calendar 13-14 m 1/27 sec 10.1-3:thermodym, phase b-1 1/29 sec 10.3: calorim, hf o, hess law b-1...
TRANSCRIPT
Calendar 13-14 M 1/27 Sec 10.1-3:Thermodym, Phase B-1 1/29 Sec 10.3: Calorim, Hfo, Hess
Law B-1 1/31 Lab M 2/3 Sec 10.4 - Entropy and S B-1 2/5 Quiz H, S Sec 11.3: G, Equil
temp calcs B-2 2/7 Review M 2/10 Test
Calendar 14-15 B-2 1/23-4 Sec 10.1-3:Thermodym Intro M 1/26 Sec 10.2-3/11.3: Calorim, Phase
Demolab w/ phase thermodynamics B-1 1/27-8 Sec 10.3 Calcs of H Hfo, Hess
Law B-2 1/29-30 Quiz, Rev of H calcs, M = B-1 2/2-3 Lab Hess Wed 2/4 Sec 10.4 - Entropy and S, G B-2 2/5-6 Quiz S, G: Review M 2/9 Test
Driving ForcesWhat causes chemical reactions to occur?
Nature drives things toward lower energy
Enthalpy = total energy content of a systemNature likes things to be random, or more disorganized
Entropy = disorder in a system
Entropy = S A measure of the disorder in a system Which has a greater entropy
A chess board before the 1st move or in the middle of a game
A jigsaw puzzle that is put together, or the same puzzle unassembled in its box
Humpty Dumpty before or after his tragedy
Nature tends to move spontaneously to a more disorganized state
ThermodynamicsStudy of energy, heat, and its flow
Energy = ability to do workPush or pull a mass
Heat = q = a form of energy Heat content is a measure of total Ek in a substance
Temp = meas of average movement (Ek)Temp and Ek are directly relation
Ek = ½mv2
Temp diff determines direction of heat flow
System vs. Surrounding System = reaction or process you
are examining Surroundings = rest of the universe
Energy is exchanged between the system and the surroundings
Endothermic vs Exothermic reactions Diagrams follow
Enthalpy = H
The total energy content of a system Sum Ek + EpCannot be measured
Can meas Ek but not Ep Ek = ½mv2
Can determine the change in heat content of a system = HChange in Enthalpy = H = Hprod-Hreact
Thermochemical Equation Exothermic CH4(g) + 1½O2(g) CO2(g) + H2O(g) + 890 KJ
CH4(g) + 1½O2(g) CO2(g) + H2O(g) H = -890 KJ
Endothermic 483.6 KJ + H2O(g) H2(g) + O2(g)
H2O(g) H2(g)+ O2(g) H = 483.6 KJ
Thermo chemical equation show the reactants and product
with the energy change that occurs
Calorimetry The measurement of heat flow =
HWe can catch the energy going into or
out of a systemUse a calorimeter
Highly insulated device used to measure the amount of heat moving into or out of the system
Water is usually used in the cal is used to absorb energy from, or give off energy to the system
Heat Capacity = Cp To measure heat flow into or out of the
system we must know how much heat the material in the calorimeter can hold
Called heat capacity Unique for each substance
Specific Heat Capacity = Cp Amount heat (q) required to increase the temp
of 1 g of a subst 1˚C Water has a HUGE heat capacity
Units of J/g°C or J/g K
Coffee Cup Calorimeter qH2O = mH2O x TH2O x CpH2O
qH2O = H = energy absorbed or given off by the water in the calorimeter
mH2O = mass (g) of the water 1 mL H2O = 1 gram H2O
TH2O = change in temp Tfinal – Tinitial
Cp = specific ht capacity of the water in the calorimeter
CpH2O = 4.18 J/g˚C
1. A reaction occurs in a calorimeter that contains 45.0 g of water. If the temp of the water goes from 25.0°C to 55.0°C, calculate the energy exchanged.
Sample Problem
How much energy is needed to increase the temperature of 50.0 g of water 15.0⁰C?
