Thermochemistry Chapter 8

Thermochemistry

First law of thermochemistry: Internal energy of an isolated system is constant; energy cannot be created or destroyed; however, energy can be converted to different forms such as potential energy, electrical energy, heat, work, light …

The total energy of a process = sum of all the different forms of energy produced. This is equal but opposite in sign to the change in potential energy of the system prior to the process

A state function is a property whose value is independent of themanner in which it is determined

State function: the straight line distance from Castroville CA to Boston MA

E = internal energy

The change in internal energy of a system depends on the heat, q, given off and any work, w, that the system does. (conservation of energy)

ΔE = q + w

Many processes that we perform are processes that occur in an open vessel

Chemical reactions are sometimes used to do work. For example a car battery turns the crankshaft of the car in order to get it started.

Consider a piston in a car engine just before a spark ignited the gasoline air mixture. The fuel + air have some internal energy. After the spark ignites the fuel, products, CO2 and H2O, also have some internal energy. In this case, the internal energy of the CO2 and H2O is less than the original fuel.

According to the first law, since energy can not be created or destroyed, the difference in energy between the fuel + air before and the CO2 and H2O after reaction, must be equal to the heat given off, q, and any work done, w. The work that is done is to cause the volume of the piston to expand. If we assume that the piston offers a constant resistance against expansion, we can relate this resistance to the pressure that must be exerted by the hot gases of combustion against the piston wall.

work = force *distance Pressure = force/unit area

If the piston moves some distance d, the volume increases;

Pressure x volume change = PΔV where volume change = ΔV

PΔV = force/l2 = force/l2x l3 = force x distance

The heat, q, and the work done must have the same sign since the system (fuel) has done work on its environment. Since q is negative (b definition, so the work must be negative

The change in internal energy, ΔE = q +w

If we designate the heat given off by the combustion as being negative, then the work done by the system in expanding the piston must also be negative since if there were no work done, more heat would have been given off.

ΔE = q - P ΔV

q = ΔE + P ΔV which is defined as H, the enthalpy of the system

We use the term enthalpy, H, to describe the heat given off in any process that occurs at constant pressure, usually atmospheric pressure.

ΔH =ΔE + P ΔV

Consider the following reaction occuring in an open vessel:

Heat is given off but also some expansion work against the atmosphere must be done since we produced more gas than we started with.

If we measure the heat given off, then we would be measuring

ΔH, or q, but not ΔE

In this case ΔE > ΔH by PΔV

Consider the following reaction:

In this case the atmosphere is doing work for us. If we measuredthe enthalpy of the reaction, ΔH, or q, this quantity would be greater the ΔE

In summary, if the volume in a chemical reaction does not change, then no work is done and ΔE = ΔH = q.

If the volume in a chemical reaction increases at constant pressure, then work is done and ΔE > ΔH = q, by the amount w or PΔV.

If the volume in a chemical reaction decreases at constant pressure, then work is done and ΔE < ΔH = q, by the amount w or PΔV.

The only difference between a volume change as shown in this diagram and a reaction run in the open, is that when the volume change occurs in an enclosure, you can use the work done to

drive a piston. Expansion work is done in both cases.

Heat capacity: energy necessary to raise a substance’s temperature

Specific heat (spht): the amount of heat needed to raise one gram one degree centigrade.

ΔH = q = spht x m x ΔT

spht (H2O) = 4.184 J/g°C

How much heat does it take to heat 1 kg of H2O from 25 to 100 °C?

ΔH = 1000*4.184*75

This is also the amount of heat that would be released into the environment when 1 kg of water at 100 °C is cooled to 25 °C.

Heat associated with phase changes

QT

Water: ΔHfus enthalpy of fusion: 6.01 kJ mol-1

ΔHvap enthalpy of vaporization: 40.7 kJ mol-1

ΔHsub enthalpy of sublimation ~46 kJ mol-1

The reason why ΔHsub ~ ΔHfus + ΔHvap is that these quantities are often measured at different temperatures.

ΔHsub(Tfus) = ΔHfus (Tfus) + ΔHvap (Tfus)

Standard states: Internal energies of are usually measured on a relative basis.

We define the enthalpy of formation of any element at 25 °C in its most stable state at 1 atm pressure as 0.00 kJ mol-1.

The enthalpy of formation of any substance is thus given by the reaction to form that substance from the elements. For example:

2C + H2 = C2H2 the heat given off by this reaction is the enthalpy of formation of this material, by definition.

ΔHf (25 °C) = 226.7 kJ mol-1

This means that acetylene is less stable than the elements by this amount.

Enthalpies of Formation: why are they of interest?

How are enthalpies of formation measured?

Bomb calorimeter

C (graphite, 25 °C) + O2 = CO2 ΔHf (25 °C) = -393.5 kJ mol-1

H2 + ½ O2 = H2O ΔHf (l, 25 °C) = -241.8 kJ mol-1

Let’s calculate the heat of formation of acetylene at 25 ° C

When 1 mol of C2H2 is burned, the heat (ΔH) given off at constant pressure is –1256 kJ mol-1

C2H2 + 2.5O2 = 2 CO2 + H2O ΔHc (25 ° C) = –1256 kJ mol-1

2 CO2 = 2 C (graphite) + 2 O2 = 2(393.5) kJ mol-1

H2O = H2 + ½ O2 = 241.8

C2H2 = 2 C + H2 -ΔHf (25 ° C) = - 226.7 kJ mol-1

2 C + H2 = C2H2 ΔHf (25 ° C) = 226.7 kJ mol-1

How are calories of food

measured?

