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174 6 CHEMICAL EQUILIBRIUM
E6A.9(b) For the reaction N2O4(g) ⇌ 2NO2(g) the following table is drawn up by
supposing that there are n moles of N2O4 initially and that at equilibrium a
fraction α has dissociated.
N2O4 ⇌ 2NO2
Initial amount n 0
Change to reach equilibrium −αn +2αn
Amount at equilibrium (1 − α)n 2αn
Mole fraction, xJ
1 − α
1 + α
2α
1 + α
Partial pressure, pJ
(1 − α)p
1 + α
2αp
1 + α
�e total amount inmoles is ntot = (1−α)n+2αn = (1+α)n.�is value is used
to �nd the mole fractions. In the last line, pJ = xJp has been used. Treating all
species as perfect gases so that aJ = (pJ/p−○), the equilibrium constant is
K =
a2NO2
aN2O4
= (pNO2/p−○)2
(pN2O4/p−○)
=
p2NO2
pN2O4 p−○
=
( 2α p
1+α )
2
( (1−α)p
1+α ) p−○
= 4α2
(1 − α)(1 + α)
p
p−○
In this case α = 0.201 and p = 1.00 bar; recall that p−○ = 1 bar.
K = 4 × 0.2012
(1 − 0.201) × (1 + 0.201)
× 1.00 bar
1 bar
= 0.168
E6A.10(b) �e relationship between K and Kc is [6A.18b–209], K = Kc × (c−○RT/p−○)∆ν .
For the reaction 3N2(g) +H2(g)⇌ 2HN3(g)
∆ν = νHN3 − νN2 − νH2 = 2 − 3 − 1 = −2 hence K = Kc × (c−○RT/p−○)−2
p−○/c−○R evaluates to 12.03 K so the relationship can alternatively be written as
K = Kc × [(12.03)/(T/K)]2.
E6A.11(b) �e following table is drawn up:
A + B ⇌ C + 2D
Initial amount, nJ,0/mol 2.00 1.00 0 3.00
Change, ∆nJ/mol −0.79 −0.79 +0.79 +1.58
Equilibrium amount,
nJ/mol
1.21 0.21 0.79 4.58
Mole fraction, xJ 0.178... 0.0309... 0.116... 0.674...
Partial pressure, pJ (0.178...)p (0.0309...)p (0.116...)p (0.674...)p