Download Lecture 5 - Thermodynamics II

1st Law of Thermodynamics
• Aka the Law of conservation of energy, Gibbs
in 1873 stated energy cannot be created or
destroyed, only transferred by any process
• The net change in energy is equal to the heat
that flows across a boundary minus the work
done BY the system
• DU = q + w
– Where q is heat and w is work
– Some heat flowing into a system is converted to work and
therefore does not augment the internal energy
Directionality from the 2nd Law
• For any spontaneous irreversible process,
entropy is always increasing
dq
dS 
T
• How can a reaction ever proceed if order
increases?? Why are minerals in the earth not
falling apart as we speak??
NEED FOR THE SECOND LAW
• The First Law of Thermodynamics tells us that
during any process, energy must be conserved.
• However, the First Law tells us nothing about in
which direction a process will proceed
spontaneously.
• It would not contradict the First Law if a book
suddenly jumped off the table and maintained itself
at some height above the table.
• It would not contradict the First Law if all the
oxygen molecules in the air in this room suddenly
entered a gas cylinder and stayed there while the
valve was open.
MEANING OF ENTROPY AND THE
SECOND LAW
• Entropy is a measure of the disorder
(randomness) of a system. The higher the
entropy of the system, the more disordered it is.
• The second law states that the universe always
becomes more disordered in any real process.
• The entropy (order) of a system can decrease,
but in order for this to happen, the entropy
(disorder) of the surroundings must increase to
a greater extent, so that the total entropy of the
universe always increases.
3rd Law of Thermodynamics
• The heat capacities of pure crystalline
substances become zero at absolute zero
• Because dq = CdT and dS = dq / T
 Cp 
Sabs    dT  Sconfig
T 
0
T
 dT 
dS  C p 

 T 
• We can therefore determine entropies of
formation from the heat capacities (which are
measureable) at very low temps
‘Free’ Energy
• Still need a function that describes reaction
which occurs at constant T, P
G = U + PV – TS = H – TS
(dH = dU + PdV + VdP)
• The total differential is:
dG = dU + PdV + VdP – TdS – SdT
G is therefore the energy that can run a process at
constant P, T (though it can be affected by
changing P and T)
Reactions that have potential energy in a system
independent of T, P  aqueous species,
minerals, gases that can react…
• Can start to evaluate G by defining total
differential as a function of P and T
dG = dU + PdV + VdP – TdS – SdT
• Besides knowing volume changes, need to
figure out how S changes with T
For internal energy of a thing:
dU = dqtot – PdV; determining this at
constant volume  dU = CVdT
where CV is the heat required to raise T
by 1°C
Increasing energy with temp?
• The added energy in a substance that
occurs as temperature increases is stored
in modes of motion in the substance
• For any molecule – modes are vibration,
translation, and rotation
– Solid  bond vibrations
– Gases  translation
– Liquid water – complex function…
Heat Capacity
• When heat is added to a phase it’s temperature
increases (No, really…)
• Not all materials behave the same though!
• dq=CVdT  where CV is a constant (heat capacity
for a particular material)
• Or at constant P: dq=CpdT
• Recall that dqp=dH then: dH=CpdT
• Relationship between CV and Cp:
C p  CV 
V

2
T
Where a and b are coefficients of isobaric thermal expansion
and isothermal compression, respectively
Enthalpy at different temps…
• HOWEVER  C isn’t really constant….
• C also varies with temperature, so to really
describe enthalpy of formation at any
temperature, we need to define C as a
function of temperature
• Maier-Kelley empirical determination:
• Cp=a+(bx10-3)T+(cx10-6)T2
– Where this is a fit to experimental data and a,
b, and c are from the fit line (non-linear)
Does water behave like this?
• Water exists as liquid, solids, gas, and
supercritical fluid (boundary between gas and
liquid disappears – where this happens is the
critical point)
• Cp is a complex function of
T and P (H-bond affinities),
does not ascribe to MaierKelley forms…
Heats of Formation, DHf
• Enthalpies, H, are found by calorimetry
• Enthalpies of formation are heats
associated with formation of any
molecule/mineral from it’s constituent
elements
Calorimetry
• Measurement of heat flow (through
temperature) associated with a reaction
• Because dH = q / dT, measuring
Temperature change at constant P yields
enthalpy
Problem
When 50.mL of 1.0M HCl and 50.mL of 1.0M NaOH are mixed in a calorimeter, the
temperature of the resultant solution increases from 21.0oC to 27.5oC. Calculate
the enthalpy change per mole of HCl for the reaction carried out at constant
pressure, assuming that the calorimeter absorbs only a negligible quantity of
heat, the total volume of the solution is 100. mL, the density of the solution is
1.0g/mL and its specific heat is 4.18 J/g-K.
qrxn = - (cs solution J/g-K) (mass of solution g) (DT K)
= - (4.18 J/g-K) [(1.0g/mL)(100 mL)] (6.5 K)
= - 2700 J or 2.7 kJ
DH = 2.7 kJ
Enthalpy change per mole of HCl = (-2.7 kJ)/(0.050 mol)
= - 54 kJ/mol
Hess’s Law
Known values of DH for reactions can be used to
determine DH’s for other reactions.
DH is a state function, and hence depends only on the
amount of matter undergoing a change and on the
initial state of the reactants and final state of the
products.
If a reaction can be carried out in a single step or
multiple steps, the DH of the reaction will be the same
regardless of the details of the process (single vs multistep).
CH4(g) + O2(g) --> CO2(g) + 2H2O(l)
DH = -890 kJ
If the same reaction was carried out in two steps:
CH4(g) + O2(g) --> CO2(g) + 2H2O(g)
DH = -802 kJ
2H2O(g) --> 2H2O(l)
DH = -88 kJ
CH4(g) + O2(g) --> CO2(g) + 2H2O(l)
DH = -890 kJ
Net equation
Hess’s law : if a reaction is carried out in a series of steps,
DH for the reaction will be equal to the sum of the enthalpy
change for the individual steps.