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Ch. 16: Energy and Chemical ChangePowerPoint Presentation

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 Î”H 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, Î”H.

Thermochemical Equations

- A subscript of Î”H will often give you information about the type of reaction or process taking place.
- For example, Î”Hcombis the change in enthalpy for a combustion reaction or, simply, the enthalpy (heat) of combustion.

Thermochemical Equations

C6H12O6(s) + 6O2(g)â†’ 6CO2(g) + 6H2O(l)

Î”Hcomb = -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 (Î”Hcomb) 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 Î”Ho.
- 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)â†’ 6CO2(g) + 6H2O(l)

Î”Hcomb = -2808 kJ/mol

- Since the Î”Hcomb 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

- Since the Î”Hcomb given is for 1 mole of glucose, you must first determine the # of moles you have.

Calculations (cont.)

- You can now use the Î”Hcomb 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 ïƒ 2SO3Î”H0 = -395.7 kJ/mol

- How much heat will be released when 11.8 g of iron react withO2 according to the equation:
3Fe + 2O2ïƒ Fe3O4 Î”H0 = -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 (Î”Hvap).
- The heat required to melt one mole of a solid substance is called its molar enthalpy (heat) of fusion (Î”Hfus).
- Both phase changes are endothermic & Î”H 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) â†’ H2O(g)Î”Hvap = 40.7 kJ/mol
- One mole of water requires 40.7 kJ to vaporize.

- H2O(s) â†’ H2O(l)Î”Hfus = 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.
Î”Hcond = - Î”Hvap = -40.7 kJ/mol

Î”Hsolid = - Î”Hfus = -6.01 kJ/mol

Practice Problems

- How much heat is released when 275 g of ammonia gas condenses at its boiling point? (Î”Hvap = 23.3 kJ/mol)
- If water at 00 C releases 52.9 J as it freezes, what is the mass of the water? (Î”Hfus= 6.01 kJ/mol)
- How much heat is required to melt 25 g of ice at its MP? (Î”Hfus= 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: Î”H = mCÎ”T. NOTE: In this expression, Î”H is found in joules or cal.
- Heat involved in a phase change is calculated using dimensional analysis. NOTE: Î”H will be in kJ.

Example

- How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (Î”Hfus= 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.
Î”H = mCÎ”T = (37.5 g)(4.184 J/g 0C)(- 20.0 0C)

= - 3138 J = -3.14 kJ

- 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.

Example

- How much heat is released when 37.5 g of water that is at 20.0 0C freezes? (Î”Hfus= 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 Î”Hsolid= -Î”Hfus= -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? (Î”Hfus= 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; Î”Hfus = 6.01 kJ/mol; Î”Hvap = 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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