Although The exact calculation of the effect of temperature changes upon an equilibrium is somewhat more involved than we can treat here, we can make several approximate relationships that are very useful. Remember that the equilibrium constant is related to the standard free energy change of the reaction
And recall that the free energy change upon reaction is related to both the enthalpy and entropy changes
Although the free energy change is strongly dependent on temperature, usually the enthalpy and entropy changes upon reaction are NOT temperature dependent (or at least not very much so). This implies that the free energy is a linear function of temperature. Combining the above two equations:
substituting in the equilibrium constant
and finally
Remember that the enthalpy and entropy changes are regarded as constants with respect to temperature, so the equilibrium constant is exponentially dependent on temperature. The relative change in equilibrium constant as a function of temperature is usually not expressed as above but as the van't Hoff equation
This equation looks very reminiscent of the Clausius-Clapeyron equation for vapor pressures and the Arrhenius equation for the temperature dependence of reaction rate constants, doesn't it? This form of temperature dependence (exponential w.r.t. 1/T) comes up over and over in chemistry.
Example:
Consider the Reaction of NO with Cl2:
The Thermodynamic quantities associated with this reaction are as follows:
At what temperature is the equilibrium constant, Kp, for this reaction equal to 1.0 x 103 ?
Sol'n:
Assuming the enthalpy and entropy values are independent of temperature, we need to
calculate the temperature at which the free energy change will imply the desired equilibrium
constant.
Using the given information
One for the Bag of Tricks... Successive Approximation
Often in the course of solving for equilibrium concentrations, the equilibrium equation
appears to be an impossibly complicated equation in x, the equilibrium advancement. Sometimes
the method of successive approximation is useful
Example
The reaction
This reaction has a KC of 7.4 x 10-3 at 1000K. If 1 mole each of methane and water are
introduced into a 10 L vessel at 1000K, what is the equilibrium number of moles of H2 in the
vessel?
Sol'n
The setup for this problem is quite straightforward. We must first parameterize the number of
moles and concentrations of each species in terms of the initial conditions and the advancement of
the reaction.
Inserting the information we are given and calling x the number of moles of reaction in the
forward direction
OK the equilibrium number of moles of H2 is just 4x, but how do we solve this equation for x?
First, lets simplify this a little by multiplying out all those powers of the volume
Now, lets assume that x and 2x are much less than 1.00, the initial number of moles of the
reactants. Then
This equation can easily be solved for an approximate value of x0 = (0.74/256)1/5 = 0.31. But 0.31
is not very much smaller than 1.00 so our initial approximation is not a very good one. So we put
this first guess back into the equation where we had assumed x was zero initially
Our next best guess for the value of x1 is ((0.74)(0.69)(0.38)2/(256))1/5 = 0.20. This is better
approximation of the value of x, but we can iterate this procedure until the value of x remains
constant to within the significant figures of the problem. Our next best guess will be the solution
to the equation.
Which yields a value of x2 = 0.24. Continuation of this method yields x3 = 0.22, x4 = 0.23,
x5 = 0.23 etc. so we may stop after the fifth iteration if we want the answer to two significant
figures. Cute, huh?
The final solution? The number of moles of H2 at equilibrium = 4 (0.23) = 0.92 moles.
