1st Law of Thermodynamics
Energy cannot be created or destroyed but only converted from one form to another
The energy of the universe is constant
State Function
A state function is independent of the pathway (does not rely on the past or future of the system
A state function depends only on the present state of the system
Ex - Elevation vs distance
EnergyThe capacity to do work or provide heat
Potential Energy (PE) = mhv
Kinetic Energy (KE) = 1/2mv2
The total energy of the system is equal to the sum of the potential and kinetic energy
Can be changed by a flow of heat/work
ΔE = q + w
Chemical Energy
Energy will flow either from the system to its surroundings (exothermic) or from the surroundings to the system (endothermic)
The system will be the reactants and products
ΔE < 0 energy flows out of the systemΔE > 0 energy flows into the system
From the system’s point of view…
q +Heat flows
into the system
q -Heat flows out of the system
w + Work done on the system
w - Work done by the system
Internal Energy
Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system
ΔE = q + wΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ
Work vs Energy flow
Work Done by or to Gases
h x F distance x force Work AF
P
A x P For F/A P since
hA x x P A x F Work hA x volumeinitial - volumefinal V
VP Work :onsubstitutiby out) flows (works VP- w
PV Work
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm
w = -PΔVΔV = 64 L – 46 L = 18 L
w = -(15 atm x 18 L)w = -270 L∙atm
Work is done by the system so w is negative
Internal Energy, Heat & Work
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. Use 1 L∙atm = 101.3 J
ΔE = q + w
q = +1.3 x 108 J
w = -PΔV
ΔV = 5.0 x 105 L
w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm
ΔE = (+1.3 x 108 J) + (-5.1 x 107 J)
ΔE = 8 x 107 J
J 10 x 5.1- atm L 1
J 101.3 x atm L 10 x 0.5 75
w
Enthalpy (H)
H = E + PV
E = internal energy
P = pressure of the system
V = volume of the system
Enthalpy is a state function
What is Enthalpy?
Consider a process at constant pressure where the only work is PV work (w = -PΔV)
ΔE = qp + w
ΔE = qp – PΔV
qp = ΔE + PΔV
H = E + PV or ΔH = ΔE + Δ(PV)ΔH = ΔE + PΔV (pressure is constant)
ΔH = qp
In other words…
the terms heat of reaction and change in enthalpy are the same so
ΔH = Hproducts - Hreactants
Enthalpy
How much heat is released when 4.03 g of hydrogen is reacted with excess oxygen?
2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ
kJ 572- H mol 2
kJ 572-
g 2.02
H mol 1 H g 03.4
2
22
Enthalpy #2
When 1 mole of CH4 is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which 5.8 g of CH4 is burned at constant pressure.
qp = ΔH = -890 kJ/mole CH4
44
44 CH mol 0.36
CH g 16.0
CH mole 1 CH g 8.5
kJ 320- CH mol
kJ 890-CH mol 36.0
44
Calorimetry
Calorimeter – used to determine the heat energy change during a reaction
Carried out under constant pressure measures enthalpy (ΔH)
Carried out under constant volume measures energy (ΔE)
Heat Capacity
Specific heat capacity – the amount of heat needed to raise one gram on a substance by 1oC (J/Co∙g or J/K∙g)
Molar heat capacity – the amount of heat needed to raise one mole of a substance by 1oC (J/Co∙mol or J/K∙mol)
in temp increase
absorbedheat C
Heat Gained or Lost Depends on…
1. The change in temperature during the reaction
2. The amount of substance present
3. The heat capacity of the substance
Tcm q
Calorimetry
A 110. g sample of copper (specific heat capacity = 0.20 J/Co∙g) is heated to 82.4oC and then placed in a container of water at 22.3oC. The final temperature of the water and the copper is 24.9oC. What was the mass of the water in the original container, assuming complete transfer of heat from the copper to the water?
Heat lost by copper =
-(heat lost by copper) = (heat gained by water)
J 1265 C)24.9 - C(82.4 Cg
J 0.20Cu g 110. oo
o
waterg 120
CgC2.6J 4.184
J 1265
Ts
q water of mass
o
o
Calorimetry #2When 1.00 L of 1.00 M Ba(NO3)2 solution at
25.0oC is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC. Assuming that the calorimeter absorbs only a negligible amount of heat, that the specific heat capacity of the solution is 4.18 J/oC∙g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.
