To determine whether the reaction behaves as a stoichiometrically ...

To determine whether the reaction behaves as a stoichiometrically simple reaction,
apply the Law of Definite Proportions by calculating ξ for each species. These
calculations are summarized in the following table:
Species ∆n i v i ξ
C 4 H 4 S -70.0 -1 70.0
H 2 145.9-410.9=265.0 -4 66.25
C 4 H 10 75.1-20.1=55.0 +1 55.0
H 2 S 95.7-25.7=70.0 +1 70.0
These calculations show that the reaction is not behaving as though it were
stoichiometrically simple. Why not The data provides some clues. The above
calculation show that the amount of hydrogen sulfide (H 2 S) being formed is in exact
proportion to the amount of thiophene being consumed. However, less hydrogen is being
consumed than predicted by the balanced stoichiometrical equation, given the
consumption of thiophene, and less butane (C 4 H 10 ) is being produced.
If we checked the elemental balances at this point, they would show that all of the
sulfur atoms were accounted for (in = out), but that there were fewer hydrogen and
carbon atoms leaving than entering. It seems likely that the analytical system in the pilot
plant has failed to detect at least one hydrocarbon species, and that this undetected
species has a lower H/C ratio than butane. If the behavior of the system cannot be
described by one stoichiometrically simple reaction, perhaps more than one reaction is
taking place. Following is one possible example that might have occurred:
Ex) Suppose that two reactions shown below were taking place in the thiophene
hydrogenolysis pilot plant mentioned in Example 1-1.
C 4 H 4 S + 3H 2 C 4 H 8 + H 2 S
(A)
C 4 H 8 + H 2 C 4 H 10
(B)
Are these reactions stoichiometrically consistent with the pilot plant material balance
data
If these two stoichiometrically-simple reactions are sufficient to account for the
behavior of the actual system, then all of the equations for ∆n i , one for each species, will
be satisfied by a single value of ξ A , the extent of Reaction A, and by a single value of ξ B ,
the extent of Reaction B.
For thiophene (T),
∆n T = v AT ξ A + v BT ξ B = -70
Since v AT = -1 and v BT = 0, ξ A = 70.
For H 2 (H),

∆n H = v AH ξ A + v BH ξ B = -265
Since v AH = -3 and v BH = -1,
(-3)(70) + (-1)ξ B = -265; ξ B = 55
These values of ξ A and ξ B must now satisfy all of the remaining equations for ∆n.
However, the equation for butene cannot be used since there is no experimental value of
∆n for butene.
For H 2 S (S),
∆n S = v AS ξ A + v BS ξ B = 70
(+1)(70) + (0)(55) = 70 Check
For butane (B),
∆n B = v AB ξ A + v BB ξ B = 55
(0)(70) + (1)(55) = 55 Check
Therefore, the data, as it exists, is consistent with Reaction A and B. This analysis does
not prove that the above two reactions are taking place.
There are other explanations that could account for the discrepancy in data. First, it may
be that the experimental data is inaccurate. Perhaps only one reaction is taking place, but
the number of moles of both H 2 and C 4 H 10 have been measured incorrectly. Perhaps more
than two reactions are taking place. Additional data is required in order to determine
which is the correct possibility. Moreover, the analysis of pilot-plant operations must be
improved so that all three species balances (carbon, hydrogen and sulfur) close to within
reasonable tolerances. What specific actions would you recommend to the team that is
operating the pilot plant