Supplemental Study Guide
Chemical Thermodynamics
Heat Capacity and Enthalpy
Laws of Chemical Thermodynamics:
- Energy cannot be created nor destroyed.
- Energy systems increase their entropy (S) – the amount of random activity or disorder. Energy spontaneously tends to be disorderly. Chemical reactions release energy and move toward a lower, more thermodynamic state. Heat moves from hot to cold (never cold to hot). Heat cannot transfer from a cold body to a hot body without converting a certain amount of work to heat energy. It is impossible for any system that operates on a cycle to receive heat from a single reservoir and produce work
Internal energy of a system – total of the kinetic energy due to the rotational, vibrational, and translational motion of molecules and potential energy associated with the vibrational energy of atoms within molecules
A thermodynamic system consists of a universe and its surroundings
A few symbols and their meanings
U = internal energy of a system
Q = heat added to a system
A positive Q means heat flowing into a system
A negative value for Q means heat flowing out of a system
W = mechanical work done ona system
W’ = energy added by all other processes
Gibbs free energy – the amount of energy in a thermodynamic system available to do work
G = H – TS
G = Gibbs energy (J)
H = enthalpy (J) the amount of heat in a system
T = temperature (K)
S = entropy (J/K) amount of randomness or disorder – entropy is a function of the thermodynamic state
Ideal gas law – the state of a gas is determined by pressure, volume, and temperature.
Spontaneous reaction – one that proceeds in a way that the end state will have a lower and more stable energy level than the initial state of the reaction
When a reaction is occurring at constant temperature and pressure the following modification of the Gibbs free energy equation can be used:
ΔG = ΔH – TΔS (Δ= change in – the change in Gibbs free energy equals the change in enthalpy minus the temperature times the change in entropy)
Use the chart on p. 2 of the Laws of Thermodynamics to see that, for example, when the ΔS is positive and ΔH is positive, the chemical process will be spontaneous only at high temperatures.
Carnot cycle – a thermodynamic cycle that converts thermal energy into work – the system goes through a series of different states and returns to the initial state – work is done on the surroundings – study the diagram and the states on p. 3 of the Laws of Thermodynamics
Heat engine – converts energy in the form of heat to mechanical work
Heat capacity – the amount of heat required to raise the temperature of 1 g of a substance 1 K (or 1 oC).
molar heat capacity – heat capacity /mol
specific heat capacity – heat capacity/g x K or heat capacity/g xoC
Equation: ΔH = Cp x m x ΔT
Cp = heat capacity (J/g x K)
M = mass (g or Kg)
T = temperature (oC or K)
How many joules of heat are required to heat 100 g of aluminum from 20 K to 100 K? The heat capacity for aluminum is 0.908 J/gK.
First we need to write down our given values.
Cp=0.908 J/gK
ΔT = 100 K – 20 K = 80 K
Mass = 100 g
The equation for calculation is: ΔH = Cp x m x ΔT.
Then we substitute the values as follows:.
ΔH = (0.908 J/gK) (100 g) x (80 K)
ΔH = 7264 Joules.
How much heat is required to heat 200 g of gold from 35 to 65 oC. The heat capacity of gold is 0.1291 j/gK?
Cp = 0.1291 J/gK
ΔT = 30 oC = 203K
m = 200 g
ΔH = 0.1291 J/gK x 200g x 203K
ΔH = 5241 J
Enthalpy (H) – the total of the internal energy of a system in addition to the energy of the work done on the surroundings – enthalpy is the amount of heat in a system.
Equation: H = U + PV
H = enthalpy – total internal energy plus work
U = internal energy (J)
P = pressure (atm – atmospheres)
V = volume (milliliters, mL)
Enthalpy can only be measured in a closed system where there is no exchange between a system and its surroundings.
Enthalpy change in a closed system: ΔH = Hfinal – Hinitial
ΔH = enthalpy change
Hfinal = final enthalpy of the products
Hinitial enthalpy of the reactants
Exothermic reaction - heat released to the surroundings - ΔH negative – products are warm to touch – the change in the enthalpy is equal to the amount of energy released by the reaction
Endothermic reaction – heat absorbed from the surroundings – ΔH positive –products are cool to the touch – the change in enthalpy is equal to the amount of energy absorbed by the reaction
Standard enthalpy change of formation (standard heat of formation)– amount of heat required to form a given chemical substance ΔHfo.
The information on the standard enthalpy table can be used to calculate the enthalpy change.
We use Hess’s law to calculate enthalpies; the enthalpy change of a chemical reaction is the same no matter how the products were obtained. What matters only are the beginning and ending states. The processes for getting there do not matter.
Isothermal system – expansion of gas/work done at a single temperature gradient
Hess’s law – enthalpy changes are additive – The ΔH for a single reaction can be calculated from the difference between the heat of formation of the products minus the heat of formation of the reactants:
ΔHo = Σ (ΔHfoproducts) - Σ (ΔHforeactants).
If we have multiple reactions, the notation changes slightly:
ΔH = Σ (ΔHoproducts) - Σ (ΔHoreactants).
Follow the model problem on p. 4 of heat Capacity and Enthalpy. Then solve the problems in the practice activity on p. 5. Check answers.
