GROUP (28)
1
SHEET (8)
THE REACTION INVOLVING
GASES
2
Question No. (1)
Calculate ΔG°(R) at 1000K for the reaction:
CH4(g) + CO2(g)  2CO(g) + 2H2(g)
At what temperature does Kp =1? In which direction
does the equilibrium shift when:
(a)
the temperature of an equilibrated CH4-CO2-CO-
H2 gas mixture is increased?
(b) the total pressure is decreased?
Given that:
C(s)+2H2(g)CH4(g) ΔG°(R) = -16,520 +12.25 T log
T -15.62T cal/mole
C(s) + 0.5O2(g) CO(g) ΔG°(R) = -26,700 -20.95 T
cal/mole
C(s) + O2(g)
 CO(g)
ΔG°(R) = -94,200 -0.2 T
cal/mole
The Answer of Question No. (1)
3
1-C+2H2CH4
(x-1)
ΔG°1 = -16,520 +12.25 T
log T-15.26T
2-C+0.5O2 CO
(x2)
ΔG°2 = -26,700 -20.95 T
3-C+O2 CO
(x-1)
ΔG°3 = -94,200 -0.2 T
ΔGº4= (ΔG°1*-1) + (ΔG°2*2) + (ΔG°3*-1)
=(-16,162*-1 -26,700*2 -94200*-1) + (15.62*-1 -20.95*2 -0.2*-1)T + (12.25*-1)T logT
=57,420-26.08T-12.25T logT
At T=100K
ΔGº4=57,420 -26.08*1000 -12.25*1000*log1000
=-5,410
=-RTlnKp
Kp=1 & ln1=0
So ΔGº4=Zero
4
ΔGº4=57,420 -26.08T-12.25*T logT=0
T=900  1,377.4
T=920  24.5
T=921  -43.19
T=920.4  -2.57
T=920.3  4.19
T=920.35 K
Case 1:
-According to Le Chatilier Principle, the
equilibrium point will move towards the
reactants & the gas mixture will increase & the
CO% , H2% will decrease.
Case 2:
-∑ no. of moles of products > ∑ no. of moles of
reactants
5
As Total pressure decrease, the equilibrium
point will move towards the products.
6
Question No. (2)
A gas mixture of 50% CO, 25% CO2 and 25% H2 (by
volume) is fed into a furnace at 900 °C. Find the
composition of the equilibrium
CO-CO2-H2O-H2 gas
if the total pressure in the furnace is
1
atmosphere. Given that:
C(s) + 0.5O2(g)  CO(g)
ΔG°(R) = -26,700 -20.95
T cal/mole
C(s) + O2(g)
 CO(g)
ΔG°(R) = -94,200 -0.2 T
H2(g) +0.5O2(g) H2O(g)
ΔG°(R) = -58,900 + 13.1 T
cal/mole
cal/mole
The Answer of Question No. (2)
H2(g) + CO2(g)  CO(g) + H2O
1- C+0.5O2 CO
2- C+O2CO2
(x-1)
ΔG°1 = -26,700 -20.95 T
(x1)
ΔG°2 = -94,200 -0.2 T
7
3- H2+0.5O2 H2O
(x-1)
ΔG°3 = -58,900 + 13.1 T
ΔGº4= (ΔG°1*-1) + (ΔG°2*1) + (ΔG°3*-1)
=-8,600-7.65T
At T= 1173K
ΔGº4= 373.45 cal
H2(g) + CO2(g)  H2O + CO(g)
No.of moles:
1

