Causes of Change Ch.11. (11-1) Governing Principles Nature favors rxns that proceed toward lower E &...
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Transcript of Causes of Change Ch.11. (11-1) Governing Principles Nature favors rxns that proceed toward lower E &...
![Page 1: Causes of Change Ch.11. (11-1) Governing Principles Nature favors rxns that proceed toward lower E & greater disorder Heat: total KE of particles –Quantity.](https://reader030.fdocuments.us/reader030/viewer/2022033106/5697bffc1a28abf838cc1ae7/html5/thumbnails/1.jpg)
Causes of Change
Ch.11
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(11-1) Governing Principles
• Nature favors rxns that proceed toward lower E & greater disorder
• Heat: total KE of particles– Quantity of thermal E– Joules
• Temp.: avg. KE of particles– Intensity of thermal E– Kelvin
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Calorimeter
• Measures heat absorbed or released in a rxn
• Exo = E released
• Endo = E absorbed
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Calorimeters
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Specific Heat Capacity
• (Cp): amt of heat needed to raise the T of 1 g of a substance by 1ºC– Units = J/g· ºC– Water = 4.18 J/g· ºC
• q = CpmΔT– m = mass (g)– ΔT = Tfinal - Tinitial = temp. change (ºC)– q = heat (J)
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Specific Heat Practice
How many joules are needed to raise the T of 300 g of Al from 20 °C to 70°C if the Cp of Al is 0.902 J/g• °C ?
1. List the eq.q = CpmΔT
2. Substitute & solveq = (0.902 J/g• °C)(300 g)(70°C - 20°C) = 13,530 J
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Law of Heat Exchange
• Heat flows from hot to cold
• The law:– Heat lost = heat gained– Heat lost by a metal will be gained by the
surrounding H2O (measured w/ calorimeter)
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Heat Exchange Practice
Find the Cp of 100 g of an unknown metal when it’s removed from H2O at 100°C & placed into 200 g of H2O at 20°C. The final T of the mixture is 23.5°C.
1. Write eq.Heat lost (by metal) = heat gained (by water)
q(metal) = q(water)
CpmΔT = CpmΔT
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Heat Exchange Practice
2. Substitute & solveCp(100 g)(100°C – 23.5°C) =
(4.18 J/g• °C)(200 g)(23.5°C - 20°C)
Cp = 0.382 J/g• °C
Note: this side must remain +, Therefore ΔT = Ti - Tf
(only in these types of problems)
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Molar Heat Capacity
• (C): heat required to inc. the T of 1 mol of a substance by 1 K– Units = J/K·mol– Table 11-1, p.389
• q = nCΔT– n = moles
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Molar Heat Capacity Practice
If C of H2O is 76 J/K·mol, calculate the amt of heat E needed to raise the T of 90.0 g of H2O from 35°C to 45°C.
1. List eq.
q = nCΔT
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Molar Heat Capacity Practice
2. Substitute (make sure to convert g to mol) & solve
q = (90 g x 1 mol ) )(76 J/K·mol)(45°C - 35°C)
18.02 g
= 3,792 J
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(11-2) Thermodynamics
• Study of E flow
• Thermo = “heat”
• Dynamics = “motion”
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Entropy
• Total disorder in a substance or system
• Molar entropy (S): quantity of entropy in 1 mol of a subst.– Units = J/K·mol
• ΔS > O (+), disorder inc.
• ΔS < O (-), disorder dec.
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Enthalpy
• E “inside” an atom or molecule• Molar enthalpy (H): total E content of a
system– Units = kJ/mol or J/mol
• ΔH > O (+), endo• ΔH < O (-), exo
• ΔH = q = nCΔT = CΔT n n
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Enthalpy Practice
How much does the molar enthalpy change when a 92.3 g block of ice is cooled from –0.2°C to –5.4°C?
