Chem 301 Problem Day #6a
Fall 2008
1. Gaseous NOCl decomposes to form NO and Cl2. 0.150 moles of NOCl were placed in a
2.00 L vessel at 598 K and allowed to react to reach equilibrium with NO and Cl2. The amount
of NO was then found to be 0.0630 moles at equilibrium.
a. Calculate Kp and ΔGorxn for this decomposition reaction at 598 K.
PNOCl = nRT/V = (0.150)(.083145)(598)/(2.00) = 3.73 bar; PNO = nRT/V = (0.0630)(.083145)(598)/(2.00) = 0.783 bar
2NOCl 2NO + Cl2 NOCl NO + (1/2)Cl2
I 3.37 0 0 I 3.37 0 0
C -2x +2x +x C -x +x +(1/2)x
E 2.16 1.57 0.783 E 2.16 1.57 0.783
Kp = (PNO)2(PCl2) /(PNOCl)2 Kp = (PNO)(PCl2)1/2 /(PNOCl)
Kp = (1.57)2(0.783)/(2.16)2 Kp = (1.57)(0.783)1/2/(2.16)
Kp = 0.414 Kp = 0.643
ΔGorxn = -RTlnKp ΔGorxn=-RTlnKp
ΔGorxn = -(8.3145)(598)(ln(0.414)) ΔGorxn = -(8.3145)(598)(ln(0.643))
ΔGorxn = 4.38 kJ ΔGorxn = 2.20 kJ
b. Using the van ‘t Hoff equation, calculate the value of Kp at 298 K and use this value to
determine the ΔGorxn at 298 K.
kJG
xG
KRTG
xK
K
TTR
H
KK
kJH
o
rxn
o
rxn
p
o
rxn
Kp
Kp
o
rxn
pp
o
rxn
9.41
)1045.4ln()298)(3145.8(
ln
1045.4
598
1
298
1
3145.8
79200
)410.0ln(ln
11
lnln
2.79)7.51(23.91(2
8
8
298,
298,
12
1,2,
kJG
xG
KRTG
xK
K
TTR
H
KK
kJH
o
rxn
o
rxn
p
o
rxn
Kp
Kp
o
rxn
pp
o
rxn
3.22
)1023.1ln()298)(3145.8(
ln
1023.1
598
1
298
1
3145.8
39600
)643.0ln(ln
11
lnln
6.397.513.91
4
4
298,
298,
12
1,2,
