Bodenstein equilibrium
Solutions
1) For the gas-phase reaction
2 HI  H2 + I2 ,
the equilibrium constant is
K = pHpI/pHI2 .
If the system contains only HI initially, at a pressure p0, and if the fraction of HI that
decomposes by the time the reaction reaches equilibrium, then, at equilibrium
pHI = (1 – )p0 ,
and
pH = p0/2 = pI .
The latter result reflects the fact that H2 and I2 are formed in equal proportions, with a half mole
of each formed for every mole of HI that decomposes. Thus the equilibrium constant is
p p (p0 /2)(p0 /2)
 2 p0 2
2
,
K  H 2I 


pHI
[(1  )p0 ]2
4(1  ) 2 p0 2 4(1  )2
as was to be proved.
2) Computing the equilibrium constants is a simple matter of plugging in numbers, such
as:
K
2
(0.2408)2
2 
2 = 0.02515 .
4(1  )
4(1 0.2408)
The computed values are shown in the table below.
temp (°C)
T(K)
1/T(K)
K

508
781
0.00128
0.2408
0.02515
487.2
760.2
0.00132
0.2340
0.02333
443
716
0.00140
0.2198
0.01984
427
700
0.00143
0.2157
0.01891
410
683
0.00146
0.2100
0.01767
393
666
0.00150
0.2058
0.01679
374
647
0.00155
0.2010
0.01582
356
629
0.00159
0.1946
0.01459
328
601
0.00166
0.1885
0.01349
302
575
0.00174
0.1815
0.01229
283
556
0.00180
0.1787
0.01184
To download a worked spreadsheet, go to:
http://webserver.lemoyne.edu/faculty/giunta/classicalcs/bodenstein.xls
ln K
–3.683
–3.758
–3.920
–3.968
–4.036
–4.087
–4.146
–4.227
–4.306
–4.399
–4.437
3) The relationship of the equilibrium constant to standard enthalpy and entropy of
reaction is
G
H S 
.
K  exp
 exp
exp
 RT 
 RT   R 
Taking the natural log of both sides gives:
ln K = –(∆H°/R)(1/T) + ∆S°/R .
If ∆H° and ∆S° are temperature independent (an assumption to which we will return below), then
a plot of ln K vs. 1/T (called a van’t Hoff plot) should yield a straight line whose slope m is equal
to –∆H°/R and whose y-intercept b is equal to ∆S°/R . Thus, we obtain ∆H° and ∆S° from the fit
of the plotted data:
∆H° = –mR and
∆S° = bR .
To make the van’t Hoff plot, we must convert the temperatures to absolute (Kelvin)
temperatures and find their reciprocals and we must take the natural log of the equilibrium
constant, all of which is done in the table displayed above with exercise 2. The plot, including
the equation of the best-fit straight line, is below
HI decomposition equilibrium
-3.6
ln K
-3.8
-4.0
-4.2
-4.4
ln K = - 1.8485 - 1474/T
R^2 = 0.979
-4.6
0.0012
0.0014
0.0016
0.0018
1/T(K)
So
∆H° = –mR = –(–1474 K)(8.3145 J mol–1 K–1) = 1.23x104 J mol–1 ,
∆H°= 12.3 kJ mol–1 ,
and
∆S° = bR = (–1.8485)(8.3145 J mol–1 K–1) = –15.4 J mol–1 K–1 .
NIST thermodynamic tables yield ∆H° = 12.3 kJ mol–1 and ∆S° = –16.0 J mol–1 K–1 at
393°C (near the mid-point of the range of Bodenstein’s data). This agreement is extremely good,
attesting to the high quality of Bodenstein’s measurements.
Finally, note that the “scatter” of experimental data around the best-fit line does not
appear to be random. The data appear to define a curve rather than a straight line. A curved
van’t Hoff plot is an indication that the standard enthalpy and entropy of reaction are not strictly
independent of temperature. Strictly speaking, ∆H° and ∆S° for a given temperature can be
derived from the straight line that is tangent to the ln K vs. 1/T curve at that temperature.