Solution (no pun intended)
Species present: Ba2+, NO3-, Na+, and
SO42-
Spectator ions: Na+ and NO3-
Net Ionic equation: Ba2+(aq) + SO4
2-(aq) BaSO4(s)
Heat evolved by reaction = heat absorbed by the solution= m x c x ΔT
Since 1.00 L of each solution is used, total volume of the mixture is 2.00 L (2.0 x 103 g)
Temperature increase =28.1oC – 25.0oC = 3.1oC
q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC) 2.6 x 104 J = ΔH
Based on…
1. State function
2. Enthalpy change is same for a reaction whether the reaction takes place in one or many steps
How to use Hess’s Law
• Manipulate equations to reach the desired reaction
• If the reaction given is reversed, so is ΔH
• If multiplying the equation to balance the coefficients also multiply ΔH by the same number
Calculate the enthapy for the following reaction:
N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = ??? kJ
N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ
2NO2(g) ---> 2NO(g) + O2(g) ΔH° = +112 kJ
Solution
N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ
2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ
+ 68 kJ
The second equation has been reversed including ΔH
Calculate ΔH° for this reaction:
2N2(g) + 5O2(g) ---> 2N2O5(g)
H2(g) + 1/2O2(g) H2O(l) ΔH° = -285.8 kJ
N2O5(g) + H2O(l) 2HNO3(l) ΔH° = -76.6 kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l) ΔH° = -174.1 kJ
Standard Enthalpies of Formation
The change in enthalpy of the formation of one mole of a compound from it elements in their standard states
ΔH° 25oC at 1 atm and 1 M
By definition, the standard heat of formation for elements in their standard states equals zero.
Example: Which of the following will have standard heats of formation equal to zero?
H2(g), Hg(s), CO2(g), H2O(l), Br2(l)
Example
Write the balanced molecular equation representing the ΔHf° for ethanol.
Answer:
2C(s) + 3H2(g) + 1/2O2(g) CH5OH(l)
Calculate the Standard Enthalpy Change for the combustion of Methane
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
1. CH4(g) C(s) + 2H2(g)
2H2(g) + C(s) CH4(g) -75 kJ
2. O2(g) 0 kJ
3. C(s) + O2(g) CO2(g) -394 kJ
4. H2(g) + ½ O2(g) H2O(l) -286 kJ
Example #1
• Calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water
• Use table 6.2 on page on Page 262
Heat of Fusion
ΔHfus = enthalpy change that occurs in melting a solid at its melting point
Example: What quantity of heat is needed to melt 1.0 kg of ice at its melting point?
ΔHfus =6.0 kJ/mol
kJ 333 mol 1
kJ 6.0
g 18.0
mol 1
kg 1
1000gkg 00.1
Heat of Vaporization
ΔHvap = the energy needed to vaporize one mole of a liquid at a pressure of 1 atm
Example: What quantity of heat is required to vaporize 130. g of water?
For liquid water ΔHvap = 43.9 kJ/mol
kJ 317 mol 1
kJ 43.9
g 18.0
mol 1 g .130
Example #2
Substance X has the following properties:
ΔHvap = 20. kJ/mol
ΔHfus = 5.0 kJ/molBoiling point = 75oCMelting point = -15oCSpecific heatSolid = 3.0 J/goCLiquid = 2.5 J/goCGas = 1.0 J/goC
Example #2 continued…
Calculate the energy required to convert 250. g of substance X from a solid at -50oC to a gas at 100oC. Assume that X has a molar mass of 75.00 g/mol.
5 Step Process
1. Heating solid
2. Melting solid
3. Heating liquid
4. Boiling liquid
5. Heating gas
Example #2 continued…
1. q = m x c x ΔT= 250.g x (3.0 J/goC) x 35oC = 26 kJ
2. mol x ΔHfus = 3.33 mol x 5.0 kJ/mol = 17 kJ3. q = m x c x ΔT
= 250.g x (2.5 J/goC) x 90oC = 56 kJ
4. mol x ΔHvap = 3.33 mol x 20. kJ/mol =67 kJ5. q = m x c x ΔT
= 250.g x (1.0 J/goC) x 25oC = 6.2 kJ172 kJ