One method for the manufacture of “synthesis gas” (a mixture of CO and H2) is the catalytic reforming of CH4 with steam at high temperature and atmospheric pressure: CH4 (g) + H2 O(g) → CO(g) + 3H2 (g) The only other reaction considered here is the water-gas-shift reaction:

Question

One method for the manufacture of “synthesis gas” (a mixture of CO and [latex]H_2[/latex]) is the catalytic reforming of [latex]CH_4[/latex] with steam at high temperature and atmospheric pressure:

[latex]CH_{4} (g) + H_{2} O(g) → CO(g) + 3H_{2} (g)[/latex]

The only other reaction considered here is the water-gas-shift reaction:

[latex]CO(g) + H_{2} O(g) → CO_{2} (g) + H_{2} (g)[/latex]

Reactants are supplied in the ratio 2 mol steam to 1 mol [latex]CH_{4}[/latex], and heat is added to the reactor to bring the products to a temperature of 1300 K. The [latex]CH_{4}[/latex] 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.

Accepted Answer

The standard heats of reaction at 25°C for the two reactions are calculated from the data of Table C.4:

[latex]CH_{4} (g) + H_{2} O(g) → CO(g) + 3H_{2} (g) ΔH^{o}_{298} = 205,813 J[/latex]

[latex]CO(g) + H_{2} O(g) → CO_{2} (g) + H_{2} (g) ΔH^{o}_{298} = −41,166 J[/latex]

These two reactions may be added to give a third reaction:

[latex]CH_{4} (g) + 2H_{2} O(g) → CO_{2} (g) + 4H_{2} (g) ΔH 2°9  8 = 164,647 J[/latex]

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:

[latex]CH_{4}(g) + H_{2}O(g) → CO(g) + 3H_{2}(g) ΔH^{o}_{298} = 205,813 J (A)[/latex]

[latex]CH_{4}(g) + 2H_{2}O(g) → CO_{2}(g) + 4H_{2}(g) H^{o}_{298} = 164,647 J (B)[/latex]

First one must determine the fraction of [latex]CH_{4} [/latex] converted by each of these reactions. As a basis for calculations, let 1 mol [latex]CH_{4} [/latex] and 2 mol steam be fed to the reactor. If x mol [latex]CH_{4} [/latex] reacts by Eq. (A), then 1 − x mol reacts by Eq. (B). On this basis the products of the reaction are:

CO: x
[latex]H_{2}: 3x + 4(1 − x) = 4 − x[/latex]
[latex]CO_{2}: 1 − x[/latex]
[latex]H_{2}O: 2  − x  −  2(1  −  x) = x[/latex]

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 [latex]CH_{4} [/latex] reacts by Eq. (A) and 0.130 mol reacts by Eq. (B). Furthermore, the amounts of the species in the product stream are:

[latex]Moles CO = x = 0.87 [/latex]
[latex]Moles H_{2}  = 4  −  x = 3.13 [/latex]
[latex]Moles CO_{2}  = 1  −  x = 0.13 [/latex]
[latex]Moles H_{2} O = x = 0.87 [/latex]

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,

[latex]ΔH = Δ H^{o}_{R}  + Δ H^{o}_{298} + Δ H^{o}_{P}[/latex]

For the calculation of [latex]Δ H^{o}_{298}[/latex], reactions (A) and (B) must both be taken into account. Because 0.87 mol [latex]CH_{4} [/latex] reacts by (A) and 0.13 mol reacts by (B),

[latex]Δ H^{o}_{298} = ( 0.87 ) ( 205,813 )  + ( 0.13 ) ( 164,647 )  = 200,460 J[/latex]

The enthalpy change of the reactants when they are cooled from 600 K to 298.15 K is:

[latex]\Delta H^{o}_{R}=\left(\sum\limits_{i}{n_{i} \left\langle C^{o}_{P_{i} } ight angle _{H} } ight) (298.15 − 600)[/latex]

where subscript i denotes reactants. The values of [latex] \left\langle C^{o}_{P_{i} } ight angle _{H} / R[/latex] are:

[latex]CH_{4} : MCPH(298.15, 600; 1.702, 9.081  imes   10^{−3} , − 2.164  imes   10^{−6}, 0.0) = 5.3272[/latex]

[latex]H_{2} O : MCPH(298.15, 600; 3.470, 1.450  imes   10^{−3} , 0.0, 0.121  imes 10^{5} ) = 4.1888[/latex]

and

[latex]\Delta H^{o}_{R} = (8.314)[ (1)(5.3272)+(2)(4.1888)] (298.15 − 600) = − 34,390 J[/latex]

The enthalpy change of the products as they are heated from 298.15 to 1300 K is calculated similarly:

[latex]\Delta H^{o}_{P} = \left(\sum\limits_{i}{n_{i} \left\langle C^{o}_{P_{i} } ight angle _{H} } ight)(1300 − 298.15)[/latex]

where subscript i here denotes products. The [latex]\left\langle C^{o}_{P_{i} } ight angle _{H} / R[/latex] values are:

[latex]CO: MCPH ( 298.15, 1300; 3.376, 0.557  imes   10^{−3} , 0.0, − 0.031  imes   10^{5} )  = 3.8131[/latex]

[latex]H_{2} : MCPH ( 298.15, 1300; 3.249, 0.422  imes   10^{−3} , 0.0, − 0.083   imes   10^{5})  = 3.6076[/latex]

[latex]CO_{2} : MCPH ( 298.15, 1300; 5.457, 1.045  imes   10^{−3} , 0.0, − 1.157 imes   10^{5} )  = 5.9935[/latex]

[latex]H_{2} O: MCPH ( 298.15, 1300; 3.470, 1.450  imes   10^{−3} , 0.0, 0.121 imes   10^{5} )  = 4.6599 [/latex]

Whence,

[latex]Δ H^{o}_{P}  = ( 8.314 ) [ ( 0.87 ) ( 3.8131 )  + ( 3.13 ) ( 3.6076 )  [/latex]
+ ( 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  [latex]W_{s} Δzand Δu^{2}/2 [/latex] are presumed negligible. Thus,

Q = ΔH = 328, 010 J

This result is on the basis of 1 mol [latex]CH_{4} [/latex] fed to the reactor.

Question 4.8: One method for the manufacture of “synthesis gas” (a mixture...

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