Thermochemistry

Chapter 5

Heat - the transfer of thermal energy between two bodies that are at different temperatures

Energy Changes in Chemical Reactions

Temperature - a measure of the thermal energy

90 °C40 °C

greater thermal energy

Temperature = Thermal Energy

greater temperature

(intensive) (extensive)

Exothermic process - gives off heat – transfers thermal energy from the system to the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

Endothermic process - heat has to be supplied to the system from the surroundings.

energy + CaCO3 (s) CaO (s) + CO2 (g)

energy + H2O (s) H2O (l)

Esystem + Esurroundings = 0

or

Esystem = −Esurroundings

C3H8 + 5O2 3CO2 + 4H2O

Exothermic chemical reaction!

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.

Internal energy

of system

ExothermicExothermic

Energy relationshipsEnergy relationshipsFig 5.4 (a)

EndothermicEndothermic

Energy relationshipsEnergy relationshipsFig 5.4 (b)

Fig 5.5

Energy Diagram for the Interconversionof H2 (g), O2 (g), and H2O (l)

Another form of the first law for Esystem

E = q + w

E is the change in internal energy of a system

q is the heat exchange between the system and the surroundings

w is the work done on (or by) the system

Table 5.1 pg 171

Sign Conventions for Heat and Work

Fig 5.6

Thermodynamics

State functions - properties that are determined by the state of the system, regardless of how that condition was achieved.

energy, pressure, volume, temperature

E = Efinal - Einitial

Fig 5.8

Work Done By the System on the Surroundings

V = Vf – Vi > 0

-PV < 0

wsys < 0

w = wfinal - winitialinitial final

Work is not astate function!

Fig 5.12

Since the opposing

pressure, P can vary:

ViVf

Expansion of a Gas:

H = E + PV

H = (q+w) − w

H = qp

• So, at constant pressure, the change in enthalpy is the heat gained or lost.

H = E + PV

Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure.

Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure.

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants Hproducts > Hreactants

Enthalpies of Reaction

H2O (s) H2O (l) H = 6.01 kJ

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 0°C and 1 atm.

H = H (products) – H (reactants) = qp

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 25°C and 1 atm.

Enthalpies of Reaction

• Enthalpy is an extensive physical property.

H for a reaction in the forward direction is equal in magnitude, but opposite in sign, to H for the reverse reaction.

2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ

H2O (s) H2O (l) H = 6.01 kJ

H2O (l) H2O (s) H = -6.01 kJ

The Truth about Enthalpy

• H for a reaction depends on the state of the products and the state of the reactants.

The Truth about Enthalpy

H2O (s) H2O (l) H = 6.01 kJ

H2O (l) H2O (g) H = 44.0 kJ

How much heat is evolved when 155 g of iron undergoes complete oxidation in air?

4Fe (s) + 3O2 (g) 2Fe2O3 (s) H = -1118.4 kJ

155 g Fe1 mol Fe

55.85 g Fex = -776 kJ

-1118.4 kJ 4 mol Fe

x

The specific heat (s) - the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius:

Heat (q) absorbed or released:

q = m·Cs·T

where T = Tfinal - Tinitial

Table 5.2

Tmq

Cs

How much heat is given off when an 869 g iron bar cools from 94.0°C to 5.0°C?

Cs of Fe = 0.45 J/(g • °C)

T = Tfinal – Tinitial = 5.0 °C – 94.0 °C = ‒ 89.0 °C

q = m·Cs·T= (869 g) · (0.45 J/(g • °C)) · (– 89.0 °C)

= ‒ 34,800 J

≈ ‒ 3.5 x 104 J

= ‒ 35 kJ

Calorimetry: Measurement of Heat Changes

• H cannot be determined absolutely, however,ΔH can be measured experimentally

• Device used is called a calorimeter

• Typically, ΔT ∝ ΔHrxn

Constant-Pressure Calorimetry

No heat enters or leaves!

qsolution = m·Cs·T = ‒ qrxn

Because reaction at constant P:

H = qrxn

Figure 5.17

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Standard enthalpy of reaction ( ) - the enthalpy of a reaction carried out at 1 atm.

rxnH

nHo (products) f= mHo (reactants) f -rxnH

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Fig 5.20 An enthalpy diagram comparing aone-step and a two-step process for a reaction.

Must I measure the enthalpy change for every reaction of interest?

The standard enthalpy of formation of any element in its most stable form is zero:

Ho (O2) ≡ 0fHo (O3) = 142 kJ/molf

Ho (C, graphite) ≡ 0fHo (C, diamond) = 1.90 kJ/molf

Establish an arbitrary scale with the standard enthalpy of formation ( ) as a reference point for all enthalpy expressions.

fH

NO!!

Standard enthalpy of formation ( ) - the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

fH

Table 5.3 Standard Enthalpies of Formation, at 298 K

rxnH

Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2. How much heat is released per mole of calcium carbide reacted? The standard enthalpy of formation of CaC2 is -62.76 kJ/mol.

CaC2 (l) + 2H2O (l) C2H2 (g) + Ca(OH)2 (s)

Ho

rxn nHo (products)f= mHo (reactants)f‒

f=Ho

rxn [Ho (C2H2) + Ho (Ca(OH)2] –[Hof(CaC2) +Ho

f (H2O)]

f

Ho

rxn= [(226.77 kJ) + (–986.2 kJ) ] – [(-62.67 kJ) + 2(-285.83 ] = -125.1 kJ

-125.1 kJ1 mol CaC2

= -125.1 kJ/mol CaC2