dt where kab and kcd are the reaction rate coefficients for forward and reverse reactions respectively. The equation simply states that the rate of buildup of C is the sum of the rates of reactive collisions of A and B minus the reverse process in which C and D react. The first term must be proportional to the respective number densities and similarly for the second (loss) term. Since for every creation of a C molecule there must be a B molecule, these rates of formation must be equal to each other and equal to the negative of the rates of formation of the A and B molecules.
In equilibrium the rates of change of the species are zero. This means
The constant K is called the equilibrium constant for the reaction. If K is known, we can determine the ratios of concentrations of the product and reactant gases in equilibrium.
For the case of a general reaction a A + b B ^ c C + d D, the equilibrium constant is (as derived in physical chemistry texts):
The rule can be generalized to cases where more than two species are on each side of the equation.
It is seen from (8.57) that when the inverse reaction rate is very small, which means that products dominate over reactants, K is large. Small K means there will be a relatively large concentration of the reactant species.
The equilibrium constant depends on collision dynamics and in principle should have a strong temperature dependence, since the intermolecular relative velocity will be an important factor in the rearrangements. To find the temperature dependence of the equilibrium constant, let us write the reaction rate coefficients for the forward and reverse reactions (8.52). Using the Arrhenius equation (8.59) the reaction rate coefficient for the forward reaction, k\, is ki (T) = Ai exp
The reaction rate coefficient for the reverse reaction, k2, is ki(T) = A2 exp
The equilibrium constant k2 -AH
Hence, knowing the standard enthalpy associated with a reaction provides information about the equilibrium concentrations of the species.
Example 8.9 Consider the recombination of OH and O:
Write the expression for the equilibrium constant. By using formula (8.60) find out whether the products will dominate over reactants at high or low temperatures.
Answer: In order for the reaction to proceed, an OH molecule has to bump into an O atom in the same tiny volume. The rate of growth term must be proportional to [OH][O]. The equilibrium constant is
To find the temperature at which the forward reaction will be favored we need to find the standard enthalpy:
AH° = AH°(O2) + AH°(H) - AH°(OH) - AH°(O) = (0 + 217.97 - 38.96 - 249.17) kJ mol-1
Since AH is negative, from (8.60) we have that the equilibrium constant K is larger with lower temperature. This explains why the products of reaction (8.61) will dominate over reactants at the lower temperatures. □
The next step is to find the connection between the equilibrium constant and the Gibbs energy. The equilibrium of the chemical reaction implies that at a given temperature there exist partial pressures of the gases involved in the reaction with which the rate of the forward reaction is equal to the rate of the reverse reaction. Let us express the equilibrium constant in terms of partial pressures of each constituent. Using the Ideal Gas Law for the molar concentrations we obtain:
(pc/R*T )c (pd/R*T ) (pa /R*T )a (Pb/R* T )b cd pcpd, \A
where thepi are partial pressures and A = (a + b) — (c + d). We can rewrite (8.64) in the form:
[equilibrium constant for ideal gases]. (8.66)
Kp is used as an equilibrium constant for chemical reactions involving species in the gaseous state.
For a reversible transformation at constant temperature the change of Gibbs energy is (see (4.97))
For 1 mol of ideal gas
(Note that the molar Gibbs energy G is also called the chemical potential.) By integrating this equation from the standard pressure level of 1 atm to some arbitrary pressure level p holding T constant we obtain
wherep is in atm. In this formula G is the Gibbs energy at pressurep and temperature T and GT is the Gibbs energy at 1 atm and temperature T.
Using (8.69) we can write the change of Gibbs energy for the general chemical reaction
AG = [cAG(C) + d AG(D)] - [a AG (A) + bAG(B)] (8.70)
in the form
AG = [cAG°T(C) + d Ag°t(D) - aAG°T(A) - bAG°T(B)]
The argument of the logarithmic function in the last formula is the equilibrium constant Kp (see (8.66)). Therefore,
At equilibrium AG = 0, and the change of Gibbs energy at a pressure of 1 atm and arbitrary temperature T, AGT, is related to the equilibrium constant at pressure p and temperature T, Kp, by the simple relation:
The equation (8.73) tells us that if AG°T is positive, Kp should be less than unity, which means that at equilibrium the concentrations of the reactants will exceed those of the products. If, on the other hand, AGT is negative and, moreover, is large, then Kp is large and the products will dominatein the equilibrium.
For standard conditions at a pressure of 1 atm and a temperature of 25 °C we obtain, in joules,
The change in Gibbs energy AG is especially useful because it simultaneously takes into account both the First and Second Laws of Thermodynamics. It does so in such a way that if the temperature and pressure are held constant (and they often are in atmospheric problems) we have a function which can be applied much more broadly.
Example 8.10 Consider the reaction of recombination of NO2 and O3:
Do the reactants or products dominate for the forward reaction at 1 atm and 25 °C? Answer: We have to find the change of the Gibbs energy for this reaction.
AG° = (115.9 + 0 - 51.3 - 163.2) kJ mol-1 = -98.6kJ mol-1. (8.76)
From (8.74) we obtain Kp = 1.9 x 1017. With such a large value of Kp the products will dominate for the forward reaction at equilibrium. □
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