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184 6 CHEMICAL EQUILIBRIUM
E6B.9(b) (i) Treating the vapour as a perfect gas so that aJ = pJ/p−○ , and noting that
pure solids have aJ = 1, the equilibrium constant for the dissociation
NH4Cl(s)⇌ NH3(g) +HCl(g) is
K =
aNH3 ,gaHCl,g
aNH4Cl,s
= (pNH3/p−○)(pHCl/p−○)
1
= pNH3 pHCl
p−○ 2
Furthermore, because NH3 and HCl are formed in a 1 ∶ 1 ratio, they each
have amole fraction of 1/2 and the partial pressure of each is half the total
pressure: pA2 = pB = 1
2 p.�e equilibrium constant at 427
○C is therefore
K =
( 12 p)(
1
2 p)
p−○ 2
= p2
4p−○ 2
= (608 kPa)2
4 × (100 kPa)2
= 9.24... = 9.24
(ii) �e standard reaction Gibbs energy at 427 ○C is obtained using ∆rG−○ =
−RT lnK [6A.15–208]
∆rG−○ = −(8.3145 JK−1mol−1) × ([427 + 273.15] K) × ln(9.24...)
= −1.29... × 104 Jmol−1 = −12.9 kJmol−1
(iii) �e variation of K with temperature, assuming that ∆rH−○ does not vary
with T over the temperature range of interest, is given by [6B.4–215]
lnK2−lnK1 = −
∆rH−○
R
( 1
T2
− 1
T1
) hence ∆rH−○ = − R ln(K2/K1)
(1/T2) − (1/T1)
Noting that the above equilibrium constant expression K = p2/4p−○ 2 im-
plies that ln(K2/K1) = ln(p22/p21), ∆rH−○ is calculated as
∆rH−○ = −
(8.3145 JK−1mol−1) × ln ((1115 kPa)2/(608 kPa)2)
[1/(459 + 273.15) K] − [1/(427 + 273.15) K]
= +1.61... × 105 Jmol−1 = +162 kJmol−1
(iv) �e standard reaction entropy is obtained from ∆rG−○ and ∆rH−○ by rear-
ranging ∆rG−○ = ∆rH−○ − T∆rS−○ [3D.9–100]:
∆rS−○ =
∆rH−○ − ∆rG−○
T
= (1.61... × 105 Jmol−1) − (−1.29... × 104 Jmol−1)
700.15 K
= +249 JK−1mol−1
Solutions to problems
P6B.2 (a) �e reaction CH4(g)⇌ C(s, graphite)+ 2H2 is the reverse of the forma-
tion reaction of methane, so
∆rH−○ = −∆fH−○(CH4 , g) = −(−74.85 kJmol−1) = +74.85 kJmol−1
∆rS−○ = −∆fS−○(CH4 , g) = −(−80.67 JK−1mol−1) = +80.67 JK−1mol−1