Unit 12: Thermodynamics

Aim: Calculating the Hf of compounds and writing formation equations

Unit 12: Thermodynamics

Aim: Calculating the Hf of compounds and writing formation equations

Energy & ChemistryEnergy & Chemistry

ENERGY is the capacity to do work or transfer heat.

HEAT is the form of energy that flows between 2 objects because of their difference in temperature.

Energy & ChemistryEnergy & Chemistry

All of thermodynamics depends on the

law of CONSERVATION OF ENERGY.

• The total energy is unchanged in a

chemical reaction.

Heat Transfer Review

• exothermic: heat transfers from SYSTEM to SURROUNDINGS.– Tempsys decreases– Tempsurr increases

• endothermic: heat transfers from SURROUNDINGS to SYSTEM.– Tempsys increases– Tempsurr decreases

Energy & Chemistry

• Burning peanuts supply sufficient energy to boil a cup of water.

• Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

• These reactions (exothermic) contain products that are favored and more stable.

• They proceed almost completely from reactants to products with little assistance.

Energy & Chemistry

https://www.youtube.com/watch?v=z7ZuBjPRFUk

Chemical Reactivity• Energy changes allows us to predict reactivity.

– In general, exothermic reactions are product favored because PEproducts<Pereactants

» Lower energy= more stability

Exothermic is favorable; ΔH-; products are more stableEndothermic is unfavorable; ΔH+; reactants are more stable

ENTHALPY (H)ENTHALPY (H)

∆H (change in heat content) = Hproducts - Hreactants

• Enthalpy is a STATE FUNCTION.

• These depend only on the state of the system and not on how the system got there.

• Absolute values of H cannot be measured due to energy being stored within bonds. Only changes of H can be measured.

Standard Enthalpy ValuesStandard Enthalpy ValuesMost ∆H values are labeled ∆Ho

Measured under standard conditions

P= 1 atm

T = usually 25 oC

with all species in standard states (most stable forms)

e.g., C = graphite and O2 = gas

To calculate the theoretical ∆H of a reaction one most know

the heat of formation; ∆Hf for all compounds involved.

Standard Enthalpy Values

∆Hfo = standard molar enthalpy of formation

Heat of formation: the enthalpy change when 1 mol

of compound is formed from elements under

standard conditions.(See Given Table)

∆Hfo = 0 for all elements in their standard states.

∆Hfo, standard molar enthalpy of formation

2 H2(g) + O2(g) --> 2 H2O(g) ∆H= -483.6 kJ

Adjust the equation to yield only 1 mole of H2O.

• ∆Hf° is for 1 mole of a compound, therefore a

balanced equation needs to have the coefficients

reduced to yield 1 mole of the compound.

• The heat of formation must be reduced by the

same amount.

∆Hfo (H2O, g)=

-241.8 kJ/mol

∆Hfo, standard molar

enthalpy of formationPractice: Write the equation for the formation of

ammonia (NH3) gas from nitrogen and hydrogen

gases.

N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH= -91.8 kJ

∆Hfo, standard molar

enthalpy of formationPractice: Write the equation for the standard formation of

ammonium chloride liquid from ammonia and hydrogen

chloride gas.

NH3 (g) + HCl (g) NH4Cl (l) Find the Hf in the Table

and insert into the equation.

Determining the Heat of Reaction (Enthalpy change)

Aim: To calculate the enthalpy change

(ΔH) using heats of formation ΔHf.

Calculating the Heat of ReactionUsing Heats of Formation

Calculating the Heat of ReactionUsing Heats of Formation

In general, when ALL enthalpies

of formation are known:

Calculate ∆H of

reaction?

∆Horxn = Σ n ∆Hf

o (products) - Σ n ∆Hfo (reactants)∆Ho

rxn = Σ n ∆Hfo (products) - Σ n ∆Hf

o (reactants)

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

Use ∆Hf°’s to calculate the total enthalpy change (∆Ho

rxn) for the following reaction:

H2O(g) + C(graphite) --> H2(g) + CO(g)

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

To convert 1 mol of water to 1 mol each of

H2 and CO requires 131 kJ of energy.

The “water gas” reaction is endothermic.

H2O(g) + C(graphite) --> H2(g) + CO(g)

∆Ho = +131 kJ

H2O(g) + C(graphite) --> H2(g) + CO(g)

∆Ho = +131 kJ

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

Calculate the heat of combustion of

methanol, i.e., ∆Horxn for

2 CH3OH(l) + 3 O2(g) --> 2 CO2(g) + 4 H2O(g)

∆Horxn = Σ n ∆Hf

o (prod) - Σ n ∆Hfo (react)

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

∆Horxn = [2 ∆Hf

o (CO2) + 4 ∆Hfo (H2O)

- [3 ∆Hfo (O2) + 2 ∆Hf

o (CH3OH)}

= [2(-394 kJ) + 2 (-242 kJ)]

- [ 3(0) + 2(-239 kJ)}

∆Horxn = -1278 kJ

2 CH3OH(l) + 3 O2(g) --> 2 CO2(g) + 4 H2O(g)

∆Horxn = Σ n∆Hf

o (prod) - Σ n ∆Hfo (react)

What is the heat of formation for methane gas given if the

heat of formation of H2O and CO2 are -242 kJ/mol and -394

kJ/mol.

CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g) ∆H=- 802.3 kJ

∆Horxn=[2 ∆Ho

f (H2O) + ∆Hof (CO2)] – [∆Ho

f (CH4) + 2 ∆Hof(O2)]

-802.3 = [2 (-242) + (-394)] – [∆Hof (CH4) + 2 (0)]

-802.3= [-484 – 394] - ∆Hof (CH4)

-802.3= -878 - ∆Hof (CH4) ; ∆Ho

f (CH4) = -75.7 kJ

∆Horxn = Σ n∆Hf

o (prod) - Σ n ∆Hfo (react)

Solving for the Heat of Formation using the Heat of Reaction