1. List eq. ΔH = q = CΔT
nCan’t usew/out q
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Enthalpy Practice
2. Find C for ice in Table 11-137.4 J/K·mol
3. Convert °C to K-0.2°C + 273 = 272.8 K-5.4°C + 273 = 267.6 K
4. Subst. & solve ΔH = (37.4 J/K·mol)(267.6 K – 272.8 K) = -194 J/mol Most ΔH are
in kJ/mol
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Properties of Matter
• Extensive property: depends on amt. of subst. – S, H, m, V, C
• Intensive property: does not depend on amt. of material – D, P, T
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(11-3) Change of State
• S & H change dramatically during a state change
• Heat of fusion (ΔHfus): heat absorbed when 1 mol of a subst. melts – Molar enthalpy of fusion
• Heat of vaporization (ΔHvap): heat absorbed when 1 mol of a liquid vaporizes– Molar enthalpy of vaporization
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TE
MP
ER
AT
UR
E (C
)T
EM
PE
RA
TU
RE
(C
)
HEAT ADDED OVER TIME HEAT ADDED OVER TIME
SOLIDSOLID
LIQUIDLIQUID
GASGAS
meltingmelting
freezingfreezing
condensationcondensation
vaporizationvaporization
Heating CurveHeating Curve
0°
100°
ΔHfus
ΔHvap
(s)
(l)
(g)
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Gibbs Energy
• Molar Gibbs E (G): “free E”; determines spontaneity of a rxn– Units = kJ
• Spontaneous rxns occur w/out outside assistance– ΔG < O (-), spont. rxn– ΔG > O (+), nonspont. rxn
• ΔG = ΔH - TΔS, (T in K)
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Gibb’s Practice
If ΔH° is 41.2 kJ/mol & ΔS° is 0.0418 kJ/K is the following rxn spontaneous at 25°C?
H2 + CO2 H2O + CO
1. List the eq.ΔG = ΔH – TΔS
2. Subst. & solveΔG = 41.2 kJ/mol – (298 K)(0.0418 kJ/K)
= 28.7 kJ, nonspontaneous
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(11-4) Hess’s Law
• Overall enthalpy change in a rxn is = to the sum of the individual steps
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
• Actually occurs in 2 steps:– CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ
– 2H2O(g) 2H2O(l) H = -88 kJ
• ΔH = -802 kJ + -88 kJ = -890 kJ
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Heat of Rxn
• E absorbed or released during a chemical rxn
• Std. heat of formation (ΔHºf): change in enthalpy when 1 mol of a cmpd is produced from free elements– Table A-13, p.802 (check state of matter)
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Standard Conditions
• (°) are generally:– Temp: 298 K or 25°C
– Pressure: 1 atm or 760 mmHg
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Equations
• ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants)
• ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants)
• ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)
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Enthalpy Practice
Calculate ΔHº for the following rxn. Is the rxn exo. or endothermic?
H2(g) + CO2(g) H2O(g) + CO(g)
1. List the eq.
ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants)
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Enthalpy Practice
2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)
ΔHº = [(1 mol)(-241.8 kJ/mol) + (1 mol)(-110.5 kJ/mol)]
- [ (1 mol)(0 kJ/mol) + (1 mol)(-393.5 kJ/mol)]
= 41.2 kJ, endothermic
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Entropy Practice
Calculate ΔSº for the following rxn. Does the rxn proceed toward a more ordered or disordered state?
H2(g) + CO2(g) H2O(g) + CO(g)
1. List the eq.
ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants)
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Entropy Practice
2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)
ΔSº = [(1 mol)(188.7 J/K•mol) + (1 mol)(197.6 J/K•mol)]
- [ (1 mol)(130.7 J/K•mol) + (1 mol)(213.8 J/K•mol)]
= 41.8 J/K, disorder
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Gibb’s Practice
Calculate ΔGº for the following rxn. Is the rxn spontaneous at 25°C?
H2(g) + CO2(g) H2O(g) + CO(g)
1. List the eq.
ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)
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Gibb’s Practice
2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)
ΔGº = [(1 mol)(-228.6 kJ/mol) + (1 mol)(-137.2 kJ/mol)]
- [ (1 mol)(0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
= 28.6 kJ, nonspontaneous