Lecture 30 4/18/07

 

Solid/Liquid

Heat of fusion Solid Liquid Endothermic ice Water (333 J/g or 6 KJ/mol)

Heat of crystallization Liquid Solid Exothermic Water ice (- 333 J/g or - 6 KJ/mol)

Liquid/Gas

Heat of vaporization Liquid Gas Endothermic water water vapor (40.7 KJ/mol)

Heat of condensation Gas Liquid Exothermic vapor Water (- 40.7 KJ/mol)

Solid/Gas

Heat of sublimation Solid Gas Endothermic

Heat of deposition Gas Solid Exothermic

What is the minimum amount of ice at 0 °C that must be added to 340 mL of water to cool it from 20.5°C to 0°C?

A rainstorm deposits 2.5 x 1010 kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms.

(∆Hcond = - 40.7 KJ/mol)

Exothermic or endothermic?

1st Law of Thermodynamics revisited

∆E = q + w

Change in Energy content

heat

work

work

work (w) = - F x d w = - (P x A) x d w = - P∆V

if ∆V = 0, then no work

State function

property of a system whose value depends on the final and initial states, but not the path driving to Mt Washington

route taken vs. altitude change

∆E is a state function q and w are not

Change in Enthalpy (∆H or qp)

equals the heat gained or lost at constant pressure

∆E = qp + w ∆E = ∆H + (-P∆V) ∆H = ∆E + P∆V

∆E vs. ∆H

Reactions that don’t involve gases 2KOH (aq) + H2SO4 (aq) K2SO4 (aq) + 2H2O (l) ∆V ≈ 0, so ∆E ≈ ∆H

Reactions in which the moles of gas does not change N2 (g) + O2 (g) 2NO (g) ∆V = 0, so ∆E = ∆H

Reactions in which the moles of gas does change 2H2 (g) + O2 (g) 2H2O (g) ∆V > 0, but often P∆V << ∆H, thus ∆E ≈ ∆H

Enthalpy is an extensive property Magnitude is proportional to amount of reactants

consumed H2 (g) + ½ O2 (g) H2O (g) ∆H = -241.8 KJ

2H2 (g) + O2 (g) 2H2O (g) ∆H = -483.6 KJ

Enthalpy change for a reaction is equal in magnitude (but opposite in sign) for a reverse reaction H2 (g) + ½ O2 (g) H2O (g) ∆H = -241.8 KJ

H2O (g) H2 (g) + ½ O2 (g) ∆H = 241.8 KJ

Enthalpy change for a reaction depends on the state of reactants and products H2O (l) H2O (g) ∆H = 88 KJ

Constant pressure calorimetry(coffee cup calorimetry)

heat lost = heat gained

Measure change in temperature of water

Constant pressure calorimetry(coffee cup calorimetry)

heat lost = heat gained

Measure change in temperature of water

10 g of Cu at 188 °C is added to 150 mL of water in a coffee cup calorimeter and the temperature of water changes from 25 °C to 26 °C. Determine the specific heat capacity of copper.

Bomb calorimetry

Mainly for combustion experiments ∆V = 0 qrxn + qbomb + qwater = 0

combustion chamber

Bomb calorimetry

Mainly for combustion experiments ∆V = 0 qrxn + qbomb + qwater = 0

Often combine qbomb + qwater into 1 calorimeter term with qcal = Ccal∆T

combustion chamber

Bomb calorimeter math

qrxn + qbomb + qwater = 0

qrxn + Cbomb∆T + Cwatermwater∆T = 0

In the lab: qrxn + qcalorimeter = 0

qcalorimeter = qbomb + qwater

qrxn + Ccalorimeter∆T = 0

empirically determined

same value

On the exam

Bond enthalpies

Enthalpies of formation

Hess’ Law

Example

A hot plate is used to heat two 50-mL beakers at the same constant rate. One beaker contains 20.0 grams of graphite (C=0.79 J/g-K) and one contains 10 grams of ethanol (2.46 J/g-K).

Which has a higher temperature after 3 minutes of heating?

Standard heat of reaction (∆H°rxn)

Same standard conditions as before: 1 atm for gas 1 M for aqueous solutions 298 K For pure substance – usually the most stable form of

the substance at those conditions

Standard heat of formation (∆H°f)

Enthalpy change for the formation of a substance from its elements at standard state

Na(s) + ½ Cl2 (g) NaCl (s) ∆H°f = -411.1 kJ

Three points An element in its standard state has a ∆H°f = 0

∆H°f = 0 for Na(s), but ∆H°f = 107.8 KJ/mol for Na(g)

Most compounds have a negative ∆H°f formation reaction is not necessarily the one done in lab

Using ∆H°f to get ∆H°rxn

2 ways to look at the problem

Calculate ∆H°rxn for:

C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)

Given:

3 C(s) + 4 H2 (g) C3H8 (g) ∆H°f = -103.85 KJ/mol

C(s) + O2 (g) CO2 (g) ∆H°f = -393.5 KJ/mol

O2 (g) + 2 H2 (g) 2H2O (l) ∆H°f = -285.8 KJ/mol

Using Hess’s Law and ∆H°f to get ∆H°rxn 1st way: Hess’s Law

C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)∆H°rxn = ∆H1 + ∆H2 + ∆H3

Reverse 1st equation:

