A gas-phase reaction between methane (A) and sulfur (B) is conducted at 600°C and 101 kPa in a PFR, to produce carbon disulfide and hydrogen sulfide. The reaction is first-order with respect to each reactant, with kB = 12 m³ mol^-1 h^-1 (based upon the disappearance of sulfur). The inlet molar flow

Question

A gas-phase reaction between methane (A) and sulfur (B) is conducted at 600°C and 101 kPa in a PFR, to produce carbon disulfide and hydrogen sulfide. The reaction is first-order with respect to each reactant, with [latex] k_B = 12 m^3 mol^{-1} h^{-1} [/latex] (based upon the disappearance of sulfur). The inlet molar flow rates of methane and sulfur are 23.8 and 47.6 mol [latex]h^{-1} [/latex] , respectively. Determine the volume (V) required to achieve 18% conversion of methane, and the resulting residence or space time.

Accepted Answer

The reaction is [latex] CH_4 + 2S_2 ightarrow CS_2 + 2H_2S [/latex] . Although this is a gas-phase reaction, since there is no change in T, P, or total molar flow rate, density is constant. Furthermore, since the reactants are introduced in the stoichiometric ratio, neither is limiting, and we may work in terms of B (sulphur), since k, is given, with [latex] f_B ( = f_A) = 0.18[/latex] . It also follows that [latex] c_A = c_B /2 [/latex] at all points. The rate law may then be written as

[latex] ( -r_B ) = k_Bc_Ac_B = k_Bc_B^2 / 2 [/latex] (A)

From the material-balance equation 15.2- 17 and (A),

[latex] V/q_o = \int{dc_A / ( -r_A)} = au [/latex] (15.2-17)

[latex] v = -q_o \int_{c_{Bo}}^{c_B}{dc_B / ( -r_B)} = - q_o \int_{c_{Bo}}^{c_{B}}{\frac {dc_B}{k_Bc_B^2 /2}} = \frac {2q_o}{k_B}\left(\frac {1}{c_B} - \frac{1}{c_{Bo}} ight) [/latex] (B)

Since [latex] F_{Bo} = c_{Bo}q_o [/latex] , and, for constant-density, [latex] c_B = c_{Bo}(1 - f_B ) [/latex] , equation (B) may be written as

[latex] V = \frac {2q_o^2}{k_BF_{Bo}} \left( \frac {f_B}{1 - f_B} ight) [/latex] (c)

To obtain [latex] q_o [/latex] in equation (C), we assume ideal-gas behavior; thus,

[latex] q_o = (F_{Ao} + F_{Bo} )RT /P = 71.4 (8.314 ) 873 / 101 , 000 = 5.13 m^3 h^{-1} [/latex]

From equation (C),

[latex] V = \frac {2 (5.13)^2 0.18}{12 (47.6) 0.82 } = 0.020 m^3 [/latex]

From equation 15.2-14, we solve for τ:

[latex] t = \int{dV /q } = V /q_o = au [/latex] (constant density) (15.2-14)

[latex] t = au = V/q_o = 0.020 /5.13 = 0.00390 h = 14.0 s [/latex]

Question 15.3: A gas-phase reaction between methane (A) and sulfur (B) is c...

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