How much energy (joules) is needed to increase the temp of 150.0 mL water from 40.0⁰C to 60.0⁰C?
Haw many grams of water can be heated from 20.0⁰C to 75⁰C using 12500.0 Joules?
A 120.0 mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.
q(H2O) = m(H2O) x t(H2O) x 4.18 J/g⁰C
A 120.0 mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.
q = m x (tf – ti) x 4.18 J/g⁰C
q = (tf – ti)
m x 4.18 J/g⁰C q
m x 4.18 J/g⁰C + ti = tf
q of the Reaction
The heat lost by the reaction is gained by the H2O in the calorimeterCalc q for the water in the calorimeter.
The value of Hrx for the reaction is = to –qH2O-qH2O = qrx = Hrx
Sample Problem A reaction is carried out in a coffee cup
calorimeter that contains 110.0 g of water at 23.5˚C. If the final temp of the water becomes 20.0˚C, determine the value of H for the reaction.
qH2O = mH2O x TH2O x CpH2O
= 110.0 g x (20.0˚C – 23.5˚C) x 4.18 J/g˚C = -1.61 x 103J
Hrx = -qH2O = +1.61 x 103 J or +1.61 KJ
Sample Problem An acid and a base are mixed in a coffee-
cup calorimeter. The calorimeter contains 125.0 mL of water at 22.0˚C. A neutralization reaction occurs that increased the temp to 28.0˚C. What is H for the reaction?
qH2O = mH2O x TH2O x CpH2O
= 125.0 g x (28.0˚C – 22.0˚C) x 4.18 J/g˚C = 3.34 x 103 J or 3.34 KJ
Hrx = -qH2O = -3.34 x 103 J or -3.34KJ
Calculation of C
Can calc the value of C for any substance through calorimetry
Heat the metal in a hot water bath and drop it in calorimeter: the heat picked up from the metal goes into the water Use the calorimetry equation to determine the
amount of energy transferred by the substance qH2O = -qPb = mPb x TPb x CpPb
A 90.0 g sample of metal is placed in a calorimeter that contains 105.0 g of water at 25.5˚C. If the temp of the water increases to 37.5 ˚C, determine the Amount of energy transferred in the process.
Hint: q(H2O) = m (H2O) x t (H2O) x C (H2O)
and t (H2O) = (tf – ti)
Phase Changes Chap 11.3
Solid GasLiquidfusion
crystallization
vaporization
condensation
sublimation
deposition
Each phase change requires a specific amount of energy
Molar Heat of fusion water = 6.01 KJ/mol
Molar Heat vaporization water = 40.7 KJ/mol
What are the molar heats of condensation and crystallization???
Blue = endoRed = exo
Lab Demo Heat Fusion Ice
How much energy does it take to melt a gram of ice? A mole of ice?
Place ice in warm water in a coffee-cup calorimeter
Measure 1- temp change for water in calorimeter (t) 2- how many grams (or mole)
Phase Change Sample Problem
How much energy is required to melt 27.0 g of ice at 0°C? q = molesH2O x Molar Ht FusionH2O
q = 1.50 mol water x 6.01 KJ = 9.01 KJ of is needed
1 mol water
Mol Htfus = 6.01 KJ/mol
Hint: 1 - calc the moles of water (mol=g/mm) 2 - multiply moles times 6.01 KJ/mol
27.0 g H2O = 1.50 mol H2O
18.0 g/mol H2O
How much energy is required to increase the temp of 27.0 g of water from zero to 100.0 degrees C? q(H2O) = m (H2O) x t (H2O) x C (H2O)
= 27.0 g x 100.0°C x 4.18 J/g°C = 11286 J = 11.3 KJ
Phase Change Sample Problem
Phase Change Sample Problem
How much energy is required to vaporize 27.0 g of water to steam, at 100.0°C? q = molesH2O x Molar Ht VapH2O
q = 1.50 mol water x 40.7 KJ = 61.1 KJ 1 mol water
Mol Htvap = 40.7 KJ/mol
Hint: - 1 calculate the moles of water (mol=g/mm) - 2 multiply moles times 40.7 KJ/mol 27.0 g H2O = 1.50 mol H2O
18.0 g/mol H2O
Phase Change Sample Problem
How much energy is required to take 27.0 g of ice at 0°C and change it to steam at 100.0°C?
