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