# How to treat stretch bond to calculate its enthalpy and entropy contribution?

##### 1 Answer

For that you'll need to calculate the vibrational partition function, which allows you to get the energy, and then the enthalpy contribution from just vibrations. The entropy comes directly from the vibrational partition function and energy.

Since you apparently want this to be done using only frequencies and a temperature...

#Theta_(vib,i) = tildeomega_i/k_B# ,#k_B = "0.695 cm"^(-1)"/K"# ,#tildeomega_i = "frequency of that stretching mode in cm"^(-1)# .

#q_(vib,i) = e^(-Theta_(vib,i)//2T)/(1 - e^(-Theta_(vib,i)//T))#

#q_(vib) = prod_(i) q_(vib,i) = q_(vib,1)q_(vib,2)cdots#

#(E-E_0)_(vib)/N = << epsilon >>_(vib) = k_B [Theta_(vib)/(2) + Theta_(vib)/(e^(Theta_(vib)//T) - 1)]#

#(H-H_0)_(vib)/N = << eta >>_(vib) = << epsilon >>_(vib) + k_BT#

#S_(vib)/N = << s >>_(vib) = k_B ln q_(vib) + (<< epsilon >>_(vib))/T#

Enthalpy units here are in

To check your work, you should get

(The degeneracy doesn't matter unless you want the heat capacities.)

Any **partition function** is written as

#q = sum_(i) g_i e^(-betaepsilon_i)# where

#g_i# is the degeneracy of state#i# ,#epsilon_i# is the energy of state#i# , and#beta = 1//k_BT# is a constant.#k_B# is the Boltzmann constant and#T# is temperature in#"K"# .

For **vibrational** energy states in a **one**-dimensional harmonic oscillator,

#q_(vib) = sum_(upsilon) e^(-ℏomega(upsilon + 1/2)//k_BT)#

Define the vibrational temperature as

#q_(vib) = sum_(upsilon) e^(-(upsilon + 1/2)Theta_(vib)//T)#

#= e^(-Theta_(vib)//2T)sum_(upsilon) e^(-upsilonTheta_(vib)//T)#

#= e^(-Theta_(vib)//2T)sum_(upsilon) (e^(-Theta_(vib)//T))^(upsilon)#

this has a nice power series solution of

#color(blue)(q_(vib) = e^(-Theta_(vib)//2T)/(1 - e^(-Theta_(vib)//T)))#

Next, the **molecular energy** is

#color(blue)(<< epsilon >>_(vib)) = -((del ln q_(vib))/(del beta))_V#

#= k_BT^2((del ln q_(vib))/(del T))_V#

#= k_BT^2((del ln [e^(-Theta_(vib)//2T)/(1 - e^(-Theta_(vib)//T))])/(del T))_V#

#= k_BT^2((del(-Theta_(vib)/(2T) - ln(1 - e^(-Theta_(vib)//T))))/(del T))_V#

#= k_BT^2 [Theta_(vib)/(2T^2) - 1/(1-e^(-Theta_(vib)//T)) cdot -Theta_(vib)/T^2 cdot e^(-Theta_(vib)//T)]#

#= k_B [Theta_(vib)/(2) + (Theta_(vib)e^(-Theta_(vib)//T))/(1-e^(-Theta_(vib)//T))]#

#= color(blue)(k_B [Theta_(vib)/(2) + Theta_(vib)/(e^(Theta_(vib)//T) - 1)])#

If we want the enthalpy,

#C_V/N = ((del << epsilon >> )/(del T))_V#

then

#C_V/N + k_B = C_P/N = ((del << epsilon >> )/(del T))_V + k_B#

and so, the **vibrational molecular enthalpy** is:

#(H-H_0)_(vib)/N = color(blue)(<< eta >>_(vib))#

#= int ((del << epsilon >> )/(del T))_V + k_B dT#

#= color(blue)(<< epsilon >>_(vib) + k_BT)#

Finally, if we want the molecular entropy, then since I'm lazy, here is the TOTAL entropy from my homework. An equilibrium distribution of particles in everyday life allows this derivation.

And the vibrational part only is for internal motions. Thus, the **vibrational molecular entropy** is:

#S_(vib)/N = color(blue)(<< s >>_(vib) = k_B ln q_(vib) + (<< epsilon >>_(vib))/T)#