The given moment of inertia for C3 are I=39.340muA2∗, 39.032muA2∗ and 0.3082muA2∗
The value of mu is 1.66054×10-27 kg.
The given vibrational wavenumbers for C3 are 63.4, 1224.5, and 2040 cm−1.
The molar mass of C3 is 36.033 g/mol.
The given temperature is 200 K.
The value of Planck’s constant is 6.626×10−34 J⋅s.
The value of speed of light is 2.998×1010 cm/s.
The expression that is used to calculate the standard molar Gibbs energy, GmΘ, is mentioned as follows.
GmΘ(2000 K)−GmΘ(0)=RTlnqm0NA..................(1)
Where,
qm0 is the molecular partition function.
R is the universal gas constant.
T is the temperature.
qm0NA=qmTNAqRqVqRqE
In the above expression, qm0NA is equal to qmTNA is Translational partition function, qR is Rotational partition function, qE is Electronic partition function and qV is Vibrational partition function.
The expression for translational partition function, qmTNA for C3 at 10 K given below.
qmTNA=kTpΘ(2πmkTh)3=(2πk)3/2pΘh3(T/K)5/2(M/gmol−1)3/2
Where,
h is the Planck’s constant.
M is the molar mass of C3.
k is the Boltzmann constant.(1.38×10−23 J K−1)
Substitute (2πk)3/2pΘh3 as 2.561×10−2, T as 10 K and M as 36.033 g/mol in the above equation.
qmTNA=2.561×10−2×(10)5/2(36.033 g/mol)3/2=2.561×10−2×(316.2278)×(216.2971)=1752
Thus the value of translational partition function for C3 at 10 K is 1752.
The expression for rotational partition function, qR for C3 which is a nonlinear molecule given below.
qR=1σkThc(πABC)1/2=1.0270σ×(T/KABC/cm−3)3/2...............(2)
Where,
k is the Boltzmann constant.(1.38×10−23 J K−1)
h is the Planck’s constant.
c is the speed of light.
ABC are the rotational constants of the molecules along three directions.
σ is the symmetry number.
The value of ABC is calculated by the expression given below.
ABC=(ℏ4πc)31IAIBIC
Substitute the values of ℏ as 1.055×10−34 J s, c 2.998×1010 cm/s and moment of inertia for C3 are IA=39.340muA2∗, IB=39.032muA2∗ and IC=0.3082muA2∗ in the above equation.
ABC=(1.055×10−34 J s4×3.14×2.998×1010 cm/s)31mu(39.340A2∗×39.032A2∗×0.3082A2∗)=(1.055×10−34 J s4×3.14×2.998×1010 cm/s)3×1060 m−6(1.66054×10−27 kg)3×A6∗(39.340A2∗×39.032A2∗×0.3082A2∗)=2.1993×10−5−102−30kg3 m6 s−3cm3 s−3×1060 m−6(1.66054×10−27 kg)3×A6∗(39.340×39.032×0.3082)=10.134 cm−3
Substitute the values of ABC as 10.134 cm−3, T=10 K and σ as 2 in equation (2).
qR=1.02702××((10)3/2(10.134)1/2)=1.02702×(31.623.18)=5.1......................(3)
Thus, the value of rotational partition function of C3 at 10 K is 5.1.
The vibrational degrees of freedom of C3 is given below.
qV=(11−e−hcv˜kT)..........................(4)
Where,
ν¯ is the vibrational wave number.
Substitute the value of h, c, k T and ν¯ as 63.4 cm−1 in equation (4).
qV=(1−exp(6.626×10−34 J s×2.998×1010 cm s−1×63.4 cm−11.38×10−23 J K−1×10 K))−1=(1−exp(−91.2×10−34+33))−1=(1−1.0945×10−4)−1=1.0001
Thus, the vibrational number for C3 at 10 K is 1.0001.
The degeneracy of electronic ground sate is qE=1 because the lowest-mode of partition function is also one.
Substitute the value of qE, qT, qR, R as 8.3145 J mol−1 K−1, T and qV for C3 in equation (1).
GmΘ(10 K)−GmΘ(0)=8.3145 J mol−1 K−1×10 Kln[(152)(5.1)(1)(1)]=756.43 J mol-1_
Therefore, the value of GmΘ(10 K)−GmΘ(0) for C3 is 756.43 J mol-1_.
For 100 K, substitute (2πk)3/2pΘh3 as 2.561×10−2, T as 100 K and M as 36.033 g/mol in the above equation.
qmTNA=2.561×10−2×(100 K)5/2(36.033 g/mol)3/2=2.561×10−2×(100000)×(216.2971)=5.54×107
Thus the value of translational partition function for C3 at 100 K is 5.54×107.
The expression for rotational partition function, qR for C3 which is a nonlinear molecule given below.
qR=1σkThc(πABC)1/2=1.0270σ×(T/KABC/cm−3)3/2......................(2)
Where,
k is the Boltzmann constant.(1.38×10−23 J K−1)
h is the Planck’s constant.
c is the speed of light.
ABC are the rotational constants of the molecules along three directions.
σ is the symmetry number.
The value of ABC is calculated by the expression given below.
ABC=(ℏ4πc)31IAIBIC
Substitute the values of ℏ as 1.055×10−34 J s, c 2.998×1010 cm/s and moment of inertia for C3 are IA=39.340muA2∗, IB=39.032muA2∗ and IC=0.3082muA2∗ in the above equation.
ABC=(1.055×10−34 J s4×3.14×2.998×1010 cm/s)31mu(39.340A2∗×39.032A2∗×0.3082A2∗)=(1.055×10−34 J s4×3.14×2.998×1010 cm/s)3×1060 m−6(1.66054×10−27 kg)3×A6∗(39.340A2∗×39.032A2∗×0.3082A2∗)=2.1993×10−5−102−30kg3 m6 s−3cm3 s−3×1060 m−6(1.66054×10−27 kg)3×A6∗(39.340×39.032×0.3082)=10.134 cm−3
Substitute the values of ABC as 10.134 cm−3 T=100 K and σ as 2 in equation (2).
qR=1.02702×((100)3/2(10.134)1/2)=1.02702×(10003.18)=0.5135×314.47=161.48
Thus, the value of rotational partition function of C3 at 100 K is 161.48.
The vibrational degrees of freedom of C3 is given below.
qV=(11−e−hcv˜kT)..................(4)
Where,
ν¯ is the vibrational wave number.
Substitute the value of h, c, k T and ν¯ as 63.4 cm−1 in equation (4).
q1V=(1−exp(6.626×10−34 J s×2.998×1010 cm s−1×63.4 cm−11.38×10−23 J K−1×100 K))−1=(1−exp(91.26×10−3))−1=(1−1.09555)−1=10.465
Thus, the vibrational number for C3 at 100 K is 10.465.
The degeneracy of electronic ground sate is qE=1 because the lowest-mode of partition function is also one.
Substitute the value of qE, qT, qR, R as 8.3145 J mol−1 K−1, T and qV for C3 in equation (1).
GmΘ(10 K)−GmΘ(0)=8.3145 J mol−1 K−1×100 Kln[(5.54×107)(161.48)(10.465)(1)]=2.10×104 J mol-1_
Therefore, the value of GmΘ(10 K)−GmΘ(0) for C3 at 100 K is 2.10×104 J mol-1_.