Temperature and Equilibrium
 Page ID
 39000
Temperature Effects on Equilibrium: A Study Guide
Discussion Questions
 How does temperature affect Gibb's energy?
 How does temperature affect and equilibrium?
Temperature Effects on Equilibrium
For convenience, we treat temperature, entropy, enthalpy, and Gibb's energy as general state functions and represent them by T, S, H, and G respectively, dropping the change symbol D, associated with them. The justification for doing these have been introduced in Gibb's energy.
Since S, H, and G depends on T, temperature affects chemical reactions and chemical equilibria. We discuss the temperature effects in general terms, and the strategy introduced here may be useful for solving complicated problems later.
How does temperature affect Gibb's energy?
Gibb's energy, G, is a general state function defined in terms of state functions entropy S and enthalpy H. It's relationship with the equilibrium constant K has also been defined using the formula below. This notations are widely used in the field of thermodynamics.
G = H  T S
=  R T ln K
For entropy, we have defined the standard or absolute entropy S^{o} as
S^{o} = ò_{0}^{298} C_{P} dT / T
The symbol C_{P} is the heat capicity at constant pressure, it also depends on T, not a constant.
Thermodynamic data usually lists standard enthalpy H^{o}, and standard entropy S^{o}. When measured at temperature T other than 298 K, their values are given by
DS = S^{o} + ò^{T}_{298} C_{P} dT / T
= S^{o} + C_{P} ln (298/T)
DH = H^{o} + ò^{T}_{298} C_{P} dT
= H^{o} + C_{P} (T  298)
The integrations given above are valid only if C_{P} is a constant between 298 and T K. If the temperature range is large, an integration process either by mathmatical means or by numerical means should be carried out. Within small ranges, the assumption is reasonable, especially if average values are used. Under the circomstance,
DG = G^{o} + C_{P} ln (298/T) + C_{P} (T  298)
=  R T ln K
The above model shows that Gibb's energy depends on temperature. This formulation is true if C_{P} is a constant between 298 and T K. In general, the heat capacity varies with temperature and the variation should not be ignored if the temperature range is very large. In some cases, an average value for the heat capacity may be used.
Gibb's free energies of a system drive both physical and chemical changes. For physical changes, the heat capacities of the system over the ranges of temperature should be considered. In cases of chemical reactions, the differences in heat capacity between products and reanctants shall be considered.
How does temperature affect equilibrium?
For chemical reactions, the differences in heat capicity between the rpoducts and the reactants DC_{P} are to be used. Using the equation derived previously,
DG = G^{o }+ DC_{P} ln (298/T) + DC_{P} (T  298)
=  R T ln K,
the effect of temperature on the equilibrium constant K can be derived,
ln K =  DG^{o} / R T  [C_{P} / (R T)] [298  T + ln (298/T)]
=  ln K^{o}  [C_{P} / (R T)] [298  T + ln (298/T)]
Recall that K^{o} is the equilibrium constant at 298 K. The equation can be simplified to give,
ln (K / K^{o}) =  [C_{P} / (R T)] [298  T + ln (298/T)]
And to generalize it, if K_{1} and K_{2} are equilibrium constants at T_{1} and T_{2} respectively, we can derive the formula,
ln (K_{2}/K_{1}) =  [C_{P} / R] [(1/T_{2})  (1/T_{1})] [T_{1}  T_{2} + ln (T_{1}/T_{2})]
By measuring the equilibrium constants at different temperatures and using this equation, we may estimate the heat capacities over ranges of temperatures. Note that
C_{P}(T_{1}  T_{2}) = DH
And if we ignore the term ln (T_{1}/T_{2}), the above equation is the same as the familiar Helmholtz equation, which is discussed in Example 1 below.
Example 1
The GibbsHelmholtz equation is
d (DG / T)/dT =  DH / T ^{2}.
Show that
DH d ln K =   d(1/T) R
SOLUTION
Since
DG =  R T ln K,
ln K =  DG / R T
Differentiate both sides with respect to (1/T) in the above equation gives,
d(ln K) / (d T) = ( 1 / R (d DG / T) / (dT)
=  DH / R T ^{2}
DISCUSSION
If K_{1} and K_{2} are the equilibrium constant at T_{1} and T_{2} respectively, show further that
ln (K_{1} / K_{2}) =  (DH / R) (1/T_{1}  1/T_{2}).
This is achieved by definite integral. This relationship indicates that the plot of ln (K versus 1/(T is a straight line, and the slope is  (DH / R). Thus, DH can be determined by measuring the equilibrium constant at different temperatures.
Example 2
The vapour pressure of water at 298 K is 23.756 mmHg. Estimate the molar standard Gibb's energy.
SOLUTION
The vapour pressure of 23.756 mmHg should be converted to a pressure in units of pascal (N m^{2}),
101,300 N m^{2} 23.756  = 3166 Pa 760 mmHg
G^{o} = 8.312 J * 298 ln(3166)
= 20.0 kJ / hr
DISCUSSION
This example illustrates the evaluation of Gibb's energy when the equilibrium constant is known.
Questions
 For a system that is at equilibrium, what is the Gibb's free energy of the system?
Skill 
Explain the physical meaning of Gibb's free energy in terms of changes. 
The standard enthalpies of H_{2}O (g) and H_{2}O (l) given in a Table of Chemical Thermodynamic Data are 242 and 286 kJ mol^{1} respectively. What is the heat of vaporization?
Skill  Construct cycles by applying Hess's law to evaluate the energy changes for a chemical or physical process, such as
H_{2}(g) + (1/2)O_{2}(g)
 
 242 
  286
¯ 
H_{2}O (g) 

 H=? 
 ¯
~~H_{2}O (l)~~~Note that the heat of vaporization so calculated is the heat of vaporization at 298 K.
Solutions
 At equilibrium, there is no net change, and G = 0.
 Heat of vaporization = 242  (286) kJ/mol.
Contributors
Chung (Peter) Chieh (Chemistry, University of Waterloo)