# Ch. 16: Energy and Chemical Change - PowerPoint PPT Presentation

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Ch. 16: Energy and Chemical Change. Sec. 16.3: 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. Review.

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Ch. 16: Energy and Chemical Change

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## Ch. 16: Energy andChemical Change

Sec. 16.3: 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.

### Review

• The 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, ΔHcombis the change in enthalpy for a combustion reaction 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

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

• That means, 2808 kJ of heat are released for every mole of glucose that is oxidized (or combusts). Two moles would release 5616 kJ:

2 mol glucose x -2808 kJ = -5616 kJ

1 mol

### Standard Enthalpies

• Standard 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.

### Calculations

• How 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  2SO3ΔH0 = -395.7 kJ/mol

• How much heat will be released when 11.8 g of iron react withO2 according to the equation:

3Fe + 2O2 Fe3O4 ΔH0 = -373.49 kJ/mol

### Changes of State

• The 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 State

• The 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 State

• The 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 Problems

• How 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.

### Example

• How 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

### Example

• How 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

### Example

• How 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 Problems

Use 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?