Ch. 15: Energy andChemical Change

Sec. 15.3: Thermochemical Equations

Objectives

Write thermochemical equation 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.

ReviewThe change in energy is an important part of chemical reactions so chemists include ΔH as part of the chemical equation.

Thermochemical Equations

A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change.The energy change is usually expressed as the change in enthalpy, ΔH.

Thermochemical Equations

A subscript of ΔH will often give you information about the type of reaction or process taking place.For example, ΔHcomb is the enthalpy change for the complete burning of one mole of a substance or, simply, the enthalpy (heat) of combustion.

Thermochemical Equations

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

ΔHcomb = -2808 kJ/mol

This is the thermochemical equation for the combustion of glucose.The enthalpy of combustion is -2808 kJ/mol.

Thermochemical Equations

Since the enthalpy of combustion (ΔHcomb) of a substance is defined as the enthalpy change for the complete burning of one mole of the substance, this means that 2808 kJ of heat are released for every mole of glucose that is oxidized (or combusts). Two moles, therefore, would release 5616 kJ:

2 mol glucose x -2808 kJ = -5616 kJ 1 mol

Standard EnthalpiesStandard enthalpy changes have the symbol ΔHo. Standard conditions in thermochemistry are 1 atm pressure and 25 0C.

**DO NOT CONFUSE THESE WITH THE

STP CONDITIONS OF THE GAS LAWS.

CalculationsHow much heat is evolved when 54.0 g of glucose (C6H12O6) is burned according to this equation?

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

ΔHcomb = -2808 kJ/mol Since the ΔHcomb given is for 1 mole of

glucose, you must first determine the # of moles you have.

54.0 g x 1 mole = 0.300 moles glucose

180 g

Calculations (cont.)

You can now use the ΔHcomb value as a conversion factor:

0.300 mol glucose x -2808 kJ = -842 kJ

1 mole

Practice Problems

How much heat will be released when 6.44 g of sulfur reacts with O2 according to this equation:

2S + 3O2 2SO3 ΔH0 = -395.7 kJ/mol

How much heat will be released when 11.8 g of iron react with O2 according to the equation:

3Fe + 2O2 Fe3O4 ΔH0 = -373.49 kJ/mol

Changes of StateThe heat required to vaporize one mole of a liquid is called its molar enthalpy (heat) of vaporization (ΔHvap).

The heat required to melt one mole of a solid substance is called its molar enthalpy (heat) of fusion (ΔHfus).

Both phase changes are endothermic & ΔH has a positive value.

Changes of StateThe vaporization of water and the melting of ice can be described by the following equations:H2O(l) → H2O(g) ΔHvap = 40.7 kJ/mol

One mole of water requires 40.7 kJ to vaporize.

H2O(s) → H2O(l) ΔHfus = 6.01 kJ/mol

One mole of ice requires 6.01 kJ to melt.

Changes of StateThe same amounts of energy are released in the reverse processes (condensation and solidification (freezing)) as are absorbed in the processes of vaporization and melting.Therefore, they have the same numerical values but are opposite in sign.

ΔHcond = - ΔHvap = -40.7 kJ/mol

ΔHsolid = - ΔHfus = -6.01 kJ/mol

Practice ProblemsHow much heat is released when 275 g of ammonia gas condenses at its boiling point? (ΔHvap = 23.3 kJ/mol)

If water at 00 C releases 52.9 J as it freezes, what is the mass of the water? (ΔHfus= 6.01 kJ/mol)

How much heat is required to melt 25 g of ice at its MP? (ΔHfus= 6.01 kJ/mol)

Combination Problems

At times, you will need to calculate the amount of heat that is absorbed or released when a temperature change AND a phase change occur in sequence. Recall, the heat involved in a temperature

change is calculated by using: ΔH = mCΔT. NOTE: In this expression, ΔH is found in joules or cal.

Heat involved in a phase change is calculated using dimensional analysis. NOTE: ΔH will be in kJ.

ExampleHow much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) Water cannot freeze at 20.0 0C. It must be

at 0 0C. Therefore, we must first determine how much energy is released when it is cooled from 20.0 0C to 0 0C.

ΔH = mCΔT = (37.5 g)(4.184 J/g 0C)(- 20.0 0C) = - 3138 J = -3.14 kJ

ExampleHow much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol)

-3.14 kJ of heat are released when the sample is cooled to 0 0C.Now we must calculate the energy released when the entire sample freezes. Recall that ΔHsolid= -ΔHfus= -6.01kJ/mol

37.5 g x 1 mole x -6.01 kJ = -12.5 kJ 18 g 1 mole

ExampleHow much heat is released when 37.5 g of water that is at 20.0 0C freezes? (ΔHfus= 6.01 kJ/mol) -3.14 kJ of heat are released when the

sample is cooled to 0 0C. -12.5 kJ of heat are released when the

sample freezes. The total heat released is -3.14 + -12.5

or -15.6 kJ.

Practice ProblemsUse Cw = 4.184 J/g0C; ΔHfus = 6.01

kJ/mol; ΔHvap = 40.7 kJ/mol.

How much heat is needed to melt 8 g of ice at 0 0C to water at 15 0C?How much heat is needed to change 28.0 g of water at 60.0 0C to steam?