Question

Challenge: A Different Approach to Calculating Equilibrium Constants

Answer 1

In this challenge we are going to calculate the equilibrium constant for the reaction:
\[Na_{2} (g) <-> 2 Na (g)\]
At 1500 K we have got the following spectroscopic data:
The dissociation energy D0 equals 70.4 kJ/mol
The rotational constant B_Na2 equals 0,1547 cm^-1
The vibrational constant v_Na2 equals 159,2 cm^-1
In addition to that sodium is 2 times degenerated in the ground state.

Answer 2

Hint to solve the problem: Rewrite Gibbs' free energy so it depends of the molecular partition function and write up the contributions to the molecular partition function for Na and Na2. We assume at 1500 K that Na and Na2 acts as a ideal gas.

Answer 3

Good luck :)

Answer 4

Shouldn't that be Na*
Where * means an off electron

Answer 5

As I have read this eq. :
\(Na_2 <-> 2Na*\)

Answer 6

Actually, I was trying for Na* only.......

Answer 7

As I haven't wrote a charge I found it unnecessary to show it contains a single electron.

Answer 8

Thanks

Answer 9

Challenge questions shouldn't be anything higher than thermodynamics. Or o-chem.

Answer 10

It is thermodynamics. :)

Answer 11

But I sure like to post the solution if people like.

Answer 12

This challenge actually test your understanding for the equilibrium constant.

Answer 13

This is like university level statistical mechanics... not a chance anyone here would be able to solve without Googling.

Answer 14

First of all,

\(K = \cfrac{(\cfrac{q^{0} Na}{N_A})^2 }{(\cfrac{q^{0} {Na}_2 }{N_A})} e^{-\beta \Delta \epsilon } \)

where, \(\Delta \epsilon = -D_0 \)

Total partition function for \(Na \cdot\) is,

\(\mathsf{{q^0}_{Na} = {q^{\mathsf{Trans}}}_{Na} \times {q^{\mathsf{elec}}}_{Na} \\
\quad \quad \space = {2q^{\mathsf{Trans}}}_{Na} = 9780.54 * 10^{34} }\)

Also, molar volume :

\(\mathsf{V =\cfrac{RT}{P} } \) = \(0.12308 m^3 \)

\(\mathsf{Since ~ P = 1.01325 * 10^{5} ~ Pa ; ~ T = 15000 K }\)

Transitional partition function for \(Na \cdot\) is ;

\(\mathsf{{q^{\mathsf{Trans}}}_{Na} = \cfrac{V \times (2\pi m_{Na} k T )^{\cfrac{3}{2}} }{h^3} }\)

\(\quad \quad \quad \mathsf{= \cfrac{0.12308 \times [2\times 3.14 \times \cfrac{23.5}{6.023 \times 10^{23} } \times 1.38 \times 10^{-23} \times 1500]^\cfrac{3}{2} }{(6.6 \times 10^{-34})^{3} }}\)

\(\mathsf{Since ~ {Na}_2 ~ has ~ no ~ unpaired ~ e^{-} . \\
Therefore ~ electronic ~ partition ~ function ~ = 1}\)

\(\mathsf{Therefore ~ Total ~ paritition ~ function ~ for ~ {Na}_2 ~ molecule ~ is, \\
{q^0}_{{Na}_2} = {q^{trans}} _ {Na_2} {q^{rot}} _ {Na_2} {q^{vib}} _ {Na_2} {q^{elec} }_ {Na_2} \\
\quad = {q^{trans}} _ {Na_2} {q^{rot}} _ {Na_2} {q^{vib}} _ {Na_2} }\)

Since,

\(\mathsf{\cfrac{{q^{trans}}_{Na_2}}{{q^{trans}}_{Na} } = (\cfrac{m_{Na_2} }{m_{Na}})^\cfrac{3}{2} = 2^\cfrac{3}{2} }\)

Therefore, \(\mathsf{{q^{trans}}_{Na_2} = 2^\cfrac{3}{2} \times {q^{trans}}_{Na} = 13829.694 \times 10^{34} }\)

Also, \(\mathsf{{q^{rot}}_{Na_2} = \cfrac{T}{2\theta _R} \\
\quad \quad = \cfrac{kT}{2ncB} \\
\quad \quad = \cfrac{1.38 \times 10^{-23} \times 1500}{2 \times 6.6 \times 10^{-34} \times 3 \times 10^8 \times 0.1547 \times 10^2 } \\
\quad \quad = 3378.974 }\)

Also,

\(\mathsf{{q^{vib}}_{Na_2} = \cfrac{1}{1-e^{\cfrac{-nc\nu }{kT}}}}\)

\(\mathsf{\quad \quad = \cfrac{1}{1-e^{\cfrac{-6.6 \times 10^{-34} \times 3 \times 10^{8} \times 1592 \times 10^2 }{1.38 \times 10^{-23} \times 1500}}}}\)

\(\mathsf{ = 1.278944 } \)

Put all the values and solve :

\(\mathsf{K = \cfrac{({q^0}_{Na_2} )^2}{N_A ({q^0}_{Na_2} )} \times e^{\cfrac{70400}{RT}}}\)

\(\mathsf{K = 751.916599 \times 10^{10}}\)

Answer 15

I never tried such questions in past. This is my first time, so pardon my mistakes (if there are).

