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What is the difference between steady state and equilibrium? Steady state implies no changes
with passage of time. Likewise, equilibrium can also imply no change of state with passage of time.
In many situations, this is the case—the system is not only at steady state but also at equilibrium.
However, this is not always the case. In some cases, where the flow rates are steady but the phase
contents, for example, are not being maintained at the equilibrium values, the system is at steady
state but not at equilibrium.
3.6 MATH OPERATIONS AND LAWS OF EQUILIBRIUM
Earlier we observed that no chemical reaction goes to completion. There are qualitative conse-
quences of this insight that go beyond the purpose of this text, but in this text we are interested in the
basic quantitative aspects of equilibria. The chemist usually starts with the chemistry of the reaction
and fully utilizes chemical intuition before resorting to mathematical techniques. That is, science
should always precede mathematics in the study of physical phenomena. Note, however, that most
chemical problems do not require exact, closed-form solutions, and the direct application of math-
ematics to a problem can lead to an impasse. Several basic math operations and fundamental laws
from physical chemistry and thermodynamics serve as the tools, blueprints, and foundational struc-
tures of mathematical models. They can be used and applied to environmental systems under cer-
tain conditions to solve a variety of problems. Many laws serve as important links between the state
of a system, its chemical properties, and its behavior. As such, some of the basic math operations
used to solve basic equilibrium problems and laws essential for modeling the fate and transport of
chemicals in natural and engineered environmental systems are reviewed in the following sections.
3.6.1 s
olving
e
quilibrium
p
roblems
In the following math operations, we provide examples of the various forms of combustion of hydro-
gen to yield water to demonstrate the solution of equilibrium problems. Let's first consider the
reaction at 1000.0 K where all constituents are in the gas phase and the equilibrium constant is
1.15 × 10
10
atm
-1
. This reaction is represented by the following equation and equilibrium constant
expression (Equation 3.12).
2H
2
(g) + O
2
(g) = 2H
2
O (g)
K = [H
2
O]
2
/[H
2
]
2
[O
2
]
(3.12)
where concentrations are given as partial pressures in atm. Observe that K is very large; conse-
quently, the concentration of water is large and/or the concentration of at least one of the reactants
is very small.
■
EXAMPLE 3.3
Problem:
Consider a system at 1000.0 K in which 4.00 atm of oxygen is mixed with 0.500 atm of
hydrogen and no water is initially present. Note that oxygen is in excess and hydrogen is the limiting
reagent. Because the equilibrium constant is very large, virtually all of the hydrogen is converted to
water, yielding [H
2
O] = 0.500 atm and [O
2
] = 4.000 - 0.5(0.500) = 3.750 atm. The final concentra-
tion of hydrogen, a small number, is an unknown, the only unknown.
Solution:
Using the equilibrium constant expression, we obtain
1.15 × 10
10
= (0.500)
2
/[H
2
]
2
(3.750)
from which we determine that [H
2
] = 2.41 × 10
-6
atm. Because this is a small number, our initial
approximation is satisfactory.
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