Transcript THERMODYNAMICS!!!!

THERMODYNAMICS!!!!
Nick Fox
Dan Voicu
Changes of State
• Order of phase changes based on temperature
(lower-higher)
• Solid
– Melting/solidification
• Liquid
– Boiling/Condensation
• Gas
• For more info on the changes of state look at
Review of Section 2
Specific Heat
• Heat capacity: amount of heat required to raise
its temperature by 1K
• Molar heat capacity Cm: heat capacity of one
mole of a substance
• Specific heat: the heat capacity of one gram of a
substance
• Formula= (quantity of heat transferred)/(grams of
substance) X (temperature change)
• CS= q/m*∆T
• Water Specific heat= 4.18 J/g*K
Heating Curves
• This is the heating curve of water (H2O)
• 1st plateau: 0 °C = 273.15 K
• 2nd plateau:
100 °C =
373.15 K
http://bhs.smuhsd.org/sciencedept/marcan/apchemistry/cool_phase_changes_explain
.html
Calorimetry
• Calorimetry: experiments measuring the
amount of heat transferred between the
system and the surroundings
• Calorimeter: measures temperature change
accompanying a process
http://www.livingston.org/153220824141414520/blank/browse.asp?a=383
&BMDRN=2000&BCOB=0&c=62270&153220824141414520Nav=%7C&Node
ID=4484
Covalent Bond Energies
• Covalent Bond: when two atoms share a pair
of valence electrons
• The energy that is released when covalent
bonds are formed=∆H (kJ)
• Forming bonds releases energy which makes
∆H negative
• Breaking bonds takes energy in which makes
∆H positive
Enthalpy of Formation/Reaction
• ∆Hf
• It is the enthalpy change for the reaction
where the substance is formed from the
reactants
• To solve the Enthalpy of a reaction you would
use this forumla:
∆Horxn= ∑n∆Hof (products) – ∑m∆Hof (reactants)
Entropy and Spontaneity
•
Three Types of Motion
• Translational motion is the movement of an entire molecule in one direction
•
Vibrational motion is the periodic movement of atoms in a molecule
•
Rotational motion is the spinning movement of a molecule about an axis
Entropy and Spontaneity (Cont.)
•
Three Main Factors Affecting Entropy
–
Temperature
•
–
Volume
•
–
When the molecules occupy a greater volume, there is more disorder to the system because they are more
spread out and have more motional energy; less volume means more order because molecules are less spread
out and have less motional energy
Number or independently moving particles
•
•
Increasing the temperature increases the kinetic energy of the molecules, therefore the there is more
movement of the molecules, increasing the randomness/disorder; decreasing the temperature has the
opposite effect
The more molecules there are present in the system, the greater the entropy because there is more motional
energy present in a greater number of molecules
The Third Law of Thermodynamics
•
The entropy of a pure crystalline substance at absolute zero is zero. S(0 K) = 0
Entropy and Spontaneity (Cont.)
• One way for a reaction to be spontaneous is for the change in
entropy to be positive (and the change in enthalpy to be
negative)
• ∆S° = ∑nS°(products) - ∑mS°(reactants) where n and m are
coefficients in the chemical equation
• We also can look at the change in entropy of the surroundings
• ∆Ssurr = -q /T = - ∆H/T
• ∆S°univ = ∆S°sys + ∆S°surr
• ∆S°univ will be positive for any spontaneous reaction
Gibb’s Free Energy and Spontaneity
• ∆G = ∆H – T∆S, under standard conditions ∆G° = ∆H° – T∆S° (standard
temperature is 298K)
• If Gibb’s free energy is negative, the reaction is spontaneous in the
forward direction
• If Gibb’s free energy is zero, the reaction is in equilibrium
• If Gibb’s free energy is positive, the forward reaction is nonspontaneous,
and work must be supplied by surroundings to make it occur, however the
reverse reaction is spontaneous
• Standard free energy of formation can be tabulated, similar to how it is
done for standard enthalpy of formation
• ∆G° = ∑n∆G°f(products) - ∑m∆G°f(reactants)
• The above equation gives the standard free energy change of a reaction
Gibb’s Free Energy and Equilibrium
• ∆G° = -RT lnQ can be used to find the reaction quotient
• To find equilibrium constant:
–
–
–
–
–
–
–
–
∆G = ∆G° + RT lnK
Gibb’s free energy is zero when reaction is in equilibrium
0 = ∆G° + RT lnK
RT lnK = -∆G°
lnK = -∆G°/RT
Raise both sides as powers of e
K = e-∆G°/RT
A positive value for lnK means K>1, therefore the more negative the standard
Gibb’s free energy, the larger the equilibrium constant
– If Gibb’s free energy is positive, then lnK is negative (note lnK is negative, not
K) which means that K<1 (K cannot equal zero or have a negative value)
Hess’s Law
• Hess’s Law states that if a reaction is carried out in a series of steps, the
enthalpy for the overall reaction will equal the sum of the enthalpy
changes for the individual steps
• C(s) + O2(g)  CO2(g)
∆H1 = -393.5 kJ
• CO(g) + 1/2 O2(g)  CO2(g)
∆H2 = -283.0 kJ
• Calculate the enthalpy for the combustion of C to CO.
– C(s) + 1/2 O2(g)  CO(g)
∆H3 = ?
– In order to cancel some of the molecules, one reaction must be reversed. The
second reaction should be reversed because the enthalpy will remain negative
if this is done
– C(s) + O2(g)  CO2(g)
∆H1 = -393.5 kJ
– CO2(g)  CO(g) + 1/2 O2(g)
∆H2 = 283.0 kJ
– C(s) + 1/2 O2(g)  CO(g)
∆H3 = -110.5 kJ