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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 185
�e equilibrium constant K is calculated using ∆rG−○ = −RT lnK [6A.15–
208], with ∆rG−○ being obtained from ∆rG−○ = ∆rH−○ −T∆rS−○ [3D.9–100]
K = e−∆rG
−○/RT = exp(−∆rH
−○ − T∆rS−○
RT
)
= exp(−(74.85 × 103 Jmol−1) − (298 K) × (80.67 JK−1mol−1)
(8.3145 JK−1mol−1) × (298 K)
)
= 1.24 × 10−9
(b) �e temperature dependence of K is given by [6B.4–215]
lnK2 = lnK1 −
∆rH−○
R
( 1
T2
− 1
T1
)
assuming that ∆rH−○ is constant over the temperature range of interest.
�e value of K at 50 ○C is therefore
lnK2 = ln(1.24... × 10−9)
− 74.85 × 10
3 Jmol−1
8.3145 JK−1mol−1
( 1
(50 + 273.15) K
− 1
298 K
)
= −18.1... hence K2 = 1.30 × 10−8
(c) �e following table is drawnup, assuming that the initial amount ofmethane
is n and that at equilibrium a fraction α has dissociated. Graphite is a solid
so is not included in the calculations.
CH4(g) ⇌ 2H2(g) + C(s)
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 in moles is ntot = (1 − α)n + 2αn = (1 + α)n. �is
value is used to �nd the mole fractions. Treating CH4 and H2 as perfect
gases, so that aJ = pJ/p−○ , and recalling that pure solids have aJ = 1, the
equilibrium constant is
K =
a2H2aC
aCH4
= (pH2/p−○)2 × 1
(pCH4/p−○)
=
p2H2
pCH4 p−○
=
( 2α p
1+α )
2
( (1−α)p
1+α ) p−○
= 4α2
(1 − α)(1 + α)
p
p−○
= 4α2
1 − α2
p
p−○