q = molesH2O x Molar Ht FusionH2O
qH2O = mH2O x TH2O x CpH2O
q = molesH2O x Molar Ht VapH2O
= 9.1 KJ + 11.3 KJ + 61.1 KJ = 81.5 KJ
Phase Change Sample Problem
How much energy is required to take 63.0 g of ice at 0°C and change it to steam at 100.0°C?
q = molesH2O x Molar Ht FusionH2O
(63.0/18.0) mol x 6.01 KJ/mol = q1
qH2O = mH2O x TH2O x CpH2O
63.0 g x 100.0°C x 4.18 J/g°C = q2
q = molesH2O x Molar Ht VapH2O
(63.0/18.0) mol x 40.7 KJ/mol = q3
Phase Change Math
Molar heats of Fusion (etc) are used to calc energy change for phase change, (at a constant temp).q = moles x Heat Fusion (etc)
Calorimetry is used to calc energy exchanged went a substance temp increasesqH2O = mH2O x TH2O x CpH2O
qice = molice x Hfusice
qH2O = mH2O x TH2O x CpH2O
qsteam = msteam x Tsteam x Cpsteam
qH2O = molH2O x HvapH2O
qice = mice x Tice x Cpice
Phase Change Sample Problem
How much energy is given off when 27.0 g of steam condenses to water at 100.0°C? q = molesH2O x Molar Ht VapH2O
q = 1.50 mol water x 40.7 KJ = 61.1 KJ is given off
1 mol water
Mol Htvap = 40.7 KJ/mol
Hint: - 1 calculate the moles of water (mol=g/mm) - 2 multiply moles times 40.7 KJ/mol 27.0 g H2O = 1.50 mol H2O
18.0 g/mol H2O
3 Ways to Determine H
1. Calorimetry . H° = -qH2O
2. H°f tables 3. Hess’s Law of Heat Summation
Rules of thermodynamics
Rules of Thermodynamics
H is proportional to the amount of reactant or product
Reverse reactions have opposite sign Hess’s Law of Heat Summation
H is proportional to the amount of Reactant or
Product Based on stoichiometric coefficients
Must have a balanced thermochemical equation
CH4 + 2O2 CO2 + 2H2O H = -890 KJ How many KJ are given off by the complete
combustion of 2 moles of CH4?2 mol CH4 x -890 KJ = -1780 KJ
1 mol CH4
Sample ProblemHow many KJ of energy are produced by burning 1.25 g of CH4 according to the rx:
CH4 + 2O2 CO2 + 2H2O H = -890 KJ
How many KJ of energy are produced by burning 1.25 g of O2?
Reverse Reactions• H for reverse reactions have the
same value, but the opposite sign• Reactants Products H = (-)
• Products Reactants H = (+)
Reverse Reactions Have Opposite Signs on H
CH4 + 2O2 CO2 + 2H2O H = -890 KJ
CO2 + 2H2O CH4 + 2O2 H = +890 KJ
Hess's Law of Heat Summation
In a multi-step reaction the sum of the H values for all the steps equals H for the overall reaction
H1 + H2 = H3
A + B --> C H1 = -20 KJ B + C ---> D H2 = -40 KJ A + 2B --> D H3 = ? = -60 KJ
Hess's Law of Heat Summation
Can use Hess's Law to measure H for a reaction that is too hard to run in the lab
C + ½O2 ---> CO H1 = ?
