Use the partition function for a harmonic oscillator, q = 1 / (1- e^-hv/kBT) , to determine the ratio of populations in the lowe
st and first excited energy levels, P0/P1 , for the indicated systems at two temperatures, 50 K and 298 K :
(a) The 1700 cm−1 stretching vibrational mode for the carbonyl (C=O) group in the peptide bond of a polypeptide.
(b) the 150 cm−1 torsional vibrational mode for the C=C bond in trans-2-butadiene.
(a) The 1700 cm−1 stretching vibrational mode for the carbonyl (C=O) group in the peptide bond of a polypeptide.
(b) the 150 cm−1 torsional vibrational mode for the C=C bond in trans-2-butadiene.
2 answers:
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Answer : The correct option is, (B) 0.11 M
Solution :
First we have to calculate the concentration and
.
The given equilibrium reaction is,
Initially 0.70 0.70 0
At equilibrium (0.70-x) (0.70-x) x
The expression of will be,
Now put all the given values in the above expression, we get:
By solving the term x, we get
From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.
Thus, the concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of at equilibrium = x = 0.59 M
Therefore, the concentration of at equilibrium is 0.11 M