Question 4.8: One method for the manufacture of “synthesis gas” (a mixture...
One method for the manufacture of “synthesis gas” (a mixture of CO and H_2) is the catalytic reforming of CH_4 with steam at high temperature and atmospheric pressure:
CH_{4} (g) + H_{2} O(g) → CO(g) + 3H_{2} (g)
The only other reaction considered here is the water-gas-shift reaction:
CO(g) + H_{2} O(g) → CO_{2} (g) + H_{2} (g)
Reactants are supplied in the ratio 2 mol steam to 1 mol CH_{4}, and heat is added to the reactor to bring the products to 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.
The standard heats of reaction at 25°C for the two reactions are calculated from the data of Table C.4:
CH_{4} (g) + H_{2} O(g) → CO(g) + 3H_{2} (g) ΔH^{o}_{298} = 205,813 J
CO(g) + H_{2} O(g) → CO_{2} (g) + H_{2} (g) ΔH^{o}_{298} = −41,166 J
These two reactions may be added to give a third reaction:
CH_{4} (g) + 2H_{2} O(g) → CO_{2} (g) + 4H_{2} (g) ΔH 2°9 8 = 164,647 J
Any pair of the three reactions constitutes an independent set. The third reaction is not independent; it is obtained by combination of the other two. The reactions most convenient to work with here are the first and third:
CH_{4}(g) + H_{2}O(g) → CO(g) + 3H_{2}(g) ΔH^{o}_{298} = 205,813 J (A)
CH_{4}(g) + 2H_{2}O(g) → CO_{2}(g) + 4H_{2}(g) H^{o}_{298} = 164,647 J (B)
First one must determine the fraction of CH_{4} converted by each of these reactions. As a basis for calculations, let 1 mol CH_{4} and 2 mol steam be fed to the reactor. If x mol CH_{4} reacts by Eq. (A), then 1 − x mol reacts by Eq. (B). On this basis the products of the reaction are:
CO: x
H_{2}: 3x + 4(1 − x) = 4 − x
CO_{2}: 1 − x
H_{2}O: 2 − x − 2(1 − x) = x
Total: 5 mol products
The mole fraction of CO in the product stream is x/5 = 0.174; whence x = 0.870. Thus, on the basis chosen, 0.870 mol CH_{4} reacts by Eq. (A) and 0.130 mol reacts by Eq. (B). Furthermore, the amounts of the species in the product stream are:
Moles CO = x = 0.87
Moles H_{2} = 4 − x = 3.13
Moles CO_{2} = 1 − x = 0.13
Moles H_{2} O = x = 0.87
We now devise a path, for purposes of calculation, to proceed from reactants at 600 K to products at 1300 K. Because data are available for the standard heats of reaction at 25°C, the most convenient path is the one that includes the reactions at 25°C (298.15 K). This is shown schematically in the accompanying diagram. The dashed line represents the actual path for which the enthalpy change is ΔH. Because this enthalpy change is independent of path,
ΔH = Δ H^{o}_{R} + Δ H^{o}_{298} + Δ H^{o}_{P}
For the calculation of Δ H^{o}_{298}, reactions (A) and (B) must both be taken into account. Because 0.87 mol CH_{4} reacts by (A) and 0.13 mol reacts by (B),
Δ H^{o}_{298} = ( 0.87 ) ( 205,813 ) + ( 0.13 ) ( 164,647 ) = 200,460 J
The enthalpy change of the reactants when they are cooled from 600 K to 298.15 K is:
\Delta H^{o}_{R}=\left(\sum\limits_{i}{n_{i} \left\langle C^{o}_{P_{i} } \right\rangle _{H} } \right) (298.15 − 600)
where subscript i denotes reactants. The values of \left\langle C^{o}_{P_{i} } \right\rangle _{H} / R are:
CH_{4} : MCPH(298.15, 600; 1.702, 9.081 \times 10^{−3} , − 2.164 \times 10^{−6}, 0.0) = 5.3272
H_{2} O : MCPH(298.15, 600; 3.470, 1.450 \times 10^{−3} , 0.0, 0.121 \times 10^{5} ) = 4.1888
and
\Delta H^{o}_{R} = (8.314)[ (1)(5.3272)+(2)(4.1888)] (298.15 − 600) = − 34,390 J
The enthalpy change of the products as they are heated from 298.15 to 1300 K is calculated similarly:
\Delta H^{o}_{P} = \left(\sum\limits_{i}{n_{i} \left\langle C^{o}_{P_{i} } \right\rangle _{H} } \right)(1300 − 298.15)
where subscript i here denotes products. The \left\langle C^{o}_{P_{i} } \right\rangle _{H} / R values are:
CO: MCPH ( 298.15, 1300; 3.376, 0.557 \times 10^{−3} , 0.0, − 0.031 \times 10^{5} ) = 3.8131
H_{2} : MCPH ( 298.15, 1300; 3.249, 0.422 \times 10^{−3} , 0.0, − 0.083 \times 10^{5}) = 3.6076
CO_{2} : MCPH ( 298.15, 1300; 5.457, 1.045 \times 10^{−3} , 0.0, − 1.157 \times 10^{5} ) = 5.9935
H_{2} O: MCPH ( 298.15, 1300; 3.470, 1.450 \times 10^{−3} , 0.0, 0.121 \times 10^{5} ) = 4.6599
Whence,
Δ H^{o}_{P} = ( 8.314 ) [ ( 0.87 ) ( 3.8131 ) + ( 3.13 ) ( 3.6076 )
+ ( 0.13 ) ( 5.9935 ) + ( 0.87 ) ( 4.6599 ) ] × ( 1300 − 298.15 )
= 161, 940 J
Therefore,
ΔH = −34,390 + 200,460 + 161,940 = 328,010 J
The process is one of steady flow for which W_{s} Δzand Δu^{2}/2 are presumed negligible. Thus,
Q = ΔH = 328, 010 J
This result is on the basis of 1 mol CH_{4} fed to the reactor.
