1. A fuel gas consisting of 94 mole% methane and 6 mole% nitrogen is burned with
ID: 516284 • Letter: 1
Question
1. A fuel gas consisting of 94 mole% methane and 6 mole% nitrogen is burned with 30%
excess air in a continuous water heater. Both fuel gas and air enter dry at 298.15
K(25
oC). Water is heated at a rate of 34.0 kg/s from 298.15 K(25°C) to 368.15
K(95°C). The flue gases leave the heater at 483.15K (210°C). Of the entering
methane, 70% burns to carbon dioxide and 30% burns to carbon monoxide. What
volumetric flow rate of fuel gas is required if there are no heat losses to the
surroundings?
2.
a) Calculate the heats of reaction H at 298 K and at 750 K for the oxidation of
sulfur dioxide at one atmosphere
SO2(g) + O2(g) = SO3(g)
b) Calculate the heat of reaction for the hydration of CaO at one atmosphere and
298 K.
CaO(s) + H2O(l) = Ca(OH)2(s)
c) Calculate the heat of reaction at one atmosphere and 750 K for the similar
reaction
CaO(s) + H2O(g) = Ca(OH)2(s)
d) Why is the percentage change in the heat of reaction between 298 and 750 K
for the oxidation of sulfur dioxide smaller than that for the hydration reactions
for calcium oxide?
Explanation / Answer
2.a) SO2(g) + O2(g) = SO3(g)
Writing what we know :
where Cp/R = A + BT + CT2 + DT-2 T in kelvin
Here X = i Xi
For example: A = ASO3 – ASO2 – AO2,
and similarly for B, C, and D.
Hrxn,298 = Hf SO3 – [Hf SO2 + Hf O2] = (-395720) – (-296830) = -98,890 J/mol (at 298K)
Hrxn,750 = Hrxn,298 + i Cpi dT
= Hrxn,298 + i R(Ai + BiT + CiT2 + DiT-2) dT
= Hrxn,298 + i R[Ai (T2–T1) + 1/2 Bi (T22 –T12 ) + 1/3 Ci (T23 –T13 ) – Di (T2-1–T1-1)]
= Hrxn,298 + R[A (T2–T1) + 1/2 B (T22 –T12 ) + 1/3 C (T23 –T13 ) – D (T2-1–T1-1)]
where,
T2 = 750 K
T1 = 298 K
R = 8.314 J/mol K
Hrxn,750 = -98890 J/mol + 8.314 J/mol K [-1.278 (750–298) + 1/2 (-0.251 x 10-3) (7502 –2982 ) +
1/3 (0) (7503 – 2983 ) – (-0.786 x 105 ) (750-1–298-1)] K
Hrxn,750 = -105,508 J/mol (at 750 K)
similarly, for 2 .b) CaO(s) + H2O(l) = Ca(OH)2(s) (at 298 K),
Hrxn,298 = Hf Ca(OH)2 – [Hf CaO + HfH2O(l)] = -65,170 J/mol (at 298K)
2 c) Now looking at the reaction at 750 K (where the H2O is now gas instead of liquid):
CaO(s) + H2O(g) = Ca(OH)2(s)
Hrxn,298 = Hf Ca(OH)2 – [Hf CaO + HfH2O(g)] = -109,182 J/mol (at 298K)
Following the same method of calculation as in part a) :
Hrxn,750 = Hrxn,298 + i Cpi dT
= Hrxn,298 + R[A (T2–T1) + 1/2 B (T22 –T12 ) + 1/3 C (T23 –T13 ) – D (T2-1–T1-1)]
Hrxn,750 = -109182 J/mol + 8.314 J/mol K [0.023 (750–298) + 1/2 (3.542 x 10-3) (7502 –2982 ) +
1/3 (0) (7503 –2983 ) – (0.926x 105 ) (750-1–298-1)] K
Hrxn,750 = -100,563 J/mol (at 750 K)
2 .d) The % change going between 298 K and 750 K for the first reaction is
(-105,508 J/mol – 98,990 J/mol) / (98,990 J/mol) = 8.4%
The % change going between 298 K and 750 K for the second reaction is
(-100,563 J/mol – 65,170 J/mol) / (65,170 J/mol) = 54.3%
The change in the second reaction(CaO one) is larger than the first one(SO2 one) because when we go to 750K (still at 1 atm), the water is in the vapor phase. Much energy is required to vaporize the water.
SO2(g) O2(g) SO3(g) Stoich. coef. -1 -1 1 Hof298 (J/mol) -296830 0 -395720 -98890 A 5.699 3.639 8.06 -1.278 B 0.801E-03 0.506E-03 1.056E-03 -0.251E-03 C 0 0 0 0 D -0.251E-03 -0.227E+05 -2.028E+05 -0.786E+05Related Questions
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