# Give the thermodynamic derivation of van't Hoff reaction isotherm, and explain its significance?

##### 1 Answer

The **van't Hoff reaction isotherm** is given by:

#((del barG)/(del xi))_(T,P) = DeltabarG^@(T) + RTlnQ# ,where:

#((delbarG)/(del xi))_(T,P)# describes theinfinitesimal changein the molar Gibbs' free energy as the reaction proceeds atconstant temperature and pressure.#xi# is theextent of reactionin terms of#"mols"# (it's represented as#x# in ICE tables you see in general chemistry).#DeltabarG^@# is thestandard change in molar Gibbs' free energy, the reference point, defined at standard pressure (#"1 bar"# ), and is a function of only temperature.#RTlnQ# is thedeviationfrom#DeltabarG^@# at the same temperature.

It allows you to find the deviation of the Gibbs' free energy **away from equilibrium**, or away from **standard conditions**, at the same temperature.

- Equilibrium is if
#((del barG)/(del xi))_(T,P) = DeltaG = 0# and#Q = K# . - Standard conditions is if
#Q = 1# , i.e. if all the activities#a_i# are#1# , so that#((del barG)/(del xi))_(T,P) = DeltaG^@# .

*An important distinction is that standard conditions has #Q = 1#, but equilibrium does not require #Q = 1#, since #K# does not have to be #1#. Equilibrium just needs #DeltaG = 0# and #Q = K#. Thus, the two ways to use this equation described above are indeed different.*

To derive this, we begin from the *definition of reaction progress* based on the **chemical potential**

#overbrace(((delbarG)/(del xi))_(T,P))^"Reaction Progress" = sum_i nu_i mu_i# ,#" "" "bb((1))# where

#nu_i# is the unitless stoichiometric coefficient of substance#i# , and is negative for reactants and positive for products.

The **deviation of the chemical potential** due to changes in *activities*

#mu_i(T,P) = mu_i^@(T) + RTln a_i# ,#" "" "bb((2))# where

#mu_i^@(T)# is the chemical potential defined at standard pressure (#"1 bar"# ).

By substituting the right-hand side of *only* temperature):

#((del barG)/(del xi))_(T,P) = sum_i nu_i (mu_i^@ + RTln a_i)#

#" "" "" "" " \ = sum_i (nu_imu_i^@ + RT nu_iln a_i)#

#" "" "" "" " \ = sum_i nu_imu_i^@ + RT sum_i nu_iln a_i#

Now, at standard pressure and the desired temperature,

#DeltabarG^@ = ul(sum_i nu_i mu_i^@)# .

You may have seen this in general chemistry as:

#DeltaG_(rxn)^@ = overbrace(sum_"products" nu_P DeltaG_(f,P)^@ - sum_"reactants" nu_R DeltaG_(f,R)^@)^"Gibbs' free energies of formation"#

Using the properties of logarithms,

#sum_i nu_i ln a_i = sum_i ln (a_i^(nu_i)) = ln (prod_i (a_i)^(nu_i))# .

The definition of the **reaction quotient**

#Q = prod_i (a_i)^(nu_i) = (prod_"Products" (a_j)^(nu_j))/(prod_"Reactants" (a_i)^(nu_i))#

So, this really means that

Therefore, we obtain the **van't Hoff reaction isotherm**:

#color(blue)(barul|stackrel(" ")(" "((del barG)/(del xi))_(T,P) = DeltabarG^@(T) + RTlnQ" ")|)#