Ch 16 energy and chemical change
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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 and chemical change

Ch. 16: Energy andChemical Change

Sec. 16.3: Thermochemical Equations


Objectives

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

Review

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


Thermochemical equations

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 equations1

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 equations2

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 equations3

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 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

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

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

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

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 state1

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 state2

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 problems1

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

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

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


Example1

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


Example2

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 problems2

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?


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