CO + ½O2 ---> CO2 H2 = -283 KJ
C + O2 ---> CO2 H3 = -393 KJ
H1 + H2 = H3
H1 = H3 - H2
H1 = (-393 KJ) - (-283 KJ)
= -110 KJ
Use the following 2 equations to determine
H for the overall reaction:
Sn (s) + 2 Cl2 (g) --- > SnCl4 (aq)
Sn (s) + Cl2(g) ----> SnCl2(s) -325 kJ/mol
SnCl2 (s) + Cl2 (g) ----> SnCl4(aq) -186
kJ/mol
Mn (s) + 02 (g) ---> MnO2 (s)
MnO2 (s) + Mn (s) --- > 2 MnO (s) -240 kJ
2 MnO2 (s) --- > 2 MnO (s) + 02 (g) +264 kJ
SnO2 (s) + 2H2 (g) --- > Sn (s) + 2 H20 (l)
Sn (s) + 02 (g) --- > SnO2 (s) -580.7 kJ
H2 (g) + ½ 02 (g) --- > H20 (l) -285.8 kJ
2 Mg (s) + SiCl4 (1) --- > Si (s) + 2 MgCl2
(s)
SiCl4 (l) --- > Si (s) + 2 Cl2 (g) +687 kJ
Mg (s) + Cl2 (g) --- > MgCl2 (s) -641 kJ
CH4 + 2 02 --- > C02 + 2 H20
C + 2 H2 ---> CH4 -74.8 kJ
C + O2 ----> CO2 -393.5 kJ
H2 + ½ O2 ----> H2O -235.8 kJ
Find H° for the reaction
2H2(g) + 2C(s) + O2(g) --> C2H5OH(l)
using the following thermochemical data.
C2H5OH (l) + 2 O2 (g) --> 2 CO2 (g) + 2 H2O (l) H = -875. kJ
C (s) + O2 (g) --> CO2 (g) H = -394.51
kJ
H2 (g) + ½ O2 (g) --> H2O (l) H = -285.8 kJ
H2(g) + Br2(g) → 2 HBr(g) ΔH = −72 kJ
H2(g) → 2 H(g) ΔH = 436 kJ
Br2(g) → 2 Br(g) ΔH = 224 kJ
Calculate the enthalpy change for
H(g) + Br(g) → HBr(g)
Enthalpies of Formation = Hf°
Cu(s) + Cl2(g) ---> CuCl2(s) H = -220 KJ The energy change H⁰, when 1 mol of a
substance is formed from its elements in their normal (standard) state (the ° means 25°C and normal pressure) Elements (including diatomic like H2, N2, Cl2) are
arbitrarily assigned Hf° = 0
Hf° = a relative measure of how stable compounds
are A(-)Hf° means compound is more stable than its elements
A(+)Hf° means compound is less stable than its elements
Enthalpies of Formation
H°rx = Hf°(products) - Hf°(reactants)
Hf° values are on table A-11 pg 833 They are used to calculate, NOT
measure, the change in enthalpy, H°, for a chemical reaction that is hard or impossible to carry out in the lab.
Sample prob E pg 356 Do pract prob #1,2 pg 356
Sample Problem Calculation H° for the reaction: C3H8(g) + 5O2 ---> 3CO2(g) + 4H2O(l)
-103.8 0 -393.5 -285.8 = Hf° KJ/mol
H°rx = Hf°(products) - Hf°(reactants)
H°rx = [3mol(-393.5)KJ/mol + 4
mol(-285.8)KJ/mol]- [1mol(-
103.8)KJ/mol + 0] = -2219.9 KJ
Entropy S
Randomness or disorder in a system Entropy is ever increasing Molar entropy is entropy possessed by 1 mole of a substance Nature favors increases in entropy
Reactions that increase entropy occur more easily
Elements have a value for entropy
Entropy S = Randomness As temp entropy Sol Liq Gas entropy goes up A solution has more entropy than its
separate components NaCl water solution vs water and solid NaCl
Reactions that produce more moles of gases increase entropy
Is S (+) or (-) for: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)
Entropy Change = S°
Calculted in the same manner as H° S°rx = S°(products) - S°(reactants)
Sample prob F pg 361
Calc S° for: 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)
S°J/molK = 127.2 205.0 213.8 188.7
S° = [2 mol (213.8KJ/molK) + 4 mol (188.7KJ/molK)] - [2 mol (127.2 KJ/molK) + 3 mol (205.0 KJ/molK)]
= 313 KJ/K Does the entropy change, S, favor a spontaneous reaction or
not???
