Transcript Exam 2 review:

Exam 2 review:
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Ag+(aq) + Cl-(aq) -------> AgCl(s) H = -65.5 kJ
Calculate H for the formation of 2.5 g of AgCl
MW. 143.319 g/mol (AgCl).
2.5 g/143.319 g/mol = 0.0174 mole
0.0174 mol(-65.5 kJ/mol AgCl) = -1.14 kJ
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Calorimetry problem:
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example problem:
10 g of NaOH is dissolved in 100 mL of water in a calorimeter, the temperature changes
from 23.6 to 47.4 °C. Calculate H for the process, assume the specific heat of the
solution is 4.184 j/°Kg, the same as water.
Chapter 5
q = specific heat(g solution)(T)
q = 4.184j/Kmol(110 g)(26.4 -43.3) = -10953 J
moles NaOH = 10g/40 g = .25 mole
H = -10953 J/.25 mol = -43812 = 40000 J/mol NaOH.
– Hess’s law:
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Given a series of reactions, rearrange to find H for the reaction in question:
Example problem:
Given the data:
N2(g) + O2(g) ----> 2NO(g)
H =180.7 kJ
2NO(g) + O2(g) -----> 2NO2
H = -113.1 kJ
2N2O(g) -----> 2N2(g) + O2(g) H = -163.2 kJ
Calculate: N2O(g) +NO2(g) -----> O2(g)
– Enthalpies of formation (the tables of enthalpies of formation):
– Hprdts - Hreactants = Hreaction
Chapter 5
– Hess’s law:
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Given a series of reactions, rearrange to find H for the reaction in question:
Example problem:
Given the data:
N2(g) + O2(g) ----> 2NO(g)
H =180.7 kJ
2NO(g) + O2(g) -----> 2NO2
H = -113.1 kJ
2N2O(g) -----> 2N2(g) + O2(g) H = -163.2 kJ
Calculate: N2O(g) +NO2(g) -----> 3NO(g)
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NO2 ---- NO(g) + 1/2O2(g)
113.1/2
N2O(g) -----> N2(g) + 1/2O2(g) H = -163.2/2 kJ
N2(g) + O2(g) ----> 2NO(g)
H =180.7 kJ
N2O(g) +NO2(g) -----> 3NO(g) H =155.6 kJ
Chapter 5
– Enthalpies of formation (the tables of enthalpies of formation):
– Hprdts - Hreactants = Hreaction
Chapter 6
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Chapter 6
– Characteristics of waves (v = ):
• What is wavelength?
• What is frequency?
– Electromagnetic radiation:
• E = h
• visible spectrum (ROYGBV)
• rest of spectrum
Chapter 6
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Chapter 6
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Characteristics of waves (v = ):
Black body radiation
Photo-electric effect
Heisenberg uncertainty: (mvx ≥ h)
Line spectra of atoms
matter waves (De Brogli)
v  
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v
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E  mv  h  h
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h
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mv
v
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Chapter 6
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Chapter 6
– Wavefunctions and quantum mechanics
• wavefunction vs. probability distribution
• orbitals and quantum numbers
– Quantum numbers
• what are the four?
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principle (energy) n = 1,2,3...
azimuthal (shape) l = 0,1, 2... n-1
magnetic (orientation) ml = -l,...0,...+l
spin (differentiates two electrons in same orbital) (±1/2)
• naming the l qm:
• l=0, s, l=1, p, l=2, d, l=3, f
– Shapes of orbitals
Chapter 6
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Chapter 6
– Many electron atoms
– Energy of orbitals in H versus other atoms with other electrons.
Chapter 6
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Chapter 6
– Pauli exclusion principle
– Hund’s rule (don’t pair until you have to)
– Electron configurations