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212 6 CHEMICAL EQUILIBRIUM
I6.12 (a) �e standard reaction enthalpy is 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, from which ∆rH−○ can be determined.�e
value of ∆rS−○ is found by combining ∆rG−○ = ∆rH−○ − T∆rS−○ [3D.9–100]
and ∆rG−○ = −RT lnK [6A.15–208]. Equating these expressions for ∆rG−○
gives
−RT lnK = ∆rH−○ − T∆rS−○ hence − lnK = ∆rH
−○
R
1
T
− ∆rS
−○
R
Assuming that ∆rH−○ and ∆rS−○ do not very signi�cantly over the temper-
ature range of interest this equation implies that a plot of − lnK against
1/T should be a straight line of intercept −∆rS−○/R, from which ∆rS−○ can
be determined; such a plot is shown in Fig. 6.6.�e plot will have a slope
of ∆rH−○/R, as already deduced above.
T/K K 1/(T/K) − lnK
233 4.13 × 108 0.004 29 −19.8
248 5.00 × 107 0.004 03 −17.7
258 1.45 × 107 0.003 88 −16.5
268 5.37 × 106 0.003 73 −15.5
273 3.20 × 106 0.003 66 −15.0
280 9.62 × 105 0.003 57 −13.8
288 4.28 × 105 0.003 47 −13.0
295 1.67 × 105 0.003 39 −12.0
303 6.02 × 104 0.003 30 −11.0
�e data fall on a reasonable straight line, the equation of which is
− lnK = −8 787 × 1/(T/K) + 17.62
∆rH−○/R is determined from the slope
∆rH−○ = R × (slope ×K) = (8.3145 JK−1mol−1) × (−8787 K)
= −73.0... kJmol−1 = −73.1 kJmol−1
−∆rS−○/R is determined from the intercept
∆rS−○ = −R × intercept = −(8.3145 JK−1mol−1) × (+17.62)
= −1.46... × 102 JK−1mol−1 = −147 JK−1mol−1
(b) �e standard reaction enthalpy for the reaction 2ClO(g)→ (ClO)2(g) is
expressed in terms of standard enthalpies of formation
∆rH−○ = ∆fH−○ [(ClO)2 , g] − 2∆fH−○(ClO, g)