Chemical Equilibrium

One of the most fundamental applications of thermodynamics is to chemical reactions at equilibrium. The underlying fundamental idea that we describe here is that DG = 0 for a system in equilibrium (at constant temperature and pressure), and the sign of DG determines whether or not a given process or chemical reaction will occur spontaneously at constant T and P.

Consider a general gas phase chemical reaction

n_AA (g) + n_BB (g) Û n_yY (g) + n_ZZ (g)

We define a quantity x, called the extent of reaction such that the numbers of moles of reactants and products are given by

n_A = n_A0 - n_Ax

n_B = n_B0 - n_Bx

n_Y = n_Y0 - n_Yx

n_Z = n_Z0 - n_Zx

where n_j0 is the initial number of moles for each species. According to the equations above x must have units of moles. As the reaction proceeds from reactants to products, x varies from 0 to some maximum value dictated by the stoichiometry of the reaction.

Differentiation of the equations gives that rate of change of the number of moles.

Reactants

dn_A = -n_Adx

dn_B = -n_Bdx

Products

dn_Y = n_Ydx

dn_Z = n_Zdx

The negative signs indicate that the reactants are dissappearing and the positive signs indicate that the products are being formed as the reaction progresses from reactants to products.

Now consider a system containing reactions and products

dG = -SdT + VdP + m_Adn_A + m_Bdn_B + m_Ydn_Y + m_Zdn_Z

At constant T and P we have

dG = m_Adn_A + m_Bdn_B + m_Ydn_Y + m_Zdn_Z

or using the above expression

dG = -m_An_Adx - m_Bn_Bdx + m_Yn_Ydx + m_Zn_Zdx

or

dG = (m_Yn_Y + m_Zn_Z -m_An_A - m_Bn_B)dx

or

(¶G/¶x) = (m_Yn_Y + m_Zn_Z -m_An_A - m_Bn_B)

We define the standard free energy of a reaction as (¶G/¶x).

In other words, (¶G/¶x) = D_rG. As we have seen the units of D_rG are J/mole. The quantity D_rG has meaning only if the balanced chemical equation is specified.

If all of the gases in the reaction are ideal then

m_j(T,P) = m_j^o(T) + RT(lnP_j/P^o)

and substituting this expression into the above equation we find

or

D_rG = D_rG^o + RT lnQ

where

D_rG^o = m_Y^on_Y + m_Z^on_Z -m_A^on_A - m_B^on_B

and

The quantity D_rG^o is the standard Gibbs energy for the reaction between unmixed reactants in their standard states at temperature T and a pressure of one bar to form unmixed products in their standard states at the same T and P. The quantity Q is called the reaction quotient. Its magnitude is dependent on the quantity of reactant and product at any given point during the chemical reaction. The pressures in Q are all referenced to P^o = 1 bar which is the standard state. Since P^o = 1 bar it need not be written explicitly and is often implied in the expression for Q. Thus, Q may have the form

where P_j is the partial pressure of the jth component.

When the system is in equilibrium, the Gibbs energy is a minimum

In this case, find that the reaction quotient has a special value known as the equilibrium constant.

We have identified the set of partial pressures consistent with equilibrium with a constant K(T) known as the equilibrium constant.

Example: Determine the equilibrium constant expression for the reaction that is represented by the equation

3H₂ (g) + N₂ (g) = 2 NH₃ (g) (form A)

.The stoichiometric coefficients appear as exponents in the expression for the equilibrium constant.

Note that if we were to express the chemical equation as

3/2 H₂ (g) + 1/2 N₂ (g) = NH₃ (g) (form B)

the Gibbs free energy would be exactly half as large as form A of the chemical equation above. The equilibrium constant in this case is

The equilibrium constant is a function of temperature only. We may start with any given initial pressures, but the equilibrium constant tells us that the ratio of the pressures products to that of reactions (raised to their respective stoichiometric coefficients) is fixed at equilibrium. When the total pressure appears in the equilibrium constant it can only change the relative proportion of the products and reactants.

Example: For the reaction

PCl₅ (g) = PCl₃ (g) + Cl₂ (g)

The equilibrium constant expression for this reaction is

Suppose that we initially have 1 mole of PCl₅. At equilibrium, x moles will have reacted so there will 1 - x moles of PCl₅. There will be x moles each of PCl₃ and Cl₂. Thus, the total number of moles will be 1 - x + x + x = 1 + x. Using Dalton's law we can express the partial pressure of each of the components at equilibrium

where P is the total pressure. The x represented here are those observed at equilibrium. Substituting these expression into the equilibrium constant we find

Although the pressure appears in this expression, the equilibrium constant K(T) is not a function of pressure. Rather it is x that is a function of pressure. The greater the total pressure, the smaller x must be to compensate. This is an example of Le Chatelier's principle: If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium, then the reaction adjusts toward a new equilibrium state in such a way as to minimize the effect of the change. In other words, if we increase the total pressure, the equilibrium shifts in the direction of fewer moles of total components (toward the left in the present reaction). This minimizes the total pressure, thus acting to minimize the effect of the change.

Example: we consider once again the reaction

NH₃ (g) = 3/2 H₂ (g) + 1/2 N₂ (g)

for which K(T) = 1.36 x 10^-3 at 298 K. Determine the extent of reaction x at equilibrium at a pressure of 1millibar and at a pressure of 1 bar.

SOLUTION: We construct a table

NH3 (g)	3/2 H2 (g)	1/2 N2 (g)
1 - x	3/2x	1/2x

The total number of moles is 1 - x + 3/2x + 1/2x = 1 + x.

Dalton's law tells that the partial pressure of each gas is

NH3 (g)	3/2 H2 (g)	1/2 N2 (g)
(1 - x)P/(1+x)	3/2xP/(1+x)	1/2xP/(1+x)

We substitute these expressions into the equilbrium constant.

Now we solve for x

As P gets large we can see that x gets small. This is in accord with Le Chatelier's principle since the number of moles of all components will be reduced as the equilibrium is shifted to the left.

If P = 1 mbar then

And if P = 1 bar

Given the relation D_rG^o = -RT ln K(T) it is also possible to calculate equilibrium constants from tabulated standard molar Gibbs energy data. We simply invert the relation to find

K(T) = exp{ -D_rG^o/RT }.