C3H8 (g) 3 C(s) + 4 H2 (g) ∆H1 = - ∆H°f = 103.85 KJMultiply 2nd equation by 3:3C(s) + 3O2 (g) 3CO2 (g) ∆H2 = 3x∆H°f = -1180.5 KJMultiply 3rd equation by 2:2O2 (g) + 4 H2 (g) 4H2O (l) ∆H2 = 2x∆H°f = -571.6 KJ

∆H°rxn = (103.85 KJ) + (-1180.5 KJ) + (-571.6 KJ)

∆H°rxn = -1648.25 KJ

Using ∆H°f to get ∆H°rxn2nd way

C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)

∆H°rxn = Σn ∆H°f (products) - Σn ∆H°f (reactants)

∆H°rxn = [3x(-393.5 KJ/mol) + 2x(-285.8 KJ/mol)] – [(-103.85 KJ/mol) + 0]

∆H°rxn = [-1752.1 KJ] – [-103.85 KJ]

∆H°rxn = -1648.25 KJ

Spontaneity

Some thought that ∆H could predict spontaneity

Sounds great BUT . . . . .

Spontaneity

Some thought that ∆H could predict spontaneity Exothermic – spontaneous Endothermic – non-spontaneous

Sounds great BUT some things spontaneous at ∆H > 0 Melting Dissolution Expansion of a gas into a vacuum Heat transfer

Clearly enthalpy not the whole story

Entropy (Measurement of disorder)

Related to number of microstates

∆Suniverse = ∆Ssystem + ∆Ssurroundings

2nd Law of Thermodynamics Entropy of the universe increases with spontaneous

reactions

Reversible reactions vs. Irreversible reaction

Standard heat of formation (∆H°f)

Enthalpy change for the formation of a substance from its elements at standard state

Na(s) + ½ Cl2 (g) NaCl (s) ∆H°f = -411.1 kJ

Key points

Entropy(Measurement of disorder)

Related to number of microstates S = klnW

∆Suniverse = ∆Ssystem + ∆Ssurroundings

2nd Law of Thermodynamics Entropy of the universe increases with spontaneous reactions

Reversible reactions ∆Suniverse = ∆Ssystem + ∆Ssurroundings = 0 Can be restored to the original state by exactly reversing the

change Each step is at equilibrium

Irreversible reaction ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 Original state can not be restored by reversing path spontaneous

3rd Law of thermodynamicsS = O at O K

S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions

3rd Law of thermodynamicsS = O at O K

T

qS rev

S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions

∆S° = ΣS°(products) - ΣS°(reactants)

Example

Sulfur (2.56 g) was burned in a bomb calorimeter with excess O2. The temperature increased from 21.25 ºC to 26.72 ºC. The bomb had a heat capacity of 923 J/ºC and the calorimeter contained 815 g of water. Calculate the heat evolved per mole of SO2 formed.

S(s) + O2 (g) SO2 (g)

Standard heat of reaction (∆H°rxn)

Same standard conditions as before:

Using ∆H°f to get ∆H°rxn

2 ways to look at the problem

Calculate ∆H°rxn for:

C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)

Given:

3 C(s) + 4 H2 (g) C3H8 (g) ∆H°f = -103.85 KJ/mol

C(s) + O2 (g) CO2 (g) ∆H°f = -393.5 KJ/mol

O2 (g) + 2 H2 (g) 2H2O (l) ∆H°f = -285.8 KJ/mol

Degrees of freedom

translational motion molecules in gas > liquid > solid

vibrational motion movement of a atom inside a molecule

rotational motion rotation of a molecule

Entropy trends

Entropy increases: with more complex molecules with dissolution of pure gases/liquids/solids with increasing temperature with increasing volume with increasing # moles of gases

Which has higher entropy?

dry ice or CO2

liquid water at 25°C or liquid water at 50°C

pure Al2O3(s) or Al2O3 with some Al2+ replaced with Cr3+

1 mole of N2 at 1 atm or 1 mol of N2 at 10 atm

CH3CH2CH2CH3 (g) or CH3CH3 (g)

Is the reaction spontaneous?

Gibbs Free Energy (∆G)

∆G° = ∆H° - T∆S°

∆G = ∆H - T∆S

∆G° = Σn∆Gf°(products) - Σn∆Gf°(reactants)

Gibbs Free Energy ∆G = ∆H - T∆S

∆H ∆S -T∆S ∆G spontaneous

? example

- + 2O3 (g) 3O2 (g)

+ - 3O2(g) 2O3 (g)

- - H2O (l) H2O (s)

+ + H2O (s) H2O (l)

Gibbs Free Energy (∆G) and equilibrium

QlnRTGG

R = 8.314 J/mol-K

Example

A hot plate is used to heat two 50-mL beakers at the same constant rate. One beaker contains 20.0 grams of graphite (C=0.79 J/g-K) and one contains 10 grams of ethanol (2.46 J/g-K).

Which has a higher temperature after 3 minutes of heating?

Example

59.8 J are required to change the temperature of 25.0 g of ethylene glycol by 1 K.

What is the specific heat capacity of ethylene glycol?