Scenario: In fireworks (See Reactions 1-3 below), the heat released during a com

Question

Scenario: In fireworks (See Reactions 1-3 below), the heat released during a combustion reaction (reaction 1) excites metal electrons in some salts (reaction 2). The photon of emitted electromagnetic radiation when an excited electron returns to a lower energy level is in the visible range (reaction 3). Strontium chloride emits a strong light at wavelength 641 nm and barium chloride salts have a strong emission around 493 nm. Assume a. all the light energy emitted comes from complete conversion of the heat energy absorbed, b. only one photon per atom is released, and c. the electron makes only one transition to emit the photon. KClO_4(s) + 3C_(graphite) + O_2(g) rightarrow KCl(s) + 3CO_2(g) + Heat Reaction 1. Heat + BaCl_2 + SrCl_2 Ba*Cl_2 + Sr*Cl_2 (*=excited states) Reaction 2. Ba*Cl_2 + Sr*Cl_2 rightarrow BaCl_2 + SrCl_2 + light Reaction 3. What color is the light emitted by each metal as it returns to ground state? Energy calculations Calculate the E absorbed by each metal ion using the photon wavelengths above. Then calculate the E released per mol of each salt when returning to ground state (leaving it in kJ/mol). Write the electron configuration for each cation. Use the answers to #2 and #3 and your knowledge of atomic energies to explain the difference in emission wavelengths for Ba and Sr ions. (You can draw an energy diagram or bore diagram to help visualize.) In Reaction 1 above, write oxidation states for each element and identify what was reduced and oxidized. Calculate the Delta H_rxn degree for Reaction 1 above. Use Delta H_f degree values found in Appendix B of the textbook. If 5 barium ions are excited for every potassium perchlorate reacted instead of 1 barium ion, How would Reaction 2 above change? Use the new reaction to determine the smallest mass (g) of potassium perchlorate needed to completely use 1 mol of BaCl_2. Give one example of how commonly used assumptions could result in any of these calculated answers being different from reality. (Sample answer: The average bond energies used to calculate Delta H_rxn degree do not represent the high temperatures seen in fireworks which could underestimate the enthalpy change.)

Explanation / Answer

1. What color is the light emitted by each metal as it returns to ground state?

Ans. It was mentioned in the question
   SrCl - 641 nm - orange color
   BaCl - 493 nm - Green color
When the heat is supplied to the atom the electrons gets excited and from ground state to excited state, and the atom is so unstable in exicted state as it falls from excited state to ground state it emits visible light.

2. Energy calculations

a. E absorbed
SrCl - 641 nm - orange color
BaCl2 - 493 nm - green color

Plancks equation E = h X
Where    h = 6.63 × 10-34 J·s

   = frequency

first you need to find frequency for the atom

= c / where c = 3.00 × 108 m/s; = wavelength

For SrCl
Find frequency = (3.00 × 108 m/s) / (641 nm X (1 m / 1 X 109 nm))
(we need to convert the wavelength into m to normalise the units)

   = 4.68 X 1014 1/s
Now we find Energy E
   E = (6.63 × 10-34 J·s) X (4.68 X 1014 1/s)
   E = 3.10 X 10-19 J/photon

For BaCl

Find frequency = (3.00 × 108 m/s) / (493 nm X (1 m / 1 X 10 9 nm))
(we need to convert the wavelength into m to normalise the units)

   = 6.08 X 1014 1/s
Now we find Energy E
   E = (6.63 × 10-34 J·s) X (6.08 X 1014 1/s)
   E = 4.0 X 10-19 J/photon

3. Electronic configuration of each cation
Sr (38)- 1s2 2s2 2p6 3s2 3p6 4s2 4p6 5s2 - ground state
Sr +2 - 1s2 2s2 2p6 3s2 3p6 4s2 4p6 (two 5s electrons were lost)

Ba (56)- 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 - ground state
Ba+2 - 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 (two 6s electrons were lost)

5. In reaction 1 above write oxidation states

KClO4 + 3C + O2 ----> KCl + 3CO2 + heat

KClO4 = K+, Cl-, O-2
O2 = -2
KCl = K+, Cl-
CO2 = C+4, O-2

KClO4 has been reduced to KCl
Carbon (graphite) is oxidised to CO2

Question number 6 required appendix.

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