PHYSICAL CHEMISTRY

Temperature and Heat1. Thermochemistry is the study of energy changes which take

place during chemical reactions.

2. Temperature is the degree of hotness.

3. Molar heat capacity: energy required to raise the temperature of one mole of the substance by one kelvin or (1oC)

4. Specific heat capacity (c): energy required to raise the temperature of one g of the substance by one kelvin or (1oC)

5. The specific heat capacity for water 4.18 J g-1K-1

6. ∆H = m c ∆T

∆H = heat absorbed or released7. When zinc powder is added to plastic cup which contains 50 cm3 of Cu2+

(aq) solution, the temperature rises by 6.2oC. The specific heat capacity for Cu2+ solution is 4.2J g-1K-1 Calculate the heat liberated during the reaction. [1.302 kJ]

1. Enthalpy Change, ∆H, is given:

The total enthalpy of products - The total enthalpy of reactants

2. Exothermic process: heat given out to surrounding. The value has a negative sign as the heat content of product is less than that of the reactant. Define the endothermic process.

Enthalpy Changes

Energy Level Diagram

1. When one mole of methane is burnt in excess oxygen at a constant pressure, the heat liberated is 890 kJ. The thermochemical equation is:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆ HӨ = -890 kJ mol-1

2. ∆ HӨ is enthalpy change of reaction under standard conditions.

a) The total amount of energy released or absorbed is directly proportional to the number of moles of reactants used.

b) The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to the enthalpy change for the forward reaction.

Thermochemical Equations

Hess's Law if a reaction is carried out in a series of steps, ∆H for the reaction

will be equal to the sum of the enthalpy changes for the individual steps

the overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

Consider the combustion reaction of methane to form CO2 and liquid H2O

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) This reaction can be thought of as occurring in two steps: In the first step methane is combusted to produce water vapor: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) ∆HӨ = -802 kJ

In the second step water vapor condenses from the gas phase to the liquid phase: 2H2O(g) -> 2H2O(l) ∆HӨ = -88 kJ

Hess’s Law

Combining these equations yields the following:CH4(g)+2O2(g)+2H2O(g) CO2(g)+2H2O(g)+2H2O(l) ∆H = (-802) kJ + (-88) kJ= -890 kJ

Determining ∆H for the reaction

is difficult because some CO2 is also typically produced.

However, complete oxidation of either C or CO to yield CO2 is experimentally pretty easy to do:

We can invert reaction number 2 (making it endothermic) and have CO(g) as a product. (This describes the decomposition of CO2 to produce CO and O2)

                                                                                                               

                                                                                                                                                                           

Combining the Equations

C(graphite) + O2(g) CO2(g) ∆H = -393.5 kJC(diamond) + O2(g) CO2(g) ∆H = -395.4 kJ

C(graphite) C(diamond)

C(graphite) + O2(g) CO2(g) ∆H = -393.5 kJCO2(g) C(diamond) + O2(g) ∆H = +395.4 kJ

C(graphite) C(diamond) ∆H = +1.9 kJ

1. The standard heat of formation, or standard enthalpy of formation (represented as ∆H of), for a substance is the enthalpy change in the formation of 1 mole of that substance from its individual elements at 1 atmosphere (1 atm) and 25o C (unless otherwise noted).

2. Elements found in many forms, or allotropes, which have different heats of formation. To standardize this process, the allotrope with the most stability is used in determining heats of formations. The most stable element will always have a heat of formation = 0.

3. The more negative the value, the more energetically stable the compound.

Standard Molar Enthalpy Change of Formation

For example, carbon (C) is found as CO, CO2, C(graphite), etc. However, C(graphite) is the most stable form of C. Therefore, any heat of formation equation using carbon will use C(graphite), which is represented for simplicity, as C(s):

C(s) + O2(g) CO2(g)

C(s) C(diamond) C(s) + 1/4 S8(rhombic) CS2(l)

C(s) + Ca(s) + 3/2 O2(g) CaCO2(s)

Example #1: Which of the following equations represents a reaction that

provides the heat of formation of ethanol (CH3CH2OH)?

A) CH2 = CH2(g) + H2O(l) CH3CH2OH(l)

B) 2C(s) + 6 H(g) + O(g) CH3CH2OH(l)

C) 2 CO2(g) + 6 H2(g) CH3CH2OH(l) + 3H2O(l)

D) 2 C(s) + 3H2(g) + 1/2 O2(g) CH3CH2OH(l)

E) 2 CO(g) + 3 H2(g) CH3CH2OH(l) + 1/2 O2(g)

Equation A: CH2 = CH2(g) is not the most stable form of C, and H2O(l) is not the

most stable form of H or O (C(s), H2(g) and O2(g) are).

Equation B: H(g) is not the most stable form of H (H2(g) is).

