From WolfWikis

LECTURE 17

Chemical reactions and the liquid problem

Many important chemical reactions -if not most- are performed in solution rather than between solids or gases. Solid state reactions are often very slow and not all chemical species can be put in the vapor phase because they decompose before evaporating.

In this course we are not concerned much with the time factor of a reaction. That can be technologically very important but it is the domain of kinetics - a different branch of Physical Chemistry - rather than classical thermodynamics.. The latter is more concerned with the endpoint. This is the thermodynamically speaking the (stable) equilibrium, but chemically it can either represent a completed reaction or a chemical equilibrium.

Unfortunately, of the three main aggregation states: gas – liquid – solid, the structure of liquids is least understood and one of the most complex liquids is also one of the most extensively used ones: water. It is vital to many branches of chemistry varying from geochemistry to environmental chemistry to biochemistry. We shall make just a small inroad into its complexity.

Extent of reaction

To describe the progress of a reaction we define the extent of reaction. It is usually denoted by the Greek letter ξ (It is the Greek version of x and the Greeks say ksee, Americans zy)

Consider a simple reaction:

v_AA + v_BB => v_YY + v_ZZ

Using stoichiometry we can define the extent by considering how the number of moles (or molar amounts) of each species changes during the reaction:

reactants

n_A= n_A,0 - v_A ξ

n_B= n_B,0 - v_B ξ

products

n_Y= n_Y,0 + v_Y ξ

n_Z= n_Z,0 + v_Z ξ

The dimension of ξ is [mol] because the stoichiometric coefficients v_i are dimensionless integers. If the reaction goes to completion for one of the reactants -the limiting reactant- n_{A or B}=n_limiting will go to zero.

If we start with n_limiting= v_limiting moles, the value of ξ starts at 0 (no products) and goes to 1 at completion (limiting reactant depleted). When approaching an equilibrium ξ will not go beyond ξ_eq.

Measuring ξ

The extent of reaction is what is the central subject of reaction kinetics. Its value is typically measured as a function of time indirectly by measuring a quantity q that is linearly dependent on ξ(t):

q( ξ ) = aξ +b

Consider the situation at the extremes ξ=0 and ξ=1:

q₀= a.0+b= b

q₁= a.1+b= a+b

q₁-q₀= a

Thus, ξ can be found from [q(t)-q₀]/[q₁-q₀]=[q(t)-b]/[a]

The nature of q can vary widely from UV/Vis absorption, conductivity, gravimetric to caloric data.

In practice, q₀ at ξ=0 is often hard to observe because it takes time to mix the reactants, particularly in solutions, and q₁ at ξ=1 may never be reached if the reaction goes to equilibrium. Nevertheless the values of a and b can often be found from the available data by fitting techniques.

In (equilibrium, static) thermodynamics we are only concerned with the endpoint:

ξ=1: the reaction runs to completion

ξ=ξ_eq: the reaction goes to a state of chemical equilibrium

Thermodynamic potentials

As we have seen we can write any change in the Gibbs free energy due to changes in the molar amounts of the species involved in the reaction (at T,P constant) as:

dG =Σ ∂G/∂n_idn_i = Σμ_idn_i

where μ is the thermodynamic potential, often called chemical potential when dealing with reactions.

From the definition of ξ we can see by differentiation that

d n_A=- v_Adξ

d n_B=- v_Adξ

d n_Y= v_Ydξ

d n_Z= v_Zdξ

This allows us to unify the changes in the molar amount of all the species into one single variable dξ.

We get

dG = [Σ-v_i,reactantsμ_i,reactants+ Σ+v_i,productsμ_j,products]dξ

or

∂G/∂ξ |_T,P = -Σv_i,rμ_i,r+ Σv_i,pμ_j,p

This quantity is also written as:

∂G/∂ξ |_T,P =Δ_rG

This quantity gives the change in Gibbs free energy for the reaction (as written!!) for Δξ=1 mole. (Units are [J/mol] therefore).

Gas reactions

Let us assume that our reaction is entirely between gas species and that the gas is sufficiently dilute that we can use the ideal gas law.

