Question 14.9: The gas-phase oxidation of SO2 to SO3 is carried out at a pr...

The reaction is:                      \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3
for which,          \Delta H_{298}^{\circ}=-98,890 \quad \Delta G_{298}^{\circ}=-70,866 \mathrm{~J} \mathrm{~mol}^{-1}

On the basis of one mole of SO2 entering the reactor,

M oles O_2 entering = (0.5)(1.2) = 0.6
Moles N_2 entering = (0.6)(79/21) = 2.257

Application of Eq. (14.4) yields the amounts of the species in the product stream:

Moles  SO_2  = 1  −  ​ ε_e
Moles  O_2  = 0.6  −  0.5  ​ε_e
Moles  SO_3  = ​ε_e
Moles  N_2  = 2.257

Total Moles = 3.857 − 0.5 ε_e

Two equations must be written if we are to solve for both ε_e and the temperature. They are an energy balance and an equilibrium equation. For the energy balance, we proceed as in Ex. 4.7:

\Delta H_{298}^{\circ} \varepsilon_e+\Delta H_P^{\circ}=\Delta H=0     (A)

where all enthalpies are on the basis of 1 mol SO2 entering the reactor. The enthalpy change of the products as they are heated from 298.15 K to T is:

\Delta H_P^{\circ}=\left\langle C_P^{\circ}\right\rangle_H(T-298.15)     (B)

where \left\langle C_P^{\circ}\right\rangle is defined as the mean total heat capacity of the product stream:

\left\langle C_P^{\circ}\right\rangle_H \equiv \sum_i n_i\left\langle C_{P_i}^{\circ}\right\rangle_H

Data from Table C.1 provide \left\langle C_{P_i}^{\circ}\right\rangle values:

SO_2 : 𝖬𝖢𝖯𝖧(298.15, T; 5.699, 0.801E-3, 0.0, −1.015E+5)

O_2 : 𝖬𝖢𝖯𝖧 (298.15, T; 3.639, 0.506E-3, 0.0, −0.227E+5)

SO_3 : 𝖬𝖢𝖯𝖧(298.15, T; 8.060, 1.056E-3, 0.0, −2.028E+5)

N_2 : 𝖬𝖢𝖯𝖧(298.15, T; 3.280, 0.593E-3, 0.0, 0.040E+5)

Equations (A) and (B) combine to yield:

\Delta H_{298}^{\circ} \varepsilon_e+\left\langle C_P^{\circ}\right\rangle_H(T-298.15)=0

Solution for T gives:

T=\frac{-\Delta H_{298}^{\circ} \varepsilon_e}{\left\langle C_P^{\circ}\right\rangle_H}+298.15     (C)

At the conditions of temperature and pressure of the equilibrium state, the assumption of ideal gases is fully justified, and the equilibrium relationship is therefore given by Eq. (14.28), which here becomes:

\prod_i\left(y_i\right)^{\nu_i}=\left(\frac{P}{P^{\circ}}\right)^{-\nu} K     (14.28)

K=\left(\frac{\varepsilon_e}{1-\varepsilon_e}\right)\left(\frac{3.857-0.5 \varepsilon_e}{0.6-0.5 \varepsilon_e}\right)^{0.5}

Because −ln K = ΔG°∕RT, Eq. (14.18) can be written:

\frac{\Delta G^{\circ}}{R T}=\frac{\Delta G_0^{\circ}-\Delta H_0^{\circ}}{R T_0}+\frac{\Delta H_0^{\circ}}{R T}+\frac{1}{T} \int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} d T-\int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} \frac{d T}{T}     (14.18)

-\ln K=\frac{\Delta G_0^{\circ}-\Delta H_0^{\circ}}{R T_0}+\frac{\Delta H_0^{\circ}}{R T}+\frac{1}{T} \int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} d T-\int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} \frac{d T}{T}

Substitution of numerical values yields:

\ln K=-11.3054+\frac{11,894.4}{T}-\frac{1}{T}(\text { IDCPH })+\text { IDCPS }       (E)

IDCPH = IDCPH ( 298.15, T; 0.5415, 0.002E-3, 0.0, −0.8995E+5 )
IDCPS = IDCPS ( 298.15, T; 0.5415, 0.002E-3, 0.0, −0.8995E+5 )

These expressions for the computed values of the integrals show parameters ΔA, ΔB, ΔC, and ΔD as evaluated from data of Table C.1.

An intuitive iteration scheme for solution of these equations for εe and T that converges fairly rapidly is as follows:

1. Assume a starting value for T.

2. Evaluate IDCPH and IDCPS at this value of T.

3. Solve Eq. (E) for K and Eq. (D) for ε_e .

4. Evaluate \left\langle C_P^{\circ}\right\rangle_H and solve Eq. (C) for T.

5. Find a new value of T as the arithmetic mean of the value just calculated and the initial value; return to step 2.

This scheme converges on the values ε_e = 0.77 and T = 855.7 K. For the product,