In a bomb calorimeter, by burning them

K

Spontaneous processes: most spontaneous processes occur because of the release of energy to the surroundings.

A spontaneous process is a process that is thermodynamically possibly, but not necessarily one that will actually occur.

Release of energy, the most common cause for a spontaneous process, it is not the only factor that can cause a process to be spontaneous. Consider the following:

H2, 1 atm O2, 1 atm

What happens when we open the valve?

Spontaneous processes: most spontaneous processes occur because of the release of energy to the surroundings.

However, release of energy, although the most common cause for aspontaneous process, it is not the only factor that can cause a process to be spontaneous. Consider the following:

H2, 0.5 atmO2, 0.5 atm

H2, 0.5 atmO2, 0.5 atm

Is this process spontaneous? Was any heat involved?

Entropy: S Processes that involve an increase in entropy can also be spontaneous, provided that there is not a large enthalpicbarrier ( a large positive enthalpy).

Entropy: is related to an increase in randomness

Gibb’s free energy = G

ΔG = ΔH - TΔS

If ΔG < 0 (negative) , a process is spontaneous

If ΔG > 0 (positive), the process will not occur but the reverse process is spontaneous

If ΔG = 0 the process is at equilibrium (net net changes occuring)

Just because a process is spontaneous, it doesn’t mean it will occur immediately (example: burning of paper)

Definition of spontaneous in a thermodynamic sense: the process can occur; not necessarily that it will occur immediately or that it will occur.

ΔG = ΔH - TΔS

If ΔG < 0 (negative) , a process is spontaneous

If ΔG > 0 (positive), the process will not occur but the reverse process is spontaneous

If ΔG = 0 the process is at equilibrium (no net change occurs)

High temperature favors processes with positive (increase in randomness) entropy

Factors favoring an increase in entropy:

increase in the number of particles

expansion of a gas

phase change from solid to liquid or from

liquid to gas

ΔG = ΔH - TΔS

High temperatures favors formation of small molecules

Properties of State Functions

Hess’s Law: the overall enthalpy of a reaction is equal to the enthalpy changes of all the individual steps that lead up to theoverall change.

C

CH3

CH3 CH2+ CH3OH = CCH3

CH3

CH3

OCH3

ΔHf (l) = -37.5 -246.4 -313.6 kJ mol-1

ΔHr = Σ ΔHf (products) - Σ ΔHf (reactants) (all in the liquid state)

ΔHr = -313.6 - (-246.4 – 37.5) = - 29.7 kJ mol-1

Is the entropy change of this reaction positive or negative?

If we mix these two reagents together will they react immediately?

isobutene (l) + CH3OH (l) = t-butyl methyl ether (l)What is the heat of this reaction?

C

CH3

CH3 CH2+ CH3OH = CCH3

CH3

CH3

OCH3

ΔHf (g) = - 37.5 -246.4 -313.6 kJ mol-1

The heat of formation of each of these compounds was determined by burning them in an excess oxygen and measuring the heat of combustion just as the enthalpy of formation of acetylene was determined. For example:

Do the following involve an increase, decrease or no change in entropy?

1. Boiling of water?

2. The freezing of water?

3. The expansion of a gas at constant temperature?

4. The burning of natural gas (CH4)?

CH4(g) + 2O2(g) = CO2(g) + 2H2O (g) + heat

5. The burning of methanol (CH3OH)?

CH3OH (l) + 2O2(g) = CO2 (g) + 2H2O (g) + heat

6. The crystallization of a solid at a constant temperature?

Which will give you a worst burn, water at 100 °C or water vapor at 100 °C?

How much energy is required to convert 100 mL of water from the liquid at room temperature (25 ° C) to a gas at 100 °C?

Heat capacity of water = 4.184 J/g °C

Vaporization enthalpy(∆Hv(373 K) =2260 J/g

1. Heat the water from 25 to 100 °C

4.184 J/g °C* (100-25) *100g = 31380 J

2. Vaporize the water

100 ml *1 g/mL*2260J/g = 226000 J

3. 31380+ 226000 = 257380 J

Which will give you a worst burn, water at 100 °C or water vapor at 100 °C?

Calculate the enthalpy of reaction for the photosynthesis of glucose by plants from CO2 and liquid water at T = 25 °C

∆Hf (25 °C)

H2O (l) = -285.8 kJ/mol

C6H12O6 = -1260 kJ/mol

CO2 = -393.5 kJ/mol

6 CO2 + 6 H2O = C6H12O6 + 6O2

6(-393.5) 6(-285.8) -1260 0

∆Hrx = ∆Hf (products) -∆Hf (reactants)

= -1260 - [6(-393.5) + 6(-285.8) ]

= -1260 - [-2361 - 1714.8] = -1260 + 4075.8 = 2815.8 kJ/mol

Suppose we have ice and water at equilibrium, what is ΔG ?

ΔG = 0

If the melting of ice requires +6.01 kJ/mol, what is ΔS for this process?

ΔG = ΔH - TΔS; ΔS = ΔH/T; ΔS = 6010/273 (J/mol K)

ΔS = 22 (J/mol K)

Suppose we have water and water vapor at equilibrium at 373 K, what is ΔG ?

The heat necessary to evaporate H2O is 40680 J/mol

ΔG = 0

ΔS = ΔH/T; ΔS = 40680/373 (J/mol K) = 109.1 J/mol K