Then we can write for each species:

μ_i= μ_i^o+RTln [P_i/P_i^o]

We can then split up the Δ_rG expression in two parts:

Δ_rG = Δ_rG^o + RTlnQ

The standard potentials:

Δ_rG^o = -Σv_i,rμ^o_i,r+ Σv_i,pμ^o_j,p

and the logarithmic terms:

RTlnQ= - v_ARTln P_A/P_A^o- v_BRTln P_B/P_B^o+ v_YRTln P_Y/P_Y^o+ v_ZRTln P_Z/P_Z^o

We can combine all the logarithmic terms into Q, called the reaction quotient. The stoichiometric coefficients become exponents and the reactants' factors will be 'upside down' compared to the products, because of the properties of logarithms:

alnx = lnx^a

- alnx = ln[1/x^a]

We have kept the standard pressures P_i^o in the expression, but often they are omitted. They are usually all 1 bar, but in principle we could choose 1 bar for A 1Torr for B an 1 psi for the products. It creates a valid (though ridiculous) definition of what ^o stands for. (Of course the value of Δ_rG^o does depend on that choice!).

We could write

RTlnQ = RTln[Q_P/Q^o]

Q^o is typically unity in magnitude but it cancels the dimensions of Q_P. That means that Q and Q_P are equal in magnitude and we can get Q from Q_P by simply dropping the dimensions. Q is dimensionless but Q_P usually is not. Often this fine distinction is simply not made and Q^o is omitted, we get:

Δ_rG = Δ_rG^o + RTln[P_Y^v_YP_Z^v_Z /P_A^v_AP_B^v_B]

Notice the difference between Δ_rG which denotes the conditions (e.g. pressures) of your reaction and Δ_rG^o denotes standard conditions.

Δ_rG versus Δ_fG

If Gibbs free energies of formation are known for all the species involved in our reaction we can calculate the Gibbs free energy of the reaction.

Δ_rG^o = v_YΔ_fG^o(Y)+ v_ZΔ_fG^o(Z)- v_AΔ_fG^o(A)- v_BΔ_fG^o(B)

If Gibbs values are not available we can calculate them from enthalpies and entropies:

Δ_rG^o =Δ_rH^o -TΔ_rS^o

Equilibrium constants

If the reaction goes to equilibrium Δ_rG goes to zero. In that case Q is usually rewritten as the equilibrium constant K and we get:

0= Δ_rG = Δ_rG^o + RTlnK

Δ_rG^o = - RTlnK

where K=[P_eq,Y^v_YP_eq,Z^v_Z /P_eq,A^v_AP_eq,B^v_B]

Again the values of both K and Δ_rG^o depend on what the standard^o was chosen to be.

Concentration

The gas law contains a hidden definition of concentration:

PV= nRT

P= [n/V]RT

P= c RT

c= P/RT

Here c stands for the molar amount per unit volume or molarity. For gaseous mixtures we do not use this fact much, but it provides the link to the more important liquid solution as a reaction medium. We can rewrite the equilibrium constant as

K=[c_eq,Y^v_Yc_eq,Z^v_Z /c_eq,A^v_Ac_eq,B^v_B]

However, if you do the substitution c= P/RT into K, you'll see that not all the factors RT cancel out. The missing term g.ln[RT] depends on the stoichiometric coefficients:

g=v_Y+v_Z-v_A-v_B

The term is generally incorporated in Δ_rG^o so that the latter now refers to a new standard state of 1 mole per liter of each species rather than 1 bar of each (or so).

Activity

The above is a general principle that can be extended to other concentration units and to liquid solutions, ideal or not. In non-ideal systems we could replace

μ_i= μ_i^o+RTln P_i/P_i^o

by:

μ_i= μ_i^o+RTln a_i

and follow the same procedure as above. In stead of an expression for K involving pressures or concentrations it now read in activities:

K=[a_eq,Y^v_Ya_eq,Z^v_Z /a_eq,A^v_Aa_eq,B^v_B]

For each species we could write the activity as:

a_i = γ_ic_i/c_i^o

Here c_i^o is unity in whatever concentration measure we wish to choose. Again its function is to cancel the dimension of c_i.

