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