C8863 Free Energy Calculations -1-
C8862
Free Energy Calculations
exercise
2. Chemical Equilibrium
Petr Kulhánek
kulhanek@chemi.muni.cz
NCBR - National Centre for Biomolecular Research
&
CEITEC - Central Institute of Technology,
Masaryk University,
Kamenice 5, 625 00 Brno
C8863 Free Energy Calculations -2Excercise
I
A + 2 B C
1. Determine composition of reaction mixture at equilibrium, which is formed from
mixing 500 ml solution of substance A with molar amount of nA = 0.001 mol and 500
ml solution of substance B with molar amount nB = 0.01 mol. The volume of the
resulting mixture will be 1 L. The substances A and B react together to form C. The
reaction is characterized by an equilibrium constant of K = 106. The volume of the
reaction mixture does not change during the reaction. Consider the reaction mixture
as an ideal solution at the standard conditions.
2. Determine the change in Gibbs energy resulting from mixing of A and B from the task
1.
C8863 Free Energy Calculations -3Excercise
II
1. Substance A is titrated with substance B at 298 K. The starting concentration of
substance A is c0,A = 4 mM and the concentration does not change during titration.
Determine the equilibrium concentrations of A, B, C, and D as a function of c0,B/c0,A in
the range of 0 to 2 (i.e., up to twice of the molar equivalent). Consider the reaction
mixture as an ideal solution.
2. Recalculate the equilibrium concentrations of A, C, and D to molar fractions of substance
A as a function of c0,B/c0,A in the range of 0 to 2 (i.e., up to twice of the molar
equivalent).
A + B C
A + 2B D
K1 = 106
K2 = 106
C8863 Free Energy Calculations -4Outline
of Exercise II Solution
1. Write a set of equations that unambiguously describe the composition of the reaction
mixture, which includes
• relations for equilibrium constants
• balance equations (take into account the law of mass conservation)
2. The number of equations must be the same (4) as the number of values to be
determined (4).
3. Rewrite the equations to form
4. where x is the concentration of interest and y is a vector that reaches zero when at a
solution of the system of equations.
5. The solution can be found using the fsolve method in octave.
6. Count the titration again in the octave program using the cycle. Use the solution x from
the previous step (previous value c0,B) for the new value c0,B. Store the resulting
concentration and molar fraction values in a file.
7. Use gnuplot to view the result.
)(xy f=
0x =)(f