Section 15.3 Thermochemical Equations

• Write thermochemical equations for chemical reactions and other processes.

• Describe how energy is lost or gained during changes of state.

• Calculate the heat absorbed or released in a chemical reaction.

Section 15-3

Writing Thermochemical Equations• A thermochemical equation is a balanced

chemical equation that includes the physical states of all reactants and products, and energy change.

• The enthalpy (heat) of combustion of a substance is the enthalpy change for the complete burning of one mole of the substance.

• Standard enthalpy changes have the symbol

• ∆H°. The zero superscript tells you that the enthalpy changes were determined with all reactants and products at standard conditions. Standard conditions are 1 atm pressure and 298 K (25°C) and should not be confused with standard temperature and pressure (STP).

Section 15-3

Writing Thermochemical Equations (cont.)

Section 15-3

Changes of State

• Molar enthalpy (heat) of vaporization refers to the heat required to vaporize one mole of a liquid substance.

• Molar enthalpy (heat) of fusion is the amount of heat required to melt one mole of a solid substance.

Section 15-3

Changes of State (cont.)

• The vaporization of water and the melting of ice can be described by the following equations: ( both are endothermic)

• H2O(l) → H2O(g) ∆H vap = 40.7 kJ

• H 2O(s) → H2O(l) ∆ H fus = 6.01 kJ

• The first equation indicates that 40.7 kJ of energy is absorbed when one mole of water is converted to one mole of water vapor.

• The second equation indicates that 6.01 kJ of energy is absorbed when one mole of ice melts to form one mole of liquid water.

• What happens in the reverse processes, when water vapor condenses to liquid water or liquid water freezes to ice? The same amounts of energy are released in these exothermic processes as are absorbed in the endothermic processes of vaporization and melting. Thus, the molar enthalpy (heat) of condensation

• (∆Hcond) and the molar enthalpy of vaporization have the same numerical value but opposite signs. Similarly, the molar enthalpy (heat) of solidification

• (∆H solid ) and the molar enthalpy of fusion have the same numerical value but opposite signs.

• ∆ H vap = -∆ H cond

• ∆ H fus = -∆ H solid

H 2 O(g) → H 2 O(l) ∆H cond = -40.7 kJ

H2O(l) → H 2O(s) ∆H solid = -6.01 kJ

See the graph in the next slide

Section 15-3

Changes of State

Section 15-3

Combustion Reactions

• Combustion is the reaction of a fuel with oxygen.

• Food is the fuel in combustion reactions in biological systems.

A. A

B. B

C. C

D. D

Section 15-3

The amount of energy required to melt one mole of a solid is called ____.

A. molar enthalpy of vaporization

B. molar enthalpy of melting

C. molar enthalpy of fusion

D. molar enthalpy of condensation

A. A

B. B

C. C

D. D

Section 15-3

A thermochemical equation specifies about changes in ____.

A. temperature

B. pressure

C. enthalpy

D. molar mass

• Q1 Write a complete thermochemical equation for the combustion of ethanol

(C2H5OH). ∆H comb = -1367 kJ/mol C2H5OH(l) + 3O2(g) → 2CO2(g)+3H2O(l) ∆Hcomb= -1367

Q2 Determine Which of the following processes are exothermic? Endothermic?

a.C2H5OH(l) → C2H5OH(g)

b.Br 2(l) → Br2(s)

c. C5H12(g) + 8O2(g) → 5CO2(g)+6H2O(l)

d. NH3(g) → NH3(l)

e.NaCl(s) → NaCl(l)

• Q3 Explain how you could calculate the heat released in freezing 0.250 mol water.

• Multiply 0.250 mol times the molar heat of fusion of water, 6.01 kJ/mol.

• Q4 Calculate How much heat is released by the combustion of 206 g of hydrogen gas? ∆ H = -286 kJ/mol

• 29,300 kJ

• Q5 The molar heat of vaporization of ammonia is 23.3 kJ/mol. What is the molar heat of condensation of ammonia?

• -23.3 kJ/mol• Q6 The reaction A → C is shown in the

enthalpy diagram below. Is the reaction exothermic or endothermic? Explain your answer.

• The reaction is exothermic because the product (C) has a lower energy than the reactant (A).

A. A

B. B

C. C

D. D

The standard enthalpy of an element in its standard state is ____.

A. 0.00 calories

B. 0.00°C

C. 0.00 kilocalories

D. 0.00 kJ

End of sec 15.3