GROUP(26)
1
SHEET (6-2)
Thermochemistry
2
Question No. (1)
Calculate the standard heat of formation at 25 °C
and 1 atmosphere pressure of the following
reaction:
3FeO + 2Al  Al2O3 + 3Fe ,
per mole of Al2O3 formed, per mole of Fe formed,
per mole of FeO reacted, per mole of Al reacted
and per gram of Fe formed. Given that:
ΔH°f(FeO) at 298K = - 63.3 kcal/mole,
ΔH°f(Al2O3) at 298K= - 400 kcal/mole, and
The atomic weight of Fe = 56
The Answer of Question No.(1)
. ΔH°(298k)= ΔH°Al2O3 - 3ΔH°FeO
= - 400000 – 3 ( -63300 )
3
ΔH°(298k)= - 210100 cal/mole of Al2O3
.ΔH°(298k)= 1/3 ΔH°Al2O3 - ΔH°FeO
= 1/3 (-400,000) + 633,00
ΔH°(298k)= - 70033 cal/mole Fe
.ΔH°(298k)= 1/3 ΔH°Al2O3 - ΔH°FeO
ΔH°(298k)= - 70033 cal/mole FeO
. ΔH°(298k)= o.5 ΔH°Al2O3 – 1.5ΔH°FeO
ΔH°(298k)= - 210100/2 = - 105 kcal/mole Al
. ΔH°(298k)= - 70.033 kcal/mole Fe
= - 70.033 * 56
= - 3921.848 kcal/gm
4
Question No. (2)
Calculate the standard heat of formation of solid
WO3 from solid W and O2 gas at 25 °C and 1
atmosphere pressure from the following data at
25° C and 1 atm:
+
O2
WO2ΔH°f at 298K= - 134
3WO2 +
O2
W3O8ΔH°fat 298K = -
W
kcal
131.5 kcal
W3O8 + 0.5 O2 3WO3ΔH°f at 298K = - 66.5
kcal
The Answer of Question No. (2)
O2
WO2Equation (A)
3WO2 + O2
W3O8Equation (B)
W
+
W3O8 + 0.5 O23WO3Equation (C)
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Sequence of solving equations
 3 (A) + (B) + (C)
3 W + 4.5 O2
3 WO3ΔH°f= - 600 kcal
Heat of formation per one mole
W + 1.5 O2
WO3
kcal
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ΔH°f= - 200
Question No. (3)
Predict the heat of fusion for LiCl at 883K, given
the followings:
Li(l) + 0.5 Cl2(g) LiCl(l)
ΔH°f at 883K
= - 92.347 kcal/mole
Li(l) + 0.5 Cl2(g) LiCl(s)
ΔH°fat 883K =
- 97.105 kcal/mole
The Answer of Question No. (3)
Li(l) + 0.5 Cl2(g) LiCl(l)
Equation (A)
Li(l) + 0.5 Cl2(g) LiCl(s)
Equation (B)
SEQUANCE OF SOLVING
 (A) – (B)
LiCl(s)LiCl(l)ΔH°f= 4758 kcal
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Question No. (4)
Calculate the standard heat of formation of
anhydrous aluminum chloride from the following
data:
2Al(s) +6HCl(aq.) Al2Cl6(aq.)+3H2(g)ΔH°f at
298K = - 240 kcal
H2(g)+ Cl2(g) 2HCl(g)
ΔH°f at 298K = - 44 kcal
HCl(g) + aq.
HCl(aq.)
ΔH°f at 298K= - 17.5 kcal
Al2Cl6 + aq.Al2Cl6(aq.)
ΔH°fat 298K= - 153.7 kcal
The Answer of Question No. (4)
2Al(s) +6HCl(aq.) Al2Cl6(aq.)+3H2(g)Equ.(A)
H2(g)+ Cl2(g) 2HCl(g)Equ.(B)
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HCl(g) + aq.HCl(aq.)Equ.(C)
Al2Cl6 + aq.Al2Cl6(aq.)
