16 May 2016
Heat Effects of Industrial Reactions
Heats of Reaction ¤
Standard heats of reaction are specified for a reference temperature T0 = 298.15 K. – Standard – Standard values are determined from summing up heats of formation, also specified at T0 = 298.15 K
Hr,298.15 i H f ,298.15 i ¤
If reaction proceeds at a temperature te mperature not equal to T 0, we use a computational path to determine the new value of the heat of reaction at T.
Computational Path ¤
Always keep in mind: standard heats of reaction are always specified at a reference temperature (T 0) – Bring down the reactants from T to T0 (sensible heating) – Allow the reaction to proceed at T 0 – Bring up the products from T0 to T (sensible heating) – The standard heat of reaction at T is the sum of all the enthalpy changes from the computational path
Computational Path
aA bB cC dD Products at T
Reactants at T
H i
H i
T 0
C dT p
T
Products at T0
Hr,T0 i H f,T0
C dT p
T 0
Reactants at T0
T
Example Calculate the standard heat of the methanol synthesis reaction at 800°C. CO( g ) 2H 2( g ) CH 3OH( g ) Heat Capacity: CP ig / R A BT CT 2 DT 2 (T in K) CO: A = 3.376, B = 0.557x10 -3, C = 0, D = -0.031x10 5 H2: A = 3.249, B = 0.422x10-3, C = 0, D = 0.083x105 CH3OH: A = 2.211, B = 12.216x10-3, C = -3.450x10-6, D = 0 Heats of formation: CH3OH: -200,660 J/mol CO: -110,525 J/mol
Solution ¤
Bring down all reactants from T = 800°C to T = 25°C: H i
298.15
C p dT
CP ig 8.314 molJ K A BT CT 2 DT 2
1073.15
: C dT 22,965.936
J CO : C p dT 24,151.083 mol
H2
p
J mol
H 1 24,151.083 2 22, 965.936 H 1 70082.955 molJ K
Solution ¤
Allow the reaction to proceed at T = 25°C:
Hr,298.15 i H f ,298.15 J H r ,298.15 200,660 (110,525) 90,135 mol
Solution ¤
Bring up the products from T = 25°C to T = 800°C: H i
1073.15
C p dT
CP ig 8.314 molJ K A BT CT 2 DT 2
298.15
J CH 3OH : C p dT 56,652.000 mol
J H 2 56,652.00 mol
Solution ¤
Summing up all enthalpy changes, the standard heat of reaction at T = 800°C is:
Hr,1073.15 H1 Hr ,298.15 H 2
H r ,1073.15 70,082.955 90135 56,652.00 J H r ,1073.15 103,565.955 mol
Heat Effects of Industrial Reactions ¤
Using the same principle of computational paths, the total heat required for an industrial process may be computed: – Perform material balances first to determine all amounts – Bring down the inlet stream from T to T 0 (sensible heating) – Allow the reaction to proceed at T 0 – Bring up the outlet stream from T0 to T (sensible heating) – Sum up all enthalpy changes in the computational path
Computational Path
Products at T
Reactants at T
T
H ni C p dT
T 0
H ni C p dT
T 0
T
Reactants at T0
Products at T0
Hr ,T0 ni H r,T0
where ni mol extent
Example One method for the manufacture of synthesis gas is the catalytic reforming of CH4 with steam at high temperature and atmospheric pressure: CH 4 g H 2O g CO g 3H 2 g
The only other reaction to be considered is the water-gas-shift reaction: CO g H 2O g CO2 g H 2 g
If the reactants are supplied in the ratio 2 mol steam to 1 mol methane, and if heat is supplied to the reactor so that the products reach a temperature of 1300 K, the CH 4 is completely converted and the product stream contains 17.4 mol% CO. Assuming the reactants to be preheated to 600 K, calculate the heat requirement for the reactor.
Example ¤
Heat Capacity: CP ig / R A BT CT 2 DT 2
K) (T in
Component A
B
C
D
CH4 H2O CO H2
1.702 3.470 3.376 3.249
9.081x10-3 1.450x10-3 0.557x10-3 0.422x10-3
-2.164x10-6 0 0 0
0 0.121x105 -0.031x105 0.083x105
CO2
5.457
1.045x10-3
0
-1.157x105
Example ¤
Heats of formation (J/mol) at T = 298.15 K:
CH4 = -74,520 ¤ H O = -241,818 2 ¤ CO = -110,525 ¤ H2 = 0 ¤ CO2 = -393,509 ¤
Solution ¤
The system can be represented by the following block diagram: 1 mol CH4 2 mol H2O
¤
REACTOR
C balance: 1 11 w 11 x 11
¤
w mol CO x mol CO2 y mol H2 z mol H2O
w x 1
H balance: 1 14 2 12 y 12 z 12
2y 2z 8
Solution ¤
The system can be represented by the following block diagram: 1 mol CH4 2 mol H2O
¤
REACTOR
O balance: 2 11 w 11 x 12 z 11
¤
w mol CO x mol CO2 y mol H2 z mol H2O
Additional:
w
0.174
w x y z
w 2x z 2
Solution ¤
Solving simultaneously: w x 1 y z 4 w 2x z 2 w
0.174
w x y z w 0.87 mol CO
y 3.13 mol H2
x 0.13 mol CO2
z 0.87 mol H2O
Solution ¤
Bringing down the reactants from T = 600 K to T = 298.15 K: H ni
298.15
CP ig 8.314 molJ K A BT CT 2 DT 2
C p dT
600
H O : C dT 10,512.128
J CH 4 : C p dT 13,369.062 mol
2
p
J mol
H 1 1 13,369.062 2 10,512.128 H1 34,393.318J
Solution ¤
Allowing the reactions to proceed at T = 298.15 K: – All heats of reaction refer to a per mol extent of reaction (mol reactant consumed, or mol product formed) CH 4 g H 2O g CO g 3H 2 g
J H r 1,298.15 1 110,525 1 74,520 1 241,818 205,813 mol
CO g H 2O g CO2 g H 2 g
J H r 2,298.15 1 393,509 1 110,525 1 241,818 41,166 mol
Solution ¤
Allowing the reactions to proceed at T = 298.15 K: – All heats of reaction refer to a per mol extent of reaction (mol reactant consumed, or mol product formed)
Hr ,T0 ni H r,T0
J H r 1,298.15 205,813 mol
where ni mol extent
J H r 2,298.15 41,166 mol
Hr ,298.15 1 205,813 0.13 41,166 200, 461.42J
Solution ¤
Bringing up the product stream from T = 298.15 K to T = 1300 K: H ni
1300
CP ig 8.314 molJ K A BT CT 2 DT 2
C p dT
298.15
: C dT 49,922.360
J CO : C p dT 31,760.652 mol
CO2
p
J mol
J H 2 : Cp dT 30,049.272 mol
H 2O : Cp dT 38,813.890
J mol
H 2 0.87 31,760.652 0.13 49,922.360 3.13 30,049.272 0.87 38,813.890 H 2 161,943.980J
Solution ¤
Calculating for the heat requirement of the reactor: – Assume: negligible KE and PE changes in reactor, no shaft work
Q nH Q H1 Hr ,298.15 H2 Q 34, 393.318 200, 461.42 161,943.980 Q 328,012.082J