1
+ 1
0.25
0.25

-
0.5
X
X

X
X
(0.25-X) (0.25-X) 
X
(0.5+X)
Inter:
Reactants:
Residue :
1 +
SO:
P(CO)= 0.25-X
P(H2)= 0.25-X
8
P(H2O)= X
P(CO2)= 0.5+X
Total Pressure = P(CO)+ P(H2)+ P(H2O)+ P(CO2)=1
ΔGº= -373.45 = -RTlnK
lnK = (373.45\1.987*1173)= 0.16
K =1.17
𝐏(𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬)
K=
𝐏(𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)
(𝟎.𝟓+𝑿)𝑿
𝟎.𝟐𝟓−𝑿)(𝟎.𝟐𝟓−𝑿)
=(
1.17(0.0625-0.5X+X2) = 0.5X2 + X2
0.073-1.085X+0.17X2 = Zero
X=
−𝒃±√𝒃𝟐 −𝟒𝒂𝒄
𝟐𝒂
9
X=
𝟏.𝟎𝟖𝟓±√𝟏.𝟎𝟖𝟓𝟐 −𝟒∗𝟎.𝟏𝟕∗𝟎.𝟎𝟕𝟑
𝟐∗𝟎.𝟏𝟕
X = 6.314% = 0.063
P(CO2) = 0.25-X = 0.25 – 0.063 = 0.187
P(H2) = 0.25-X = 0.187
P(H2O) = 0.063
P(CO) = 0.5 + 0.063 = 0.563
Therefore:
 18.7% CO
 18.7% H2
 56.3% CO2
 6.3% H2O
10
Question No. (3)
How much heat is evolved when 1 mole of SO2 and
0.5 mole of O2, each at 1 atmosphere pressure, react
to form the equilibrium
SO3-SO2-O2 mixture at
1000K and 1 atmosphere pressure? Given that:
SO2(g) + 0.5O2(g)  SO3(g)
ΔG°(R) = -22,600
+21.36 T cal
The Answer of Question No. (3)
ΔGº = -22,600+21.36T = ΔH-TΔS
ΔHº = -22,600 cal
ΔGº = -RTlnK
K = 1.87
K = 1.87 =
𝐏(𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬)
𝐏(𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)
11
SO2(g) + 0.5O2(g)  SO3(g)
No.of moles:
1 +
0.5

1
Reactants:
X
0.5 X

X
(1-X) (0.5-0.5X) 
X
Residue :
N(Total) = 1-X+0.5-0.5X+X = 0.5(3-X)
P(SO2) =
P(O2) =
𝟐(𝟏−𝑿)𝑷
(𝟑−𝑿)
(𝟏−𝑿)𝑷
(𝟑−𝑿)
P(SO3) =
𝟐𝑿𝑷
(𝟑−𝑿)
K = P(SO3)/ P(O2)0.5 * P(SO2)
K2 = P(SO3)2 / P(O2) * P(SO2)2
(1.87)2 = (3-X)X2 / (1-X)3P
,P = 1 atm
3.48 = 3X2-X3 / 1-X3+3X2-3X
12
3.48+3.48X3-10.48 X2 = 3 X2- X3-10.48X
(1-3.48) X3+(10.48-3) X2-10.48X+3.48=0
-2.48X3+7.48X2-10.48X+3.48=0
X = 0.463
For equilibrium we need to react 0.463 moles
So:
ΔHº = nΔHº
= 0.463*-22,600
= -10,463.8 cal
13
Question No. (4)
A CO2-CO-H2-H2O gas mixture at a total pressure that
exerts
an
oxygen
partial
pressure
of
10-7
atmospheres at 1600 °C. In what ratio must CO2 and
H2 be mixed to produce this value of P(O2)? What
oxygen partial pressure is exerted by the equilibrium
gas mixture produced by mixing CO2 and H2 in the
ratio 3:1? Given that:
C(s) + 0.5O2(g)  CO(g)
ΔG°(R) = -26,700 -20.95
T cal/mole
C(s) + O2(g)
 CO(g)
ΔG°(R) = -94,200 -0.2 T
H2(g) + 0.5O2(g)  H2O(g)
ΔG°(R) = -58,900 + 13.1
cal/mole
T cal/mole
The Answer of Question No. (4)
CO2(g) + H2(g)  H2O + CO(g)
14
1- C+0.5O2 CO
2- C+O2CO2
(x-1)
ΔG°1 = -26,700 -20.95 T
(x1)
3- H2+0.5O2 H2O
ΔG°2 = -94,200 -0.2 T
(x-1)
ΔG°3 = -58,900 + 13.1 T
ΔGº4= (ΔG°1*-1) + (ΔG°2*1) + (ΔG°3*-1)
=-8,600-7.65T
At T= 1873 K
ΔGº4= -22928.45 cal
H2(g) + CO2(g)  H2O + CO(g)
No.of moles:
1 +
a