Equ.(D)
Sequence of solving
(A) + 3 (B) + 6 (C) – (D)
2Al + 3Cl2Al2Cl6
ΔH°f= - 240000 + 3(- 44000) + 6 (-17500) +153700
ΔH°f = - 323300 cal
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Question No. (5)
The enthalpy changes for the following reactions
are as follows:
2B+3H2+ 3O2 + aq.  2H3BO3(aq.) ΔH°fat
298K= - 512.8 kcal
B2O3+3H2O(l) + aq. 2H3BO3(aq.) ΔH°fat
298K = - 4.12 kcal
H2
+ 0.5 O2H2O(l)
ΔH° fat
298K= - 68.73 kcal
Calculate the standard heat of formation of
B2O3 per mole of B2O3, and per gram of B2O3.
The atomic weight of B and O are 11 and 16,
respectively.
The Answer of Question No. (5)
2 B2O3+3H2O(l) + aq.  2H3BO3(aq.) x (+1)
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B2O3+3H2O(l) + aq.  2H3BO3(aq.) x(-1)
H2 + 0.5 O2H2O(l) x(-3)
2B +1.5 O2B2O3
ΔH°f=-512.8+4.12+3×68.73
=-302.49 kcal/mole B2O3
ΔH°f=-302490[2×11+3×16]
=211744300 cal/gm B2O3
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Question No. (6)
Calcium carbide is considered to be a potential
fuel in the basic oxygen converters and would be
expected to burn into CaO and CO or CO2,
depending upon the conditions. If the heat
required to raise the steel scrap to 1600 °C is 333
kcal/Kg of scrap, calculate how many Kg of steel
scrap can be charged per 1000 Kg of CaC2 when
(a) all CaC2 is consumed to form CO
(b) all CaC2 is consumed to form CO2
(c) 60% of CaC2 is utilized to produce CO2 and
the rest produces CO.
Assuming that the reactions take place at 25 °C,
given that:
ΔH°f(CaC2) at 298K= - 14.1 kcal/mole
ΔH°f(CaO) at 298K = - 151.8 kcal/mole
ΔH°f(CO) at 298K = - 26.42 kcal/mole
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ΔH°f(CO2) at 298K = - 94.05 kcal/mole
The atomic weights of Ca and C are 40 and 12,
respectively.
The Answer of Question No. (6)
a. CaC2+O2CaO+2CO
ΔH°f at
298K=ΔH°f(CaO)+2ΔH°f(CO)-ΔH°f(CaC2)=-151.8+
2×-26.42+14.1
=-190.54 kcal/mole
ΔH°f at 298K=-190.54×64
=-12194.6kcal/1000kg
ΔH°f(steel) at 298K=333kcal/kg
N=12194.6÷333=36.4 kg of steel
b. CaC2+1.5O2CaO+2CO2
ΔH°f at 298K=ΔH°f(CaO)+2ΔH°f(CO2)
-ΔH°f(CaC2) =-151.8-2×94.05-14.1
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=-325.8kcal/mole
ΔH°f at 298K=-325.8×64
=-20851.2kcal/1000kg
N=20851.2÷333=62.6 kg of steel
c. CaC2+1.4O2CaO+0.8CO2+1.2CO
ΔH°f at 298K=ΔH°f(CaO)+1.2ΔH°f(CO)
+0.8ΔH°f(CO2) -ΔH°f
(CaC2)=-151.8+1.2×-26.42+0.8×-94.5+14.1
=-244.644kcal/mole
ΔH°f at 298K=-244.644×64
=-15657.16kcal/1000kg
N=15657.16÷333=47 kg of steel
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Question No. (7)
Calculate
the
heat
required
to
raise
the
temperature of Pb and O2 to 227 °C from 25 ᴼC;
you may use the following data:
ΔH°f(PbO) at 298K = - 52.4 kcal/mole
Cp (PbO)= 10.6 + 4.00 x 10-3 T cal/(deg-mole)
Cp (Pb) = 5.63+ 2.33 x 10-3 T
cal/(deg-gm atom)
Cp (O2) = 7.16 + 1.00 x 10-3 T – 0.4 x 105 T2
cal/(deg-mole)
Also calculate ΔH°f(PbO) at 227 °C.
