From charlesreid1

Lecture: Chemical Equilibrium

Pieces to Cover:

1st Law and 2nd Law of Thermodynamics

E in terms of \mu

G in terms of \mu (constant T, constant P)

  • What is G? (c.f. Wikipedia)

What is \mu?

What is equilibrium? (Criteria)

  • Phase equilibrium analogy
  • Chemical reaction equilibrium condition
  • Minimization of Gibbs free energy (subject to element conservation)

NASA CEA (Chemical Equilibrium Analysis) Program

  • T, P - methane combustion example
  • Adiabatic flame temperature
  • Condensed phase combustion (coal)

Laws of Thermodynamics

To begin a discussion of chemical equilibrium, we can start with the 1st Law of Thermodynamics:

dE = \delta Q + \delta W

(Can someone remind me of the difference between E and W/Q?)

(Why dE and not \delta E? Why \delta Q, \delta W and not dQ, dW?)

E = system property, state property

Q, W = path-dependent

For equilibrium chemical systems, how can we simplify \delta W?

Are we considering shaft work? Electrical work?

\delta W = p dV

So now the 1st Law becomes:

dE = \delta Q - p dV

We can also simplify \delta Q, by using the 2nd Law of Thermodynamics:

T dS \geq \delta Q

and plugging this into the 1st Law gives:

dE \geq T dS - p dV

or, for reversible processes,

dE = T dS - p dV

Now, E is a state function

Meaning, it is completely characterized by S and V

E = E(S,V)

But what about multicomponent systems? Does the energy change if the mixture changes?

Now E needs to be characterized with composition, too:

E = E(S, V, N_{i})

Recall the Gibbs Phase Rule

So if we differentiate this expression, we get:

dE = \left( \frac{\partial E}{\partial S} \right)_{V,N_{i}} dS 
   + \left( \frac{\partial E}{\partial V} \right)_{S,N_{i}} dV
   + \displaystyle{ 
    \left( \frac{\partial E}{\partial N_{i}} \right)_{S,V,N_{j \neq i}}

So now let's use the other identity:

dE = T dS - p dV

So what can we say about the relationship between T and \left( \frac{\partial E}{\partial S} \right)_{V,N_{i}}?

T = \left( \frac{\partial E}{\partial S} \right)_{V,N_{i}}

Same with P:

P = \left( \frac{\partial E}{\partial V} \right)_{S,N_{i}}

Now I'm going to define an arbitrary variable, that I'll call \mu_{i}, to be equal to the last partial derivative:

\mu_{i} = \left( \frac{\partial E}{\partial N_{i}} \right)_{S,V,N_{j \neq i}}

and I'm going to call \mu_{i} the chemical potential of species i.

Okay, so, that was a lot of work - what was it for? Why did I want to get this expression for dE? Why did I want it in terms of dS and dV and dN_{i}?

Let me ask that another way... What does the following equation actually tell us?

dE = \left( \frac{\partial E}{\partial S} \right)_{V,N_{i}} dS 
   + \left( \frac{\partial E}{\partial V} \right)_{S,N_{i}} dV
   + \displaystyle{ 
    \left( \frac{\partial E}{\partial N_{i}} \right)_{S,V,N_{j \neq i}}

What does the first term tell us?

How much the internal energy E will change, if we change the entropy of the system by a very small amount.

What about the second term?

How much the internal energy E will change, if we change the volume of the system by a very small amount.

What about the third term?

How much the internal energy E will change, if we change the composition of the system by a very small amount.

Thermodynamic Relationships

Let's say we want to go into the lab and actually measure values for these terms. We fill a piston with a gas mixture. How do we measure the first term?

How do you change the entropy a fixed amount? Can we go to the engineering supply store and buy an enthalpy meter?

No! - so, it would be nice if we could rewrite this state function in terms of quantities that we can actually measure in a lab, and control.

H &=& E + PV \\
A &=& E - TS \\
G &=& H - TS

That is,

H &=& H(S,P,N_{i}) \\
A &=& A(T,V,N_{i}) \\
G &=& G(T,P,N_{i}) 

Which one do you think is the easiest to deal with?

