔH

Calculate ΔH° for this reaction:

2N2(g) + 5O2(g) ---> 2N2O5(g)

H2(g) + 1/2O2(g) H2O(l) ΔH° = -285.8 kJ

N2O5(g) + H2O(l) 2HNO3(l) ΔH° = -76.6 kJ

1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l) ΔH° = -174.1 kJ

Solution

Standard Enthalpies of Formation

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

ΔH° 25oC at 1 atm and 1 M

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) + 1/2O2(g) CH5OH(l)

For any reaction…

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

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

2. O2(g) 0 kJ

3. C(s) + O2(g) CO2(g) -394 kJ

4. H2(g) + ½ O2(g) H2O(l) -286 kJ

Example #1

• Calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water

• Use table 6.2 on page on Page 262

Heating Curves

Heat of Fusion & Heat of Vaporization

Heating Curves

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?

For liquid water ΔHvap = 43.9 kJ/mol

kJ 317 mol 1

kJ 43.9

g 18.0

mol 1 g .130

Example #2

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

Example #2 continued…

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

Example #2 continued…

1. 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 kJ3. q = m x c x ΔT

= 250.g x (2.5 J/goC) x 90oC = 56 kJ

4. mol x ΔHvap = 3.33 mol x 20. kJ/mol =67 kJ5. q = m x c x ΔT

= 250.g x (1.0 J/goC) x 25oC = 6.2 kJ172 kJ