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188 6 CHEMICAL EQUILIBRIUM
P6B.8 �e following table is drawn up for the I2(g)⇌ 2I(g) equilibrium:
I2 ⇌ 2I
Initial amount nI2 0
Change to reach equilibrium −αnI2 +2αnI2
Amount at equilibrium (1 − α)nI2 2αnI2
Mole fraction, xJ
1 − α
1 + α
2α
1 + α
Partial pressure, pJ
(1 − α)p
1 + α
2αp
1 + α
�e total amount in moles is ntot = (1− α)nI2 + 2αnI2 = (1+ α)nI2 .�is value
is used to �nd the mole fractions. Treating all species as perfect gases, so that
aJ = pJ/p−○ , the equilibrium constant is:
K = a
2
I
aI2
= (pI/p−○)2
(pI2/p−○)
= p2I
pI2 p−○
=
( 2α p
1+α )
2
( (1−α)p
1+α ) p−○
= 4α2
(1 − α)(1 + α)
p
p−○
�e total amount in moles is found from the pressure using the perfect gas law
[1A.4–8], pV = ntotRT . With this, α can then be determined.
pV = ntotRT = (1 + α)nI2RT hence α = pV
nI2RT
− 1
�ese expressions are used to calculate α and hence K from the given data; the
results are shown in the table below. �e standard enthalpy of dissociation is
then found using the van ’t Ho� equation [6B.2–214]:
d lnK
dT
= ∆rH
−○
RT2
which can also be written − d lnK
d(1/T)
= ∆rH
−○
R
�e second form implies that a graph of − lnK against 1/T should be a straight
line of slope ∆rH−○/R; the plot is shown in Fig. 6.2.
T/K 100p/atm 104nI2/mol α K 104/(T/K) − lnK
973 6.244 2.4709 0.0846 1.824 × 10−3 10.277 6.307
1 073 7.500 2.4555 0.1888 1.123 × 10−2 9.320 4.489
1 173 9.181 2.4366 0.3415 4.911 × 10−2 8.525 3.014
�e data fall on a good straight line, the equation of which is
lnK = 1.880 × 104/(T/K) − 13.02
∆rH−○/R is determined from the slope
∆rH−○ = R × (slope ×K) = (8.3145 JK−1mol−1) × (1.880 × 104 K)
= +156 kJmol−1