Answer 16

Well written mathslover i got to say I'm impressed, however your expression for the equilibrium I'm unsure how you got to:
Lets observe the following expression for the equilibrium constant:
\[\ln(K)=-\frac{ \Delta _{r}G ^{\Theta} }{ RT }\]
For a ideal gas you can show that:
\[G _{m}-G _{m}(0)=-RT \ln(\frac{ q }{ N_{A} })\]
But the standard reaction Gibbs free energy is defined as:
\[\Delta _{r}G ^{\Theta}=\sum_{J}^{}v _{j}G_{m,j}^{\Theta}\]
Then it follows that we have to use standard molecular Gibbs energies of each component j:
\[G_{m,j}^{\Theta}=G _{m,j}^{\Theta}(0)-RT \ln(\frac{ q _{m,j}^{\Theta} }{ N_{A} })\]
Here is q_m,j,theta the standard molecular partition function of the component j.
So the relationship between the standard reaction Gibbs free energy and the standard molecular partition function:
\[\Delta _{r}G ^{\Theta}=\sum_{J}^{}v _{J}G _{m,J}^{\Theta}=\sum_{J}^{}v _{J}G _{m,J}^{\Theta}(0)-RT \sum_{J}^{}v _{j}\ln(\frac{ q _{m,J}^{\Theta} }{ N _{A} })\]
But for a ideal gas it follow at T=0 K that G(0)=A(0)=U(0) and thereby:
\[\sum_{J}^{}v _{J}G _{m,J}^{\Theta}(0)=\sum_{J}^{}v _{J}U _{m,J}^{\Theta}(0)=\Delta _{r}U ^{\Theta}(0)\]
But is different from the internal energy at T=0 and therefor is nothing else than the difference in energies in the lowest state of all components:
\[\Delta _{r}U ^{\Theta}(0)=N _{A}\sum_{J}^{}v _{J} \epsilon _{0, J} \equiv \Delta _{r}E _{0}\]
Put everything together so we for the reaction Gibbs free energy get:
\[\Delta _{r}G ^{\Theta}=\Delta _{r}E _{0}-RT \sum_{J}^{}v _{J} \ln(\frac{ q _{m,J}^{\Theta} }{ N _{A} })\]
Compair with out first expression for the equilibrium constant and we get the equation:
\[K=\exp(-\frac{ \Delta _{r}E _{0} }{ RT }+\sum_{J}^{}v _{J}\ln(\frac{ q _{m,J}^{\Theta} }{ N _{A} }))\]
We can rewrite that using a power calculation rule (the a^(x+y)=a^x*b^y):
\[K=\exp(-\frac{ \Delta _{r}E _{0} }{ RT })\prod_{J}^{}\left( \frac{ q _{m,J}^{\Theta} }{ N _{A} } \right)^{v _{J}}\]
So think you used the the Avogadro constant one time to much.

Answer 17

You frightened me , I got this equation in a formula bank and I used it .

Answer 18

This is out of my syllabus, in 10th standard, I never learnt such things, so sorry that I will not be able to explain you that formula. But thanks for the medal :)

Answer 19

I think you did a awesome try! :D

Answer 20

Thanks for the quest.

Answer 21

The molecular partition functions:

Answer 22

Why Na is two times degenerated is due to the unpaired electron can have spin up or down if I forgot to tell that.

Answer 23

And the results: :)

Answer 24

But I'm a undergraduate @abb0t :( on his first year :(

Answer 25

Ok.

Answer 26

how the hell are you a first year undergraduate in statistical mechanics?

Answer 27

It's like asking how to fly a plane even though you haven't learned to drive a car yet

Answer 28

@oldrin.bataku I can show you this is what the course description say:
"After the course, students should be able to:
• define and explain the Laws of Thermodynamics
• define and apply the thermodynamic variables T and p, the statistical thermodynamic partition functions q and Q and the thermodynamic functions U, S, H, A and G and their partial derivatives and the electrochemical potential E.
• calculate the activity of an ion in a dilute electrolyte solution by using the Debye-Hückel expression.
• specify the conditions for spontaneous processes, and explain the work (on the environment) and heat generation by their course
• specify the condition for a chemical reaction is spontaneous and derive eqilibrium
• apply the thermodynamic functions to describe the phase equilibria and phase transitions in physico-chemical systems
• apply the thermodynamic functions of mixtures and chemical reactions
• apply the thermodynamic functions of the electrolyte and electrochemical reactions
• write the redox process of a component of an electrochemical reaction and its electrochemical potential using the Nernst's law
• apply the thermodynamic functions of biochemical processes: membrane potentials and kolloidosmotisk pressure
• apply the statistical thermodynamic partition functions q and Q to calculate the thermodynamic functions U, S, H, A, G and equilibrium constant of phase equilibria or chemical reactions
• write the expression of a molecule's speed and the mean free path length in a gas
• define and explain the spontaneous processes time courses, including chemical kinetics
• Explain in simple enzyme-controlled reactions"