Equation C: CO2(g) is not the most stable form of C or O (C(s) and O2(g) are).

Equation D: Correct.

Equation E:

CO is not the most stable form of C (C(s) is).

Example #2: The heat of formation of CO2(g) is -394 kJ/mole and that of

H2O(l) is -286 kJ/mole. The heat of combustion of C5H12 is -3534 kJ/mole. What is the heat of formation of C5H12?

C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l) Calculate the heats of formation and combustion for the individual elements of the reaction: ∆ Ho

c C5H12 = 1 x (-3534) = -3534 kJ/mole

∆  Hof O2 = 8 x 0 = 0 kJ/mole

∆ Hof CO2 = 5 x (-394) = -1970 kJ/mole

∆ Hof H2O = 6 x (-286) = -1716 kJ/mole

  (-1970) + (-1716) - Hof + 0   = -3534    Ho

f   = -152 kJ/mole

Example #3: Calculate the heat evolved (kJ) for the reaction of 2.40 g of CH4.

(Atomic weights: C = 12.01, H = 1.008)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) Ho = -890.3 kJ/mole

0.149 x (-890.3) = -133 kJ/mole

1. The heat of combustion, or enthalpy of combustion (represented as Ho

c), is the amount of energy released in the combustion of one mole of a substance.

The standard enthalpy of formation of propyne, C3H4, is +185.4 kJ/mole. Calculate the heat of combustion of one mole of C3H4. The heats of formation of CO2(g) and H2O(l) are -394 kJ/mole and -285.8 kJ/mole respectively.

1) Write out and balance the equation based on the information given:

C3H4 + 4 O2 3CO2 + 2 H2O ∆Hoc = -1939 kJ/mole

Standard Molar Enthalpy Change of Combustion

Practice:

1. The standard enthalpy of combustion (in kJ/mol) for ethanol and ethanal (CH3CHO) are -1371 and -1167 respectively. Calculate the standard enthalpy change for the oxidation of ethanol to ethanal.

x

CH3CH2OH + ½ O2 CH3CHO + H2O

-1371 -1167

2CO2 + 3H2O

1. 2,000.0 g of water in calorimeter has its temperature raised 3.0°C by an exothermic chemical reaction. How much heat was transferred? [or 25 kJ]

Q = mcΔT note: since the temperature increased, ΔT will be a positive number

= (2000.0 g) × 4.184 J g ·°C ×3.0°C   =25,000 J or 25 kJ (to the correct number of significant digits)

Simple Calorimeter

Bomb Calorimeter

1. A weighed sample in a platinum crucible is placed in the “bomb”.

2. A volume of water sufficient to cover the bomb is added and the entire apparatus is closed.

3. Pure oxygen is pumped and initial temperature recorded.

4. The sample is ignited. The maximum temperature rise is recorded.

5. The heat capacity of calorimeter is the heat required to raise the temperature of the whole apparatus by 1 K.

Practice:1. The combustion of 0.625 g of benzoic acid causes the temperature of the

water in the bomb calorimeter to increase by 1.572 K while the combustion of 0.712 g of an organic compound causes an increase of 1.54 K. Calculate the standard enthalpy change of combustion of the organic compound.

[ ∆HӨC benzoic acid is -3230 kJmol-1. The RMM for benzoic acid and

organic compound is 122 and 60.]

SolutionSolution:

Heat released of burning of acid: 0.625/122 x 3230 = 16.55 kJ

1.572 K 16.55 kJ

1.54 K ? kJ …… (16.2 kJ) released by burning of organic compound.

Standard enthalpy change of combustion of the organic compound:

16.2 kJ/ no. of mol organic compound =

16.2 /(0.712/60) = 1361 kJmol-1

1. Enthalpy change when one mole of acid reacts with one mole of alkali to form one mole of water under standard condition.

2. When strong acid is neutralized by strong base, heat of neutralization is constant at -57.2 kJmol-1.

3. ∆HӨneut for diprotic strong acid > -57.1 kJ/mol due to dilution of acid

when alkali is added.

H2SO4 + NaOH NaHSO4 + H2O ∆HӨneut = -62.0 kJ/mol

NaHSO4 + NaOH Na2SO4 + H2O ∆HӨneut = -71.0 kJ/mol

Average = -66.5 kJ/mol

4. When weak acid or weak alkali is used, further heat change is involved in ionizing the weak acid. Thus, heat of neutralization is less exothermic (less negative) than -57 kJ/mol.

Standard Molar Enthalpy Change of Neutralization

1. Enthalpy change when one mole of gaseous atoms is formed from the element in its standard condition.

This process requires energy (endothermic)

Na (s) Na (g) ∆ HatӨ = +107 kJmol-1

½ Cl2(g) Cl (g) ∆ HatӨ = +242 kJmol-1

Standard Enthalpy Change of Atomization

1. The first electron affinity: enthalpy change when one electron is added to each gaseous atom in one mole to form one mole of x-1..