With this split in three factors we can write K as three factors as well:

K =K_γK_c/K_o

K_γ=[γ _eq,Y^v_Yγ _eq,Z^v_Z /γ_eq,A^v_Aγ_eq,B^v_B]

K_c=[c_eq,Y^v_Yc_eq,Z^v_Z /c_eq,A^v_Ac_eq,B^v_B]

K_co=[c_o^v_Yc_o^v_Z /c_o^v_Ac_o^v_B]

The last factor is unity, it cancels the dimensions of K_c and is often omitted. The factor K_γ is unity if the solution is ideal. Obviously, for ionic solutions that is seldom the case.

activities of pure condensed phases

Sometimes one of the reactants or products is a pure solid (precipitate) or liquid (more solvent e.g.). What activity should we assign in such a case?

We start by choosing a suitable standard state, say the pure compound at 1 bar and temperature of interest, we then have:

μ = μ^o

but also:

μ = μ^o +RTlna

So a=1 at standard conditions

Any change can be written as

dμ = RTdlna

We can study the pressure dependence by considering:

∂ μ /∂P |_T = V_{partial molar} ( V_bar)

For a solid or liquid V_bar is a relatively small and constant value. Thus we can write:

dμ = V_bardP

RTdlna = V_bardP

dlna = V_bardP/RT

Upon integration to a different pressure P' we find

lna' = (P'-1) V_bar/RT

Example 26-12 shows that for graphite the activity is only 1.01 at 100 bars, so the activity is not very pressure dependent. Mostly if pure condensed compounds are involved in reactions the activity can taken as unity.

This is also in line with what we said previously about the solvent following Raoult's law. In the limit of the solvent going to pure solvent we have that its P goes to P*. As the activity is defined as P/P* this converges to unity. If a reaction produces more solvent molecules we can usually consider their activity equal to one in very good approximation for dilute solutions, even if they are already non-ideal.

The fact that a=1 for pure condensed phases has an important consequence for reactions (in general: processes) that only involve such phases. If all activities are unity, Q=1 and lnQ=0 which means that Δ_rG = Δ_rG^o + 0. Thus Δ_rG can only be zero -i.e. an equilibrium achieved- if Δ_rG^o happens to be zero, which is generally not the case. In fact there can only be an equilbrium at one specific temperature:

Δ_rG^o = 0

Δ_rH^o-TΔ_rS^o

T_equilibrium= Δ_rH^o/Δ_rS^o

If the process is the transformation from a solid to a liquid this is the well-known melting point. At temperature other than 0^oC only one phase can exist: either ice or water. If the other is present, that is an unstable condition and it will transform entirely to the stable form. In other words the process will go to completion, not equilbirium. Only at 0^oC can the two coexist in equilibrium. This holds for all melting points but it also holds for e.g. a solid-solid chemical reactions only producing, say, another solid.

Another way of expressing the above is to say that in order to have equilibrium at a series of temperatures, one needs at least one species involved for which activity depends on composition, e.g. a dilute solute or a gas.

Le Chatelier

In the ideal gas reaction case K only depends on T (just like U) not on the total pressure. This leads to the well-known principle of le Chatelier. Consider a gas reaction like:

A => B + C (e.g. PCl₅ => PCl₃ + Cl₂)

In pressures, the equilibrium constant becomes:

K = P_BP_C/P_A

If initially n_A =1 we have at an extent ξ:

n_A = 1- ξ

n_B= ξ

n_C= ξ

n_Total = 1+ ξ

The partial pressures are given by Dalton's law:

P_A= [1- ξ /1+ ξ] P

P_B [ξ / 1+ ξ ] P

P_C [ξ /1+ ξ ] P

The equilibrium constant becomes:

K = P[ξ_eq²]/[1-ξ_eq²]

Even though the total pressure P does occur in this equation, K is not dependent on P. If the total pressure is changed (e.g. by compression of the gas) the value of ξ_eq will change (the equilibrium shifts) in response. It will go to the side with the fewer molecules. This fact is known as le Châtelier's principle

If the system is not ideal we will also get a Châtelier shift, but the value of K may change a little because the value of the activity coefficients (or fugacity) is a little dependent on the pressure too.

In solution the same thing holds. In ideal solutions K is only T dependent, but as we saw these systems are rare. Particularly in ionic solutions equilibrium constants will be effected by other things than just temperature, e.g. changes in the ionic strength and we need to find the activity coefficients to make any predictions.