Entropy and Gibbs Free Energy
Entropy (S) - a thermodynamic state function used to determine the free energy in a thermodynamic system used to determine whether or not a reaction is spontaneous can be used to estimate the amount of energy not available to do work the amount of energy equal to the temperature of the surroundings must be given off to equalize the system entropy increases as reactions progress over time
Think about a glass of ice in a room. Over time, the entropy in the room will decrease in order to equalize the temperature of the room with the temperature of the glass of ice.
ΔE = TSurrx ΔS
where
ΔE is the energy given up by the system.
TSurr is the temperature of the surroundings (which may not be the same as the system's current temperature).
ΔS is the entropy of the system.
The reaction cannot proceed if heat energy is not released.
Entropy:
where,
S = Entropy expressed in the units joules per Kelvin (JK−1)
Q = Heat quantity expressed in joules (J)
T = Temperature at which the work is being done expressed in Kelvins (K)
Entropy change
ΔS = SFinal – SInitital
We rearrange the equation to give us the following:
and finally, we factor out the heat quantity (Q) to yield the final form of the equation:
What is the ΔS given the following?
ΔSfinal = -37.2 kJ/mol
ΔSinitial = -32.2 kJ/mol
ΔS = SFinal – SInitital
-37.2 kJ/mol - -32.2 kJ/mol = -5 kJ/mol
What is the entropy change, given the following information?
Q = 1.1 x 104 J/mol
T1 = 12° C
T2 = 5° C
T1 = 12° C = 285 K
T2 = 5° C = 278 K
ΔS = 1.1 x 104 J/mol (3.60 x 10-3 K-1 – 3.51 x 10-3K-1)
ΔS = 1.1 x 104 J/mol (0.09 x 10-3 K-1)
ΔS = 0.099 x 10-1 K-1
ΔS = 1.0 J/Kmol
Solids will have a smaller entropy value than liquids and liquids will have a smaller entropy value than gases.
Gibbs free energy (G) – energy in a system available to do work
H - TS
where
G = The Gibb’s free energy expressed in joules (J)
H = Enthalpy expressed in joules (joules)
T = Temperature expressed in Kelvins (K)
S = Entropy (joules/Kelvin)
Gibbs free energy change
ΔG = ΔH - TΔS
where
ΔG = change in Gibbs energy expressed in kilojoules (kJ)
ΔH = change in enthalpy expressed in kilojoules per mole (kJ/mol)
T = absolute temperature in Kelvins (K)
ΔS = change in entropy expressed in kilojoules per Kelvin mole (kJ/Kmol)
used to determine the voltage of an electrolytic cell and equilibrium constants for reversible reactions
when ΔG is negative, a reaction will be spontaneous and energy will be released when the ΔG is positive, the reaction will not be spontaneous and energy will have to be added to make the reaction proceed
Gibbs free energy is useful for thermochemical processes that are at constant temperature (isothermal) and constant pressure (isobaric)
Gibbs free energy of formation - ΔGf° - change in energy that accompanies the formation of 1 mole of a substance from its component elements at their STP
Gas laws
Ideal Gas Law – an equation for a gas in which collisions are perfectly elastic and there are no intermolecular attractive forces most accurate for monatomic gases the state of a gas is determined by pressure, volume, and temperature
PV = nRT
where
P is the pressure, measured in atmospheres (atm) or millimeters of mercury (mm Hg)
V is the volume, measured in Liters (L) or milliliters (mL)
n is the number of moles of a gas (mol)
R is the universal gas constant (8.3145 J/mol K) or (0.0820 L atm/mol K)
T is the temperature, measured in kelvins (K)
This is a very important formula to learn and memorize. .Problems are not difficult if you write down all of the variables and the units and solve algebraically. Study example problems and do all problems in the practice activities.
Boyles law – deals with volume and pressure
P1V1 = P2V2
where
V1 is the volume of the gas (before), measured in Liters (L) or milliliters (mL)
P1 is the pressure of the gas (before), measured in atmospheres (atm) or millimeters of mercury (mm Hg)
V2 is the volume of the gas (after), measured in Liters (L) or milliliters (mL)
P2 is the pressure of the gas (after), measured in atmospheres (atm) or millimeters of mercury (mm Hg)
Charles law – deals with volume and temperature
V1T2 = V2T1
where
V1 is the volume of the gas (before), measured in Liters (L) or milliliters (mL)
T1 is the temperature of the gas (before), measured in kelvins (K)
V2 is the volume of the gas (after), measured in Liters (L) or milliliters (mL)
T2 is the temperature of the gas (after), measured in kelvins (K)
Combined Gas Law combination of Charles’ and Boyles’ laws temperature, pressure, and volume must remain constant
Where,
P is the pressure, measured in atmospheres (atm) or millimeters of mercury (mm Hg)
V is the volume, measured in Liters (L) or milliliters (mL)
T is the temperature, measured in kelvins (K)
The volume of a gas at 1.85 atm and 15°C is 400 mL. At what temperature does the gas occupy a volume of 150 ml and 1.35 atm?
Rearrange the formula to
T2 = P2V2T1 P1V1
V1 = 400 mL
P1 = 1.85 atm
T1 = 15 oC = 288 K
V2 = 150 mL
P2 = 1.35 atm
T2 = ??
T2 = 1.35 atm x 150 ml x 288 K
1.85 atm x 400 mL
T2 = 78.8K
Dalton’s law of partial pressures – total pressure equals the sum of the partial pressures
Ptotal = Pgas1 + Pgas2 + Pgas3 + …………
where
Pgas x represents the partial pressure of each component gas.
Study all practice problems and do all problems in the practice activity.