Gibbs function - it's in terms of easy-to-keep-constant and easy-to-measure variables, T and P

So, let's write dG in the same form as dE:

dG &=& \left( \frac{\partial G}{\partial T} \right)_{P,N_{i}} dT 
     + \left( \frac{\partial G}{\partial P} \right)_{T,N_{i}} dP
     + \displaystyle{ 
       \left( \frac{\partial G}{\partial N_{i}} \right)_{T,P,N_{j \neq i}}
     } \\
dG &=& - S dT + V dP + \displaystyle{ \sum_{i=1}^{N} \mu_{i} dN_{i} }

So, can someone tell me what \left( \frac{\partial G}{\partial T} \right)_{P,N_{i}} is equal to?

By analogy, it's equal to S

And what about \left( \frac{ \partial G }{\partial P} \right)_{T,N_{i}}?

By analogy, it's equal to V

OK, and what about \left( \frac{ \partial G}{ \partial N_{i} } \right)_{T,P,N_{j \neq i}}?

Non-intuitive - but it's equal to the chemical potential... which means:

= \left( \frac{\partial E}{\partial N_{i}} \right)_{S,V,N_{j \neq i}} 
= \left( \frac{\partial G}{\partial N_{i}} \right)_{T,P,N_{j \neq i}}

And can someone remind me of the physical meaning of chemical potential? In terms of internal energy? In terms of Gibbs energy?

The amount by which the internal energy/gibbs energy of the system changes when we change the composition by a differential amount, holding S and V/T and P constant

So does everyone see the significance of this quantity? Do we have to change the compositions? Or can they change on their own?

Chemical reactions can allow the system to change its composition on its own. This means the system can respond dynamically.

So now let's talk about the relationship between Gibbs energy and equilibrium.


What is equilibrium? Anyone remember from their thermodynamics/phase equilibria class?

Let me pose an analog question. What is the condition for two phases, e.g. vapor and liquid, to be in equilibrium?

\mu_{liquid} = \mu_{gas}

What is the condition for a chemical reaction to be in equilibrium?

\mu_{products} = \mu_{reactants}

In general, the equilibrium condition is

dG = 0

For a mixture of different gas species, at constant temperature and constant pressure (i.e. the first two partial derivatives in the expression for dG above equal to zero),

\displaystyle{ \sum_{i=1}^{N} \mu_{i} dN_{i} } = 0

For any mixture, we can solve this equation numerically and determine the amount of each species at equilibrium.

What restriction do we have - what do we have to conserve?

Are the number of moles conserved?

No - number of moles can change

Atoms must be conserved. We can make or break whatever chemical bonds - but we cannot create or destroy atoms.

NASA CEA Program Tutorial


Installation only works on Windows, it does not work on Mac OS X.

Gas Phase Reactions

Example: Reacting CH4 and air

Set (end) thermodynamic state by setting T, P

Example: adiabatic flame temperature calculation

Specify hp - we want the enthalpy change to be zero

We can specify a range of equivalence ratios to construct a plot of adiabatic flame temperature vs. equivalence ratio

Why is this plot commonly used? Why is it useful?

Tells us about the energy content of the fuel

It tells us the amount of enthalpy that goes toward heating other reactants (e.g. nitrogen - very important factor in air combustion!)

Affect of the C/H ratio in the fuel

Stoichiometric mixture required to reach the peak flame temperature

Condensed Phase Reactions

What information does CEA require for a condensed phase fuel like coal?

First: what information do you have about coal?

(Elemental composition - C, O, H, S, N)

(HHV/LHV from lab analysis)

Tuesday's lecture - elemental composition + HHV/LHV --> heat of formation of coal

So the process for condensed phase calculations is:

(MOLAR Composition and Heat of Formation) --> (NASA CEA) --> (Equilibrium products composition)

by using the equation

\displaystyle{ \sum_{i=1}^{N} \mu_{i} dN_{i} } = 0

The molar composition tells us how many atoms we're starting with

The heat of formation tells us how much energy we're starting with