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172 6 CHEMICAL EQUILIBRIUM
E6A.5(b) �e equilibrium constant is de�ned by [6A.14–207], K = (∏J a
νJ
J )equilibrium.
�e ‘equilibrium’ subscript indicates that the activities are those at equilibrium
rather than at an arbitrary stage in the reaction; however this subscript is not
usually written explicitly. In this case
K = a−1CH4(g)a
−3
Cl2(g)aCHCl3(l)a
3
HCl(g) =
aCHCl3(l)a
3
HCl(g)
aCH4(g)a
3
Cl2(g)
�e activity of CHCl3(l) is 1 because it is a pure liquid. Furthermore if the
gases are treated as perfect then their activities are replaced by aJ = pJ/p−○ .�e
equilibrium constant becomes
K = (pHCl/p−○)3
(pCH4/p−○)(pCl2/p−○)3
=
p3HClp
−○
pCH4 p
3
Cl2
E6A.6(b) �e standard reaction Gibbs energy is given by [6A.13a–207]
∆rG−○ = ∑
Products
ν∆fG−○ − ∑
Reactants
ν∆fG−○
�e relationship between ∆rG−○ and K, [6A.15–208], ∆rG−○ = −RT lnK, is then
used to calculate the equilibrium constant.
(i) For the reaction of mercury with chlorine
∆rG−○ = ∆fG−○(HgCl2 , s) − {∆fG−○(Hg, l) + ∆fG−○(Cl2 , g)}
= ∆fG−○(HgCl2 , s) = −178.6 kJmol−1
�en
K = e−∆rG
−○/RT = exp(− −178.6 × 103 Jmol−1
(8.3145 JK−1mol−1) × (298 K)
) = 2.02 × 1031
(ii) For the reduction of copper ions by zinc
∆rG−○ = ∆fG−○(Zn2+ , aq) + ∆fG−○(Cu, s)
− {∆fG−○(Zn, s) + ∆fG−○(Cu2+ , aq)}
= ∆fG−○(Zn2+ , aq) − ∆fG−○(Cu2+ , aq)
= (−147.06 kJmol−1) − (+65.49 kJmol−1) = −2.12... × 105 Jmol−1
�en
K = e−∆rG
−○/RT = exp(− −2.12... × 105 Jmol−1
(8.3145 JK−1mol−1) × (298 K)
) = 1.80 × 1037
Neither reaction has K