Page 206 - Physical Chemistry

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(b) If the reaction is run at fixed T and V, use P n RT/V to express each P in Section 6.4
i i i
terms of j . Thus: Ideal-Gas Equilibrium Calculations
eq
n i n RT
i
P x P P if P is known; P if V is known
i i i
n tot V
5. Substitute the P ’s (expressed as functions of j ) into the equilibrium-constant
i eq
expression K° (P /P°) and solve for j .
n i
i
i
eq
P
6. Calculate the equilibrium mole numbers from j and the expressions for n in
eq
i
step 3.
As an example, consider the reaction
N 1g2 3H 1g2 ∆ 2NH 1g2
2
2
3
with the initial composition of 1.0 mol of N , 2.0 mol of H , and 0.50 mol of NH . To
2 2 3
do step 3, let z j be the equilibrium extent of reaction. Constructing a table like
eq
that used in general-chemistry equilibrium calculations, we have
N 2 H 2 NH 3
Initial moles 1.0 2.0 0.50
Change z 3z 2z
Equilibrium moles 1.0 z 2.0 3z 0.50 2z
where n n j [Eq. (4.95)] was used to calculate the changes. The total number of
i i
moles at equilibrium is n 3.5 2z. If P is held fixed, we express the equilibrium
tot
partial pressures as P x P [(1.0 z)/(3.5 2z)]P, etc., where P is known. If
N 2 N 2
V is held fixed, we use P n RT/V (1.0 z)RT/V, etc., where T and V are
N 2 N 2
known. One then substitutes the expressions for the partial pressures into the K° expres-
P
sion (6.18) to get an equation with one unknown, the equilibrium extent of reaction z.
One then solves for z and uses the result to calculate the equilibrium mole numbers.
The equilibrium extent of reaction might be positive or negative. We define the
reaction quotient Q for the ammonia synthesis reaction as
P
P 2
Q NH 3 (6.41)
P 3
P P
N 2 H 2
where the partial pressures are those present in the mixture at some particular time, not
necessarily at equilibrium. If the initial value of Q is less than K [Eq. (6.19)], then
P P
the reaction must proceed to the right to produce more products and increase Q until
P
it becomes equal to K at equilibrium. Hence if Q K then j 0. If Q K ,
P
P
P
eq
P
P
then j 0.
eq
To find the maximum and minimum possible values of the equilibrium extent of
reaction z in the preceding NH example, we use the condition that the equilibrium
3
mole numbers can never be negative. The relation 1.0 z 0 gives z 1.0. The re-
2
lation 2.0 3z 0 gives z . The relation 0.50 2z 0 gives z 0.25. Hence
3
4
0.25 z 0.667. The K° equation has z as the highest power of z (this comes from
P
P P 3 in the denominator) and so has four roots. Only one of these will lie in the
N 2 H 2
range 0.25 to 0.667.
EXAMPLE 6.4 Equilibrium composition at fixed T and P
Suppose that a system initially contains 0.300 mol of N O (g) and 0.500 mol of
4
2
NO (g), and the equilibrium
2
N O 1g2 ∆ 2NO 1g2
2 4 2