Problem 2.6.5

 

Estimate the pressure required to make methanol formation feasible.  Feed is comprised of stoichiometric amounts of carbon monoxide and hydrogen.  The reaction is to be conducted at 500 degrees centigrade.  Heat capacity data is missing.  One mole of carbon monoxide reacts with two moles of hydrogen to produce one mole of methanol.  Energy of formation (delta G) and enthalpy of formation (delta F) values could be found from the literature. 

 

What is the value of equilibrium constant at this temperature?

 

Solution:

 

Delta G is equal to minus RT log (K), where delta G is given by delta H minus T delta S.  In the absence of heat capacity data, both delta H and delta S are considered constant.  These values for the reaction mixture can be calculated considering the value positive for the products (as being produced) and negative for the reactants (as being consumed). 

 

                                    Delta G298          Delta H298                                  nu

Carbon monoxide           -32,810             -26,420 calories per mole            -1

Hydrogen                      0                      0                                              -2

Methanol                       -38,690             -48,080                                     1

 

 

Delta G298 is equal to minus 1 time (-32,810) plus (-2 time 0) plus 1 time (-38,690).  This results in delta G298 value of minus 5,850 calories per mole.

Delta H298 is equal to minus 1 time (-26,420) plus (-2 time 0) plus 1 time (-48,080).  This results in delta H298 value of minus 21,660 calories per mole.

Delta S is given as (delta H minus delta G)/T.  Its value is equal to minus 52.95 calories per mole per degree Kelvin. 

 

Delta G at any temperature, T, is equal to delta H minus T delta S.  This value at 500 degrees centigrade is equal to minus 21,660 minus 773×52.95 or 19,280.6 calories per mole.  This gives a value of K, exp (minus delta G/(RT)), as 3.55×10-6.

 

If we consider a conversion of x, then for a stiochiometric feed mixture, the product composition could be calculated as follows:

 

                                    Feed                 Reaction           Product             mole fraction                              partial pressure

Carbon monoxide           1                      minus x             1 minus x          (1 minus x)/(3 minus 2 x)            (1 minus x)/(3 minus 2 x)*PT

Hydrogen                      2                      minus 2 x          2 minus 2 x       2(1 minus x)/(3 minus 2 x)          2(1 minus x)/(3 minus 2 x)*PT

Methanol                       0                      x                      x                      x/(3 minus 2 x)                          x/(3 minus 2 x)*PT

Total                             3                      minus 2 x          3 minus 2 x                                                      

 

If reaction is conducted at a pressure of PT, the equilibrium constant, K, can be expressed as

 

K is equal to x (3 minus 2 x)2/[4(1 minus x)3PT2] = 3.55×10-6

 

For a 10 percent conversion, the above equation could be solved to yield a required system pressure of 275 atmospheres.  For a 90 percent conversion, pressure requirement is 9,600 atmospherers.