PHYSICAL CHEMISTRY IN BRIEF

CHAP. 8: CHEMICAL EQUILIBRIUM [CONTENTS] 247

we obtain

lnK(T 2 ) = lnK(T 1 )−

1

R

[

A

( 1

T 2

−

1

T 1

)

−∆aln

T 2

T 1

−

∆b

2

(T 2 −T 1 )−

∆c

6

(

T 22 −T 12

)

−

∆d

2

(

1

T 22

−

1

T 12

)]

, (8.21)

whereT 1 denotes the temperature at which we know the value of the equilibrium constant,

A= ∆rH◦(To)−∆aTo−

∆b

2

To^2 −

∆c

3

To^3 +

∆d

To

,

∆rH◦(To) is the value of the standard reaction enthalpy at temperatureToa ∆a=

∑k
i=1νiai,
∆b=

∑k

i=1νibi,∆c=

∑k

i=1νici,∆d=

∑k

i=1νidi.

Example

Calculate the equilibrium constant of butane(1) isomerization to 2-methylpropane(2) at 500 K if

the equilibrium constant for this reaction at 300 K is 4.5 (standard state: ideal gas at the

temperature of the system and a pressure of 101.325 kPa).

Input data: ∆rH◦=−8.368 k J mol−^1 atT = 298 K,Cpm,◦ 1 = 123.1 J mol−^1 K−^1 andCpm,◦ 2 =

123.37 J mol−^1 K−^1.

Solution

∆Cp= 0.27 J mol−^1 K−^1 ,

∆rH◦(T) =∆rH◦(298) +∆Cp(T−298) =−8448.46 + 0.27T. Substitution to (8.19) and

integration yields

lnK(500) = lnK(300) +

8448. 46

R

( 1

500

−

1

300

)

+

0. 27

R

ln

500

300

= 1.504 + 1016. 2

(

1

500

−

1

300

)

+ 0.0325 ln

500

300

.

From this we obtainK(500) = 1. 180.