0
Reactants:
X
X

X
X
Residue :
(a-X)
(1-X) 
X
X
CO2/H2 = a
nT = a-X+1-X+X+X = a+1
15
+ 0
For: CO+0.5O2CO2
K1 = P(CO2)/ P(O2)0.5 * P(CO)
For: H2+0.5O2H2O
K1 = P(H2O)/ P(O2)0.5 * P(H2)
P(CO)=
P(H2)=
𝒂−𝑿
𝒂+𝟏
𝟏−𝑿
𝒂+𝟏
P(H2O)=
P(CO2)=
Pt
Pt
𝑿
𝒂+𝟏
𝒂−𝑿
𝒂+𝟏
Pt
Pt
Pt = 1
Total Pressure = P(CO)+ P(H2)+ P(H2O)+ P(CO2)=1
ΔGº= -22928.45 = -RTlnK
lnK = (22928.45\1.987*1873)= 6.16
K = 473.82
K=
𝐏(𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬)
𝐏(𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)
16
𝑿∗𝑿
𝒂−𝑿)(𝟏−𝑿)
=(
Get X,a
P(CO2)/P(H2) = 1.294
Part 2:
H2(g) + CO2(g)  H2O + CO(g)
No.of moles:
1 +
3

0
Reactants:
X
X

X
X
Residue :
(3-X)
(1-X) 
X
X
nT = 3-X+1-X+X+X = 4
𝑿
P(CO)= Pt
𝟒
P(H2)=
𝟏−𝑿
𝟒
Pt
𝑿
P(H2O)= Pt
𝟒
P(CO2)=
𝟑−𝑿
𝟒
Pt
17
+ 0
Pt = 1
Total Pressure = P(CO)+ P(H2)+ P(H2O)+ P(CO2)=1
ΔGº= -22928.45 = -RTlnK
lnK = (22928.45\1.987*1873)= 6.16
K = 473.82
𝐏(𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬)
K=
𝐏(𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)
(𝑿∗𝑿)
𝟏−𝑿)(𝟑−𝑿)
=(
X = 0.26
P(O2) = (P(CO)*K)/P(CO2)
=(0.065*473.82)/0.685
=1.08*10-6 atm
18
Question No. (5)
By establishing the equilibrium: PCl5  PCl3 + Cl2
In a mixture of PCl5 and PCl3, it is required to obtain
a partial pressure of Cl2 of 0.1 atmosphere at 500K
when the total pressure is 1 atm. in what ratio must
PCl5 and PCl3 be mixed? Given that:
PCl3(g)+ Cl2(g) PCl5(g) ΔG°(R)=-22,850– 4.37T log
T+ 56.22 T cal
The Answer of Question No. (5)
PCl3(g)+ Cl2(g) PCl5(g)
ΔGº = -637.2
PCl5  PCl3 + Cl2
ΔGº = 637.2
637.2 = -RTlnK
K = 0.527 = P(PCl3)*P(Cl2) / P(PCl5)
P(PCl3) / P(PCl5) = 5.265
19
P(PCl3) = P(PCl5) * 5.265
PT = P(PCl3) + P(Cl2) + P(PCl5) = 1
1 = P(PCl3) + 5.265 P(PCl5) + 0.1
P(PCl5) = 0.171 atm
P(PCl3) = 0.8999 atm
P(Cl5) / P(PCl3) = 0.19
20
Question No. (6)
Calculate the partial pressure of monatomic
hydrogen in hydrogen gas at 2000K and 1
atmosphere pressure; given that, for
0.5
H2(g) = H(g)
ΔH° at 298K = 52,102 cal and ΔS° at 298K = 11.795
e.u.
Assume the heat capacity of monatomic hydrogen to
be that of an ideal gas.
The Answer of Question No. (6)
𝟐𝟎𝟎𝟎
ΔHº at 2000 K = ΔHº298 + ∫𝟐𝟗𝟖 𝑪𝒑𝒅𝑻
= 52,102 + 4.9675 (2000-298)
= 60,556.7 cal
𝟐𝟎𝟎𝟎 𝑪𝒑
ΔSº at 2000 K = ΔSº298 + ∫𝟐𝟗𝟖
21
𝑻
𝒅𝑻
= 11.795 + 4.9675 ln(2000/298)
= 21.252 e.u
ΔGº = ΔHº - TΔSº
= 18,052.4 = -RTlnK
K = 0.0106 = P(H)/(P(H2))0.5
1.1234*10-4(1-P(H)) = P(H)2
8,900 P(H)2 + P(H) -1 = 0
P(H) = 0.0108 atm
22