Problem 4 Molecular Partition Function [60 points]. Consider the fluoromethane (

Question

Problem 4 Molecular Partition Function [60 points]. Consider the fluoromethane (CH3F) molecule in its quartet electronic ground state confined to a vesel of volume V- LLAssume that each of its center of mass and internal degrees of freedom are inde- pendent from all others a. Above a certain temperature, the molecule is in the high temperature limit with respect to both translations and rotations. What conditions must be satisfied to be in the high temperature limit for translations and rotations? b. Each vibrational degree of freedom j of CHjF is dscribed by the energy where w is the natural frequency anxis the quantum number of the jth vibrational mode. By starting from the vibrational partition function for one such mode, () ? c-o exp(-?), surn the series to find a closed-fonn expression for the partition function. Be sure to account for zero-point motion. c. Is the result from Part b valid in the high or low temperature limit, or is valid for all temperatures? Explain

Explanation / Answer

Answer:

a)
mass of Fe= 15 g
molar mass of Fe = 56 g/mol
number of moles of Fe = mass/molar mass
      = 15/56 mol
      = 0.268 mol

from reaction equation:
moles of Fe2O3 = moles of Fe/2
   =0.268/2
   =0.134 mol
molar mass of Fe2O3 = 159.7 g/mol
mass of Fe2O3 = number of moles * molar mass
   =0.134*159.7
   = 21.4g

moles of Al = moles of Fe
   = 0.268 mol
molar mass of Al = 27 g/mol
mass of Al = number of moles * molar mass
   = 0.268*27
   = 7.2 g

b)
From given reaction
moles of Al2O3 = moles of Fe/2
   =0.268/2
   =0.134 mol
molar mass of Al2O3 = 102 g/mol
mass of Fe2O3 = number of moles * molar mass
   = 0.134*102
   = 13.7g

c)
mass Al2O3 = 93% of maximum
   = 0.93*13.7
   = 12.7 g

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