# PastTripos2006Question2 ChemicalEquilibrium

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```					Problem:
A rigid container (constant volume) holds 1 mole of carbon monoxide and 1 mole of oxygen at 25 oC
and 1 bar(a) (this value must be an absolute pressure, and we can denote this by ‘a’ to avoid
confusion). The reactive mixture is ignited and after some time chemical equilibrium is reached, with
the following species present: carbon dioxide, carbon monoxide and oxygen (which we are told can
be assumed to be ideal gases) at 2600 K and 6.625 bar(a).

Chemistry:
The reaction of complete combustion of carbon monoxide with pure oxygen is:

For complete combustion of 1 mole of carbon monoxide with 1 mole of pure oxygen we have:

where the coefficients in front of each reactant (and product) are proportional to the number of
moles of that reactant (or product).

At the high temperatures associated with combustion, carbon dioxide may dissociate (see
Thermofluids Data Book, Reaction 7):

or,

or,

or:

Combining Eqns. 1 and 3 we have,

or:
Equilibrium Reaction Thermodynamics:
Consider the final chemical reaction equation, Eqn. 4. After ignition we only have the products on
the right-hand-side present at Tp = 2600 K and pp = 6.625 bar(a).

Note, that for perfect gases, the number of moles of that gas is proportional to its volume, with 1
kmol of any perfect gas occupying approximately 22.4 m3 at standard temperature and pressure, or
s.t.p. (0oC and 1 bar). Hence, the mole and volumetric fractions of carbon monoxide and oxygen in
the reactants in Eqn. 4 are both 1/2, whereas the mole and volumetric fractions of carbon dioxide,
carbon monoxide and oxygen in the products in Eqn. 4 are (with XCO2 + XCO + XO2 = 1):

where np = (3+α)/2 is the total number of moles of all the products.

The equilibrium constant for the relevant equilibrium reaction, Eqn. 2, is given in the Data Book as,

where ln(Kp)bar at the equilibrium temperature of Tp = 2600 K for this reaction is equal to 2.8, po is a
constant equal to 1 bar, and:
νi
(number of moles in Eqn. 2;                          pi’
i              +ve for Reactants / -ve for products)          (partial pressure in bar)
CO2                                 1                                     XCO2pp
CO                                 -1                                     XCOpp
O2                               -0.5                                    XO2pp
where the equilibrium pressure was given, pp = 6.625 bar(a).

Hence, substituting these results into Eqn. 5,

into which we can substitute the given value of α = 0.03884 to verify that this is indeed a solution.
Now, we know that for each 2 moles of reactants there will be,

moles of products.

Since we can treat all gases (both reactant and product) as ideal,

and given that this is an isochoric (constant volume) process:

Finally, in the last part of this question we consider the energy balance in this process. During the
combustion process, chemical energy contained in the bonds of the reactants is released, and
transformed into other forms of energy. Depending on the type of process, there are two ways to
approach this balance of energy.

The most common is a constant pressure combustion process occurring due to the steady flow of
reactants into a combustor, out of which there is a steady flow of products. In such a process, it is
convenient to draw a control volume around the combustor, and the relevant statement of energy
conservation is the Steady Flow Energy Equation:

Since there is no shaft work in the combustor, and ignoring changes in kinetic and potential energy:

The second type of process that you may encounter is a constant volume combustion process
occurring inside a combustion chamber of fixed volume, initially filled with reactants and afterwards
filled with products. In this process, it is convenient to consider a system of constant mass inside the
combustion chamber, and the relevant statement of energy conservation is the First Law of
Thermodynamics:
Since there is no displacement work (dW = pdV) in the combustion chamber, this being a constant
volume process:

Note that we can get both Eqns. 6 and 7 to contain either enthalpy or internal energy, since these
two quantities are related via H = U + pV and h = u + pv. We chose the former:

Eqns. 8 do not actually contain a term to account for changes in ‘chemical’ energy. In deriving the
Steady Flow Energy Equation and First Law it was assumed that there was no change in chemical
composition and hence no consequent release (combustion is always exothermic and results in
energy release) of energy. One way to get around this, is to define the so-called ‘sensible’ enthalpy,
such that total enthalpy change equals sensible enthalpy change plus formation enthalpy change,

where the sensible enthalpy change accounts for the change in enthalpy due to a change in
temperature, and the change in formation enthalpy (always negative in combustion) accounts for
the release of any energy due to combustion, arising from the fact that the chemical energy
contained in the bonds of the reactants is higher than that contained in the bonds of the products.

We can drop the subscript ‘sens’ with the understanding that it is the sensible enthalpy we mean:

As we can see from Eqns. 9, the released chemical energy (-ΔHf) may either escape from the control
volume or the system boundaries in the form of heat (Q), or remain inside the control volume or
system in the form of an increase in (sensible) enthalpy (H), thus manifesting itself as an increase in
temperature, or both. Three examples (isothermal, adiabatic, general) are shown graphically below:
H or U                            H or U                            H or U
Reactants
Reactants
Reactants                             Q              1
Products                         Products
-ΔHf
-ΔHf
Iso-
o
25 C         thermal T                  25oC           T            25oC                 T
(a) Isothermal                   (b) Adiabatic                      (c) General
In any case, we can use our knowledge of combustion thermodynamics to evaluate the various
changes of enthalpy in Eqns. 9. The problem at hand is of the general type (c), but with the reactants
already being supplied at 298 K. Specifically, the enthalpy changes are shown in the Figure below:

H or U

Reactants
1 -Q        Products
2

-ΔHf

25oC                T

Hence, we may write:

Enthalpy values for various gases are tabulated in the Thermofluids Data Book (per unit mole) at
various temperatures. Having obtained the value of α, we know the number of moles of each
product and reactant species, and can hence evaluate the total change in (sensible) enthalpy in the
products and reactants. In this case, the change in (sensible) enthalpy relating to the reactants is
zero, since the reactants are already supplied at 298 K, and hence we need only consider the change
in (sensible) enthalpy in taking the products from 298 K to 2600 K. Clearly, since ΔHf is negative as
stated before, and by looking that the magnitude of the (sensible) enthalpy change in the products
versus that of ΔHf, we can see that the heat flow is negative, signifying a heat loss from the system
to the surroundings.

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