In this problem we will deal with vibrations in hydrogen fluoride (HF). a. Use t
ID: 890575 • Letter: I
Question
In this problem we will deal with vibrations in hydrogen fluoride (HF).
a. Use the harmonic oscillator model for 1H19F to determine its force constant, if the harmoic frequency (vobs) is 4138.52 cm-1
b. Use your results from (a) to predict the harmonic frequency of 2H19F
c. It turns that the experimentally observed frequency (vobs) for 2H19F is 2998.25 cm-1. Provide possible sources for the discrepancy between your result in part (b) and this value.
In this problem we will deal with vibrations in hydrogen fluoride (HF) a. Use the harmonic oscillator model for H1'F to determine its force constant, if the harmonic frequency (Vobs) is 4138.52 cm It turns that the experimentally observed frequency (vobx) for H"F is 2998.25 and this value. 21 T19 b. Use your result from (a) to predict the harmonic frequency of HF c. It turns that the experimentally observed frequency (Vobs) for F is 2998.25 e. cm". Provide possible sources for the discrepancy between your result in part (b) cm"'. Provide possible sources for the discrepancy between your result in part (b)Explanation / Answer
a) The force constant of the HF molecule is
K = (2 f) 2 µ
Where µ = reduced mass = 1/m1 + 1/m2 = m1+m2 / m1m2 = ( 19 + 1) 6.66 × 1027 kg / 19X1 (6.66 × 1027 )2
µ = reduced mass = 20 / 6.66 × 1027
Now wavenumber is given as = 4138.52 cm-1
So frequency / wavenumber = speed of light
Frequency = 3X 10^8 X 4138.52 X 100 = 12415 X 10^10 = 12.415 X 10^13
(2 ×12.415 × 1013) 2 × 19 × 1.6605 × 1027 / 20 = 956.75 N/m
b) Frequency
K = (2 f) 2 µ
(956.75 / µ )1/2 / 2= frequency
Frequency = (956.75 X 6.66 × 1027 / 20 )1/2 / 2 X 3.14
Frequency = 1.784 X X 10^-13 / 6.28 = 0.2840 X 10^13
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