K as a function of temperature

We can use Gibbs-Helmholtz to get the temperature dependence of K

∂[Δ_rG^o/T]/∂T |_P = -Δ_rH^o /T²

At equilibrium we can equate Δ_rG^o to -RTlnK so we get:

∂[lnK]/∂T |_P = -Δ_rH^o /T²

We see that whether K increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative.

If temperature is changed little enough that Δ_rH^o can be considered constant, we can translate a K value at one temperature into another by integrating the above expression, we get a similar derivation as with melting point depression:

ln [K(T₂)/K(T₁) ] = -Δ_rH^o/R[1/T₂-1/T₁]

If more precision is required we could correct for the temperature changes of Δ_rH^o by using heat capacity data.

Spontaneity

We developed the Gibbs free energy function to be able to predict which changes could occur spontaneously. If we start with a set of initial concentrations we can write them in a reaction quotient

Δ_rG = Δ_rG^o + RTlnQ

if we subtract the equilibrium version of this expression:

0= Δ_rG^o + RTlnK

we get

Δ_rG = RTln[Q/K]

That gives us the sign of Δ_rG. If this is negative the reaction will spontaneously proceed from left to right as written, if positive it will run in reverse. In both case the value of Q will change until Q=K and equilibrium has been reached.

Note that 'your' Δ_rG determines that, not the sign of Δ_rG^o (because that represents the decidedly non-equilibrium standard state!)

Weak electrolytes

We have seen that strong electrolytes are non-ideal even at tiny concentrations and that even the Debye-Hückel theory only allowed us to work at very small concentrations of strong electrolytes. The same problems are encountered with weak electrolytes but they are compounded by the equilibrium that is inherent to their solutions. Take acetic acid as in household vinegar:

HAc + H₂O <=> Ac^- + H₃O⁺

We can write the equilibrium constant as:

K = (a_Ac^- a_H3O⁺)/(a_HAca_H2O)

As

a_H2O=1

K = (a_Ac^- a_H3O⁺)/(a_HAc)

At an initial concentration of say 0.1 mol/l the activity coefficient for HAc (being a neutral species) is essentially one, but for the other species we should write:

(a_Ac^- a_H3O⁺)= [Ac^-][H₃O⁺]γ_±²

The value of γ_±² is not unity and this will affect the equilibrium. In first approximation we will ignore that fact and write:

K = [Ac^-][H₃O⁺]/[HAc] = 1.74 10^-5. (We can either write [mol.l] as dimensions or drop them, depending on whether we are talking about K or K_c)

At equilibrium we would get:

K = x.x/(0.1-x) =1.74 10^-5 which yields x=1.31 10^-3 mol/l

We can now use Debye-Hückel theory to estimate the means ionic activity coefficient but the concentrations are already to high for that. Instead we will use one of its extensions:

lnγ_± = - 1.173|z₊z_-|√I]/[1+√I]

The ionic strength is ½ ([Ac^-]+[H₃O⁺]) =x

Although we do not know x precisely the value we have is at least a starting point. We can use it to calculate a first approximation for γ_±². We find a value of 0.921. We then divide the value of K by this value and recalculate x. It changes from x=1.31 10^-3 mol/l to x=1.365 10^-3 mol/l. Now we can repeat this process until the value of x does not change appreciably anymore (converges). This process is called iteration. The final value is about x=1.37 10^-3 mol/l.

As can be seen the non-ideality does change the values from the ones you would have calculated before entering this course, but the difference is not staggering, at least if no other solutes are present.

Solubility products

For solubility products the differences can be more important. Take BaF₂ in water K= 1.7 10^-6.

K = γ_±³ [Ba²⁺][F^-]²

Let's start by assuming ideality and say that γ_±³=1 and say that :[Ba²⁺]=x

[F^-]=2x

Thus K= x (4x²) so x= 7.52 10^-3 mol/lit

The ionic strength is I = ½((+2)²x+(-1)²x) = 3x

The extended Debye-Hückel theory gives γ_± =0.736 this raises x to 0.0102. Repeating the process a few times we find x=0.011. This means an increase of about 30% due to non-ideal behavior for this sparingly soluble salt.

Again the presence of other solutes may induce larger effects because they add to the ionic strength.

This concludes the lecture series for CH 431