The flames of Romance The flames of Romance Candlelight and Chemistry Molecular spectroscopy and...
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Transcript of The flames of Romance The flames of Romance Candlelight and Chemistry Molecular spectroscopy and...
The flames of Romance
Candlelight and Chemistry
Molecular spectroscopy and reaction dynamics
Arnar HafliðasonApril 10th 2015
The beauty of science
• Lets begin with Richard Feynman
• The story about the flower and his artist friend
You, light up my life
• What is a candle made of..?
• What is needed for it to burn..?
• What is, “to burn”..?
• Why is fire, yellow.. Why is fire, blue?
• Let’s gaze into the flames and see what’s cookin’?
What is candle made of?
• Paraffin wax C31H64 • Or actually a mixture of long hydro-
carbon molecules, i.e. CnH2n+2, ranging from n=25-40.
• Paraffin wax is a white or colorless soft solid derivable from petroleum
• Candle wick, a braided cotton that holds the flame of a candle
What is needed for it to burn? • Prerequisite is OXYGEN
• Oxygen around is usually enough, though you might want to add more oxygen to spice things up
• You need the “spark” for the chemistry to happen
• The process needs to be exothermic to sustain itself
• The rate of the chemical reactions needs to be fast enough to keep the process going
• Some source of hydrocarbons (gas, wax) for the oxygen to react with
What is, “to burn”?
• When the hydrocarbons start “to burn”, a chemical reaction occurs. The light that we see emitted is caused by the reaction of hydrocarbons with oxygen, a rapid oxidation process, also know as combustion:
+ Energy
Exothermic Reaction Energy
Energy from exothermic reaction
Divided into 2 main groups
1. Kinetic Energy (Vibration, Rotation, translation)• Ekin = nkBT (Increase in temperature => more energy)HEAT
2. Radiation Energy (emission following e- transfer)• E = h n = h(c/ )l (shorter wavelength => more energy)LIGHT
What is, “to burn”?
• When the hydrocarbons start “to burn”, a chemical reaction occurs. The light that we see emitted is caused by the reaction of hydrocarbons with oxygen, a rapid oxidation process, usually called combustion:
+ Energy
Exothermic Reaction
→(…+𝑪𝑯+𝑶𝑯+𝑪𝟐+…)→?????
Could this explain the different colors in the fire??
Energy
Reactive radicals are formed
Why is fire, yellow.. Why is fire, blue?
… it depends on what you’re burning … and what the heat is when it’s burning
1) Just gas, no extra oxygen (candle)
2) Gas, and little bit of oxygen
3) Gas, and oxygen
4) Gas, and a lot of oxygen
Lets connect a gas-burner: propane gas cylinder and oxygen cylinder
Heat from steps 1) – 4) estimated around 1000 K – 3000 K.
1) Yellow because of incomplete com-bustion caused by lack of oxygen. What we see as yellow/white is soot (Cn(s)) that is glowing
Less combustion – Less heat – less blue
Propane/Oxygen
4) Blue because of “complete” combustion caused by abundance of oxygen. Notice the flame is almost clear above the blue inner core
More combustion – More heat – more blue
Gaze into the flames and see what’s cookin’
• Monochromator• separates light
into wavelengths
Experimental setup
/Mono-chromator /
gas burner
/PMT/inlet slit
Diffraction grating Source about 4 cm from slit
Opening is 5x5 mm
Slit settings: 10, 30 and 50 μm
Measured emission spectra
Radicals and radiation
• C2 and OH radicals• Emission at 516 nm, the C2 radical is in
excited electronic state, it relaxes to a lower energy state, d3Π → a3Π, giving of radiation equal to that energy-difference
• Emission at 308.6 nm, the OH radical is in excited state and is relaxed to the ground state, A2Σ+ → X2Π, giving of radiation equal to that energy-difference
Radicals and radiation
• CH radicals• Emission at 430.7 nm, the CH radical is in
excited electronic state and relaxes to the ground state, A2Δ → X2Π, giving of radiation equal to that energy-difference
• Emission at 389.1 nm, transition: B2Σ- → X2Π• Emission at 314.7 nm, transition: C2Σ+ → X2Π
A = 430.7 nm
What do I mean by “energy-difference”?
h = 430.7 nm
𝐸=h𝜈=h𝑐𝜆
h: Planck constantc: speed of light constant: wavelength of light
CH
Potential curves for the CH molecule
B = 389.1 nm
C = 314.7 nm
= 389.1 nm = 314.7 nm
Radicals and radiation
• Emission observed is found at: 314.7, 389.1 and 430.7 nm for CH, 308.6 for OH and 516 nm for C2
Human color vision
Measured emission spectra
Vibrational and rotational structure
v’’
01
23
4
01
23
v’A2Δ
X2Π
Measurement
Location of vibrational bands
nm
Measured Calculated
Total
v’=3v’’=3
v’=2v’’=2
v’=0v’’=0
v’=1v’’=1
Rotational structure for v’=0v’’=0 transition.
4 68
10
14J’=20
Measurement
Spectral simulation at T=3000K
Simulation with PGOPHER
MeasurementSimulation
Lambda doubling
e-
Simulation on A2Δ → X2Π for v‘=0 → v‘‘=0
• B’’=14.17• A’’=29.75• D’’=0.00142
• B’=14.56• A’=-1.1• D’=0.00152
• B’’=14.19• A’’=27.95• D’’=0.00148
• B’=14.57• A’=-1.1• D’=0.00146
Calculated Constants from NISTMinimal adjustment of constants from NIST (National Institute of Standards and Technology).
B: Rotational constantD: Centrifugal distortionA: Spin-orbit coupling
cm-1
X2Π
A2Δ
Few measurements under different circumstances
1. Different quantity of oxygen burned with propane gas, 5 different settings
2. 6 different height settings of the slit from the source of the flame
3 cm3 mm
Slit
2x O2
1900 K
4x O2
3000 K3x O2
2200 K
Difference in rotational distribution. More heat is observed with increase in use of oxygen.
Simulations:
Comparing population distribution. Experiment vs. Boltzmann distribution
J=7
J=6
T = 2200 K Jmax = 6.74
3x O2
0 mm
50 mm
15 mm
3x O2
0 mm
50 mm
15 mm
Thank you all for coming
Some equations
𝐽𝑚𝑎𝑥=12 √ 2𝑘𝑇
h𝐵 𝑐−
12
𝑛 𝐽
𝑁=(2 𝐽+1)𝑒
(−𝐵𝐽 ( 𝐽 +1)h𝑐𝑘𝑇 )
Boltzmann distributionHönl-London factors
Frank-Condon factors. Can calculate transition probability between vibrational states v’ and v’’
v’’
v’
0
01
12
23
34
4
Voltage and slit
• Intensity for different settings was measured
10 15 20 25 30
0.4
0.5
0.6
0.7
0.8
0.9
1
micrometer
Inte
nsity
Change of slit opening
y = 0.033*x + 0.0033
Slit change
Fitting
900 920 940 960 980 1000
0.4
0.5
0.6
0.7
0.8
0.9
1Voltage change
Voltage
Inte
nsity
y = 0.0064*x - 5.4
Change in voltage
linear fitting
Slit settings: 10, 30 and 50 μm
Voltage settings: 900 and 1000 V
156 cm-1