The Answer of Question No. (7)
at T=273+227=500 K
Pb
+ 0.5O2
 PbO
𝟓𝟎𝟎
ΔH°f=ΔH°f at 298K+∫𝟐𝟗𝟖 𝐂𝐩(𝐑) dT
Cp(R)=Cp (PbO)-0.5Cp (O2) -Cp (Pb)
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=[10.6-0.5×7.16-5.63]+[4-0.5×1-2.33]×10-3 T
+[0.4×0.5×105 T-2]
=1.39+1.17×10-3 T+0.2×105 T-2
𝟓𝟎𝟎
ΔH°f (500K)=-52400+∫𝟐𝟗𝟖
( 1.39+1.17×10-3
T+0.2×105 T-2)dT
=-52400+1.39(500-298)+(1.17÷2)× 103
(5002-2982)-0.2×105(500-1- 298-1)=-52123.9 cal
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Question No. (8)
18 gm ice at 0 °C is heated to 100 °C until vapor is
formed against an external pressure of 1 atm.
(a) Calculate the thermal energy needed for the
process
(b) Calculate the work done by the system on the
surroundings
(c) The internal energy change throughout the
process.
Given that: ΔH°f(H2O) at 0 °C = 1.4 kcal/mole
ΔH°vap(H2O) at 100 °C = 9.7 kcal/mole
Cp(H2O)= 18 kcal/(deg-mole)
The Answer of Question No. (8)
H2O(l)H2O(v)
𝟑𝟕𝟑
ΔH°T=ΔH°1+∫𝟐𝟕𝟑 𝐂𝐩(𝐑) H2OdT+ΔH°2
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𝟑𝟕𝟑
=1400+∫𝟐𝟕𝟑 𝟏𝟖dT+9700
=1400+18(373-273)+9700
a. ΔH°T =12900cal/mole
ΔU=q+w
ΔH=ΔU+pv
V1 =R T1÷P1 =0.082×273=22.4 lit
V2 =R T2÷P2 =0.082×373=30.6 lit
W=P(V2-V1)=1(30.6-22.4)=8.2 lit .atm
W=8.2×101.325×(1÷4.18)=198.78 cal
b. Wsys=198.78 cal
ΔH=ΔU+Δpv
ΔU=ΔH-Δpv
ΔU=12900-198.8=12701 cal
c. ΔU=12701 cal
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Question No. (9)
For the reaction:
CO + H2O  CO2 +H2
ΔH°f at 298K= - 10 kcal
Calculate the standard change in enthalpy at 1000K
knowing that all chemical species of the reaction are in their
gases form.
