Thermochemistry

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Thermochemistry - PowerPoint PPT Presentation

Thermochemistry. Part 4: Phase Changes & Enthalpies of Formation. Specific Heat. Specific heat: The amount of heat that must be added to a stated mass of a substance to raise it’s temperature, with no change in state.

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Thermochemistry

Part 4: Phase Changes & Enthalpies of Formation

Specific Heat
• Specific heat: The amount of heat that must be added to a stated mass of a substance to raise it’s temperature, with no change in state.

Example: How much heat is released by 250.0 g of H2O as it cools from 85.0oC to 40.0oC? (Remember, specific heat of water = 4.18 J/goC)

q = mcT

q = (250.0 g)(4.18 J/goC)(40.0-85.0)

q = -47,000 J = -47.0 kJ

LATENT HEAT OF FUSION, Hfus
• Definition: the enthalpy change (energy absorbed) when a compound is converted from a solid to a liquid without a change in temperature.
• “Latent” means hidden; the heat absorbed/released during a phase change does not cause the temperature to change.
• Note:Hfusfor water is 334 J/g
LATENT HEAT OF FUSION, Hfus

A = solid

B = melting (solid + liquid)

C = liquid

D = boiling (liquid + gas)

E = gas

LATENT HEAT OF VAPORIZATION, Hvap
• Definition: the enthalpy change (energy absorbed) when one mole of the compound is converted from a liquid to a gas without a change in temperature.
• Note: for water Hvapis 2260 J/g

LATENT HEAT OF VAPORIZATION, Hvap

A = solid

B = melting (solid + liquid)

C = liquid

D = boiling (liquid + gas)

E = gas

Example 1:How much heat is released by 250.0 g of H2O as it cools from 125.0C to -40.0C?

Five steps…

• Cool the steam m∙csteam∙T
• Condense m(-Hvap)
• Cool the liquid water m∙cwater∙T
• Freeze m(-Hfus)
• Cool the solid ice m∙cice∙T
Example 1:How much heat is released by 250.0 g of H2O as it cools from 125.0C to -40.0C?

When substances change state, they often have different specific heats:

• cice= 2.09 J/goC
• cwater = 4.18 J/goC
• csteam = 2.03 J/goC
Example 1:How much heat is released by 250.0 g of H2O as it cools from 125.0C to -40.0C?

cooling = exothermic → negative heat values

qsteam = mcT = (250.0g)(2.03J/goC)(100.0–125.0) = -12,700 J

qvap = mHvap = (250.0g)(-2260J/g) = -565,000 J

qwater = mcT = (250.0g)(4.18J/goC)(0-100) = -105,000 J

qfus= mHfus= (250.0g)(-334J/g) = -83,500 J

qice= mcT = (250.0g)(2.09J/goC)(-40.0-0) = -20,900 J

qtotal = -787,000J

-787 kJ

Now YOU try a few…

Example 2: How much heat energy is required to bring 135.5 g of water at 55.0oC to it’s boiling point (100.oC) and then vaporize it?
• Water must be heated to it’s boiling point.

q = mcT = (135.5 g)(4.18J/goC)(100-55)

q = 25.5 kJ

2. Water must be vaporized:

q = mHvap= (135.5 g)(2260 J/g)

q = 306 kJ

• Add them together:

q = 25.5 kJ + 306 kJ = 332 kJ

Example 3: How much energy is required to convert 15.0 g of ice at -12.5oC to steam at 123.0oC?
• Heat the ice
• Melt the ice
• Heat the water
• Vaporize the water
• Heat the steam.
Example 3: How much energy is required to convert 15.0 g of ice at -12.5oC to steam at 123.0oC?

qice= mcT = (15.0g)(2.09J/goC)(0.0 - -12.5) = 392 J

qfus = mHfus = (15.0g)(334J/g) = 5,010 J

qwater = mcT = (15.0g)(4.18J/goC)(100-0) = 6,270 J

qvap = mHvap = (15.0g)(2260J/g) = 33,900 J

qsteam = mcT = (15.0g)(2.03J/goC)(123.0-100.0) = 700. J

qtotal = 392 J + 5,010 J + 6,270 J + 33,900 J + 700. J = 46,300J

46.3 kJ

Enthalpies of Formation

enthalpy

standard conditions

change (delta)

formation

Enthalpies of Formation
• usually exothermic
• see table for Hf value
• enthalpy of formation of an element in its stable state = 0
• these can be used to calculate H for a reaction
Standard Enthalpy Change

Standard enthalpy change, H, for a given thermochemical equation is = to the sum of the standard enthalpies of formation of the product – the standard enthalpies of formation of the reactants.

sum of (sigma)

Standard Enthalpy Change
• elements in their standard states can be omitted:

2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s)

ΔHrxn = (ΔHfproducts) - (ΔHfreactants)

ΔHrxn = ΔHfAl2O3 - ΔHfFe2O3

ΔHrxn = (-1676.0 kJ) – (-822.1 kJ)

ΔHrxn = -853.9 kJ

Standard Enthalpy Change
• the coefficient of the products and reactants in the thermochemical equation must be taken into account:

2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s)

ΔHrxn = (ΔHfproducts) - (ΔHfreactants)

ΔHrxn = 2ΔHfAl3+ - 3ΔHfCu2+

ΔHrxn = 2(-531.0 kJ) – 3(64.8 kJ)

ΔHrxn = -1256.4 kJ

Standard Enthalpy Change
• Example: Calculate H for the combustion of one mole of propane:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

ΔHrxn = (ΔHfproducts) - (ΔHfreactants)

ΔHrxn = [3ΔHfCO2+4ΔHfH2O] - ΔHfC3H8

ΔHrxn = [3(-393.5kJ)+4(-285.8kJ)]–(-103.8 kJ)

ΔHrxn = -2219.9 kJ

Standard Enthalpy Change
• Example: The thermochemical equation for the combustion of benzene, C6H6, is:

C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l)

H = -3267.4 kJ

Calculate the standard heat of formation of benzene.

-3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–ΔHfC6H6

-3267.4kJ = -3218.4–ΔHfC6H6

-49.0kJ = –ΔHfC6H6

ΔHfC6H6 = +49.0 kJ

Standard Enthalpy Change
• Example: When hydrochloric acid is added to a solution of sodium carbonate, carbon dioxide gas is formed. The equation for the reaction is:

2H+(aq) + CO32-(aq) CO2(g) + H2O(l)

Calculate H for this thermochemical equation.

ΔH = [(-393.5kJ)+(-285.8kJ)]–[2(0 kJ)+(-677.1kJ)

ΔH = (-679.3kJ)–(-677.1kJ)

ΔH = -2.2 kJ