2. The second electron affinity: enthalpy change when one electron is added to each gaseous anion in one mole to form one mole of x-2.

O (g) + e O- (g) ∆ HEA1Ө = -144 kJmol-1

O- (g) + e O2- (g) ∆ HEA2Ө = + 791 kJmol-1

The second electron affinity is endothermic because energy is required to overcome the repulsive force between the electron and the x-.

Electron Affinity

1. The lattice energy: energy given out when one mole of an ionic crystalline solid formed from its ions.

K+ (g) + Cl- (g) KCl (s) ∆ HlatticeӨ = -701 kJmol-1

KCl (s) K+ (g) + Cl- (g) ∆ Hlattice dissociationӨ = +701 kJmol-1

2. Factors that affecting the magnitude of lattice energy:

a) The attractive forces between the positive and negative ions

b) The distance between the ions/ ionic radius.

3. a) The larger the charge of the ions, the more exothermic is the lattice energy

b) The smaller the ionic radius, the more exothermic is the lattice energy

Energy changes in Forming Ionic Substances

1. Consider the reaction between Li and fluorine gas to form lithium fluoride, LiF. The reaction can take place in several steps:

Atomization enthalpy of metal (in this case lithium) Ionization enthalpy of metal Atomization enthalpy of non-metal (in this case fluorine) Electron affinity of non-metal Lattice enthalpy

a) Li (s) Li (g) ……. atomization ∆Hat = +107 kJ/mol

b) Li (g) Li+ (g) …… first ionization ∆HIE = +496 kJ/mol

c) ½ F2 (g) F (g) …… atomization ∆Hat = +122 kJ/mol

d) F + e F- (g) …….. Electron affinity ∆HEA = -349 kJ/mol

e) Li+(g) + F- (g) LiF(s) .. Lattice energy ∆HLattice = ??? kJ/mol

f) Li(s) + ½ F2 (g) LiF (s) … enthalpy formation = -411 kJ/mol

Born-Haber Cycle

a) Li (s) Li (g) ……. atomization ∆Hat = +107 kJ/mol

b) Li (g) Li+ (g) …… first ionization ∆HIE = +496 kJ/mol

c) ½ F2 (g) F (g) …… atomization ∆Hat = +122 kJ/mol

d) F + e F- (g) …….. Electron affinity ∆HEA = -349 kJ/mol

e) Li+(g) + F- (g) LiF(s) .. Lattice energy ∆HLattice = ??? kJ/mol

f) Li(s) + ½ F2 (g) LiF (s) … enthalpy formation = -411 kJ/mol

Lattice Energy = -411 –(107 +496 +122 -349)

= -787 kJ mol-1

Practice:

1. Construct a Born-Haber cycle for the formation of Na2O and use the data below to calculate the lattice energy of sodium oxide:

Na(s) Na(g) ……………………∆HӨ = + 108 kJ mol-1

Na(g) Na+ (g) ……………………∆HӨ = + 496 kJ mol-1

Na+(g) Na2+ (g) ……………………∆HӨ = + 4560 kJ mol-1

½ O2(g) O (g) ……………………∆HӨ = + 249 kJ mol-1

O(g) + e O- (g) ……………………∆HӨ = - 141 kJ mol-1

O-(g) + e O2-(g) ……………………∆HӨ = + 798 kJ mol-1

2Na(s) + ½ O2(g) Na2O ……………………∆HӨ = -602 kJ mol-1

[-2716]

1. The enthalpy change of hydration energy ∆Hhyd, is the enthalpy change that takes place when one mole of a gaseous ion dissolve in water to give an infinitely dilute solution.

2. The enthalpy change of hydration energy ∆Hhyd is always negative.

3. a) The smaller the ionic radius, the more exothermic the hydration energy.

b) The higher the charge of the ion, the more exothermic the hydration energy.

Hydration Energy

HYDRATED IONS

1. When an ionic compound dissolves in water, two enthalpy are involved.

a) The ions must be separated into gaseous ions. This requires lattice dissociation energy, an endothermic process.

b) The gaseous ions interacts with water molecules. Hydration energy (exothermic) is involved.

Heat of soultion

NaCl (s) (in water) Na+ (aq) + Cl- (aq)

lattice dissociation energy hydration energy

Na+(g) + Cl-(g)

Heat of solutions

1. Decreases down the group:

2. Energy

Solubility of Group 2 Sulphates

BeSO4 MgSO4 CaSO4 SrSO4 BaSO4

Solubility 0.34 0.18 0.0047 0.0071 0.00009

Lattice energy

Hydration energy

The hydration enthalpies of the cation decreases rapidly in magnitude than the lattice energy. Heat of solution become more positive. Solublity decerases