What determines if a reaction or process occurs spontaneously? What energy change favors a
spontaneous reaction?(-) change in energy
What entropy change favors a spontaneous reaction?(+) positive change in entropy
Neither H or S alone, can predict if a rx will be spontaneous
Gibb's Free Energy = G
Determines whether a reaction is spontaneous or not spontaneous at a specific temperature
If G is (-) reaction is spontaneous If G is (+) reaction is NOT
spontaneous If G = 0 reaction is just beginning
Called equilibrium
Gibb's Free Energy = G Can be calculated in 2 ways 1-From free energy tables Must be at standard conditions to use:
25˚C and pressure at surface of the earthG°rx = G°(products) - G°(reactants)
2-From H and S values at temperatures other than 25˚CG = H – TS
Gibb's Free Energy = G
Calc G˚ for H2(g) + CO2(g) H2O(g) + CO(g)
0 -394.4 -228.6 -137.2 (KJ/mol)
G°rx = G°(products) - G°(reactants)
= [1 mol (-228.6KJ/mol) + 1 mol (-137.2KJ/mol] – [0 + 1mol (-394.4 KJ/mol)]
= 28.6 KJ
Is the rx spontaneous???
G at Temp other than 25˚C
Calc G at 100˚C for the reaction: H2(g) + CO2(g) H2O(g) + CO(g)
H 0 -393.5 -241.8 -110.5
S 130.7 213.8 188.7 197.2
H˚ = [(1 x -241.8) + (1 x -110.50)] – [(1 x -393.5) + 0] = 41.2 KJS˚ = [(1 x 188.7) + (1 x 197.2)] – [(1 x 130.7) + (1 x 213.8)] = 41.4 J/K
= 0.041 KJ/K(Move decimal 3 places to convert J/K to KJ/K)
G = H – TS
Is a reaction spontaneous at 100˚C if H = 41.2 KJ and S = 41.4 J/K
G = H – TS= 41.2KJ – (100 + 273 K) 0.0414 KJ/K
= 25.8 KJ is the reaction spontaneous?
Determination of Equilibrium Temp
The temp (K) at which the rx just beginsEquilibrium
T = H = 41.2 KJ = 995 K S 0.0414 KJ/K
ºC + 273 = K or ºC = K – 273 995 – 273 = 722 ºC
Gibb’s Free Energy
At what temp does H2O(l) H2O(g) becomes spontaneous
H = (-241.8 KJ/mol) – (-285.8 KJ/mol) = 44.0 KJ/mol S = (188.7 J/mol K) – (70.0 J/mol K) = 118.7 J/mol K
T = H = 44.0 KJ/mol = 370 K S .1187 KJ/mol K
ºC = K - 273 = 370 K – 273 = 97 ºC
Affects of Signs of H and G
If H is (-) and S is (+)Rx is always spontaneous
If H is (+) and S is (-)Rx is never spontaneous
If H is (+) and S is (+) Increase temp to make rx spontaneous
If H is (-) and S is (-)Decreases temp to make rx spontaneous
Initial water temp 42.3 ºC final water temp 8.1ºC Change in water temp ºC Final water volume 194.5 mL Initial water volume 130.0 mL Volume of melt mL mass ice melted g Heat released by cooling water J J/g ice melted (heat of fusion) J/g kJ/mol ice melted (molar heat of fusion) kJ/mol
CaO + CO2 --------> CaCO3 ΔHº = -178 kJ
CaO + H2O(l) --------> Ca(OH)2 ΔHº = -65 kJ
Ca(OH)2 + CO2 -----> CaCO3 + H2O(l)
SOCl2 + NiO --------> SO2 + NiCl2 150 kJ
SOCl2 + H2O --------> SO2 + 2HCl -27 kJ
NiO + 2HCl -----> NiCl2 + H2O(g)