Data: Cp (CO)g = 6.6+1.0x10-3 T cal/(deg-mole)
Cp (H2)g= 6.6+1.0x10-3 T cal/(deg-mole)
Cp (H2O)g= 7.3+2.0x10-3 T cal/(deg-mole)
Cp(CO2)g=7.3+3.0x10-3T
cal/(deg-mole)
The Answer of Question No.(9)
1000
ΔH°(1000k)= ΔH(298k) +∫298 Cp dT
Cp =Cp(H2)+ Cp(CO2)- Cp(CO)- Cp(H2O)
=(6.6+1.0x10-3 T) +(7.3+3.0x10-3T) - (6.6+1.0x10-3 T)-(7.3+2.0x10-3 T)
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Cp = 1x10-3 Tcal/(deg-mole)
ΔH° (298) =ΔH(CO2)-ΔH(H2O)-ΔH(CO)
=(-94.05 + 68.73 +26.42)x103 =1100 cal
1
∴ΔH°(1000k)=1100 + x 10-3 x (10002 -2982)
2
∴ΔH°(1000k)=1556 cal
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Question No.(10)
Calculate the standard heat of formation of solid PbO from liquid
Pb and O2 gas at 527 °C. The melting point of lead is 327 °C and
its latent heat of fusion is 1.15 kcal/mole. The molar heat capacities
at constant pressure are given below:
ΔH°f (PbO)s at 298K = -52,400 cal/mole
Cp (Pb)l= 7.75 – 0.74 x 10-3 T cal/(deg-gm atom)
Cp (Pb)s= 5.63+ 2.33 x 10-3 T cal/(deg-gm atom)
Cp (PbO)s= 10.6 + 4.00 x 10-3 T cal/(deg-mole)
Cp (O2)g= 7.16 + 1.00 x 10-3 T – 0.4 x 105 T-2cal/(deg-mole)
The Answer of Question No.(10)
1
Pb(l) + O2(g) → PbO
2
T=527 +273 = 800 K
T(m)=327 +273=600
ΔHf=1150 cal.mole
600
ΔH800= ΔH298 +∫298 Cp(s) dT
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800
+∫600
Cp(𝑙) dT +ΔHf
600
ΔH(800k) =-52400 +∫298 (Cp(pbo)-Cp (O2)-Cp (pbs))dT
800
+1150+∫600 (Cp(pbo)-Cp (O2) -Cp (pbl))dT
∴ΔH(800k)= -52400 +(-502.97) +1150 +(522754.67)
=471001.7 cal/mole
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Question No. (11)
Calculate the heat of reaction of chlorination of zirconium oxide at
25 °C and 777 °C; the process occurs according to the following
reaction:
ZrO2 + 2Cl2 + C  ZrCl4 + CO2
Given that: ΔH°f(ZrCl4) at 298K= - 234.7 kcal/mole
ΔH°f(CO2) at 298K = - 94.05 kcal/mole
ΔH°f(ZrO2) at 298K= - 259.5 kcal/mole
Cp (ZrCl4)= 31.92 – 2.91 x 105 T-2cal/deg-mole
Cp (CO2)= 10.55 + 2.16 x 10-3 T – 2.05 x 105 T-2cal/(deg-mole)
Cp (ZrO2)= 16.64 + 1.80 x 10-3 T – 3.36 x 105 T-2cal/(deg-mole)
Cp (Cl2)= 8.82 - 0.06 x 10-3 T – 0.68 x 105 T-2cal/(deg-mole)
Cp (C)= 4.10 + 1.02 x 10-3 T – 2.10 x 105 T-2cal/(deg-mole)
The Answer of Question No.(11)
T=777+273 = 1050 k
ΔH(298k)= ΔH(ZrCl4) + ΔH(CO2) - ΔH(ZrO2)
= - 234.7 +(- 94.05 ) –(- 259.5 )
ΔH(298k)= -69.25 kcal/mole
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1050
ΔH(1050k)=ΔH(298k)+∫298 CpdT
Cp= Cp (CO2)+Cp (ZrCl4) - Cp (C)-2Cp (Cl2) -Cp
(ZrO2)
=(10.55 + 2.16 x 10-3 T – 2.05 x105 T-2)+(31.92–2.91
x 105 T-2 ) –(4.10 + 1.02 x 10-3 T – 2.10 x 105 T-2 ) –
2(8.82 - 0.06 x 10-3 T – 0.68 x 105 T-2 ) –(16.64 +
1.80 x 10-3 T – 3.36 x 105 T-2)
Cp=4.09–5.4x10-4 T +1.86x105 T-2cal/(deg-mole)
1050
ΔH(1050k)= -69.25 x 103 +∫298 (4.09 – 5.4x10-4 T
+1.86x105 T-2 )
∴ΔH(1050k)= - 66 kcal/mole
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