Question

Predict the vibrational contribution to the standard molar entropy of methanoic acid (formic acid, HCOOH) vapour at (i) 298 K (ii) 500 K. The normal modes occur at wavenumbers 3570, 2943, 1770, 1387, 1229, 1105, 625, 1033, 638 cm-1.

Answer 1

the vibrational contribution to the standard molar entropy of methanoic acid :

For this first , we calculate Vibrational partition function at given temperature :

at 298 K

Vibrational partition function , q_v

T = 298 K

h₀​​​​​​​/kT = hcv/kT

For each vibrational frequency :

1. 3570 cm-1

hcv/kT = 6.626*10^-34 J-s * 3*10^10 cm/s * 3570 cm-1 / 1.38*10^-23 J/K *298 K = 17.25

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-17.25)) = 1.000

2. 2943

hcv/kT = 6.626*10^-34 J-s * 3*10^10 cm/s * 2943 cm-1 / 1.38*10^-23 J/K *298 K = 14.22

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-14.22)) = 1.000

3. 1770

6.626*10^-34 J-s * 3*10^10 cm/s * 1770 cm-1 / 1.38*10^-23 J/K *298 K = 8.55

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-8.55)) = 1.000

4. 1387

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-6.7)) = 1.001

5. 1229

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-5.94)) = 1.003

6. 1105

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-5.33)) = 1.005

7. 625

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-3.02)) = 1.051

8.1033

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-4.99)) = 1.007

9. 638 cm-1

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-3.08)) = 1.048

Overall q_v = 1.000*1.000*1.000*1.001*1.003*1.005*1.051*1.007*1.048 = 1.119

Similarly :

at 500 K

1. 3570 cm-1

hcv/kT = 6.626*10^-34 J-s * 3*10^10 cm/s * 3570 cm-1 / 1.38*10^-23 J/K *500 K = 10.28

q_v = 1/ (1- exp(-h₀​​​​​​​/kT)) = 1 / (1- exp(-10.28)) = 1.000

at 2943 : 1.000 , 1770 : 1.015, 1387: 1.037, 1229 : 1.056,   1105 : 1.076 , 625 : 1.289 , 1033 : 1.092 , 638 : 1.277.

Overall q_v = 1.000*1.000*1.015*1.037*1.056*1.076*1.289*1.092*1.277 = 2.15

From generalized equation :

Vibrational contribution to standard Molar entropy :

S_v = R ln q_v

standard molar entropy of methanoic acid (formic acid, HCOOH) vapour

at (i) 298 K   : S_v = R ln 1.119 = 0.935 J/K-mol

(ii) 500 K.     :S_v = R ln 2.15 = 6.364 J/K-mol