Question 9.5.2: Energy Balance on a Methane Oxidation Reactor Methane is oxi......

Basis: 100 mol CH_{4} Fed
Since the component amounts of all streams are known, we may proceed directly to the energy balance. We choose as references the elemental species that form the reactants and products at 25°C and 1 atm (the state for which heats of formation are known) and the nonreactive species—N_{2}(g)—also at 25°C and 1 atm (the reference state for Table B.8). The inlet–outlet enthalpy table is shown below.

Calculate Unknown Enthalpies
All of the quantities shown can be retrieved easily using the appropriate functions in APEx. In the following calculations, values of Δ\hat{H}^{°}_{f} come from Table B.1, formulas for C_{p}(T) come from Table B.2, and values of \hat{H}(T) for O_{2} and N_{2} are specific enthalpies relative to the gaseous species at 25°C taken from Table B.8.
Effects of any pressure changes on enthalpies are neglected.

CH_{4}(25°C):   \hat{H}_{1}= (\Delta \hat{H}^{°}_{f})_{CH_{4}} = – 74.85  kJ/mol \\ O_{2} (100°C):   \hat{H}_{2}= \hat{H}_{O_{2}}(100°C) = 2.235  kJ/mol \\ N_{2} (100°C):   \hat{H}_{3}= \hat{H}_{N_{2}} (100°C)= 2.187  kJ/mol \\ CH_{4}(150°C):   \hat{H}_{4}= (\Delta \hat{H}^{°}_{f})_{CH_{4}}+ \int_{25°C}^{150°C}{(C_{p})_{CH_{4}}  dT} \\= (- 74.85+ 4.90)  kJ/mol = – 69.95  kJ/mol \\ O_{2} (150°C):   \hat{H}_{5}= \hat{H}_{O_{2}}(150°C) = 3.758  kJ/mol \\ N_{2} (150°C):   \hat{H}_{6}= \hat{H}_{N_{2}} (150°C)= 3.655  kJ/mol \\ HCHO(150°C):   \hat{H}_{7}= (\Delta \hat{H}^{°}_{f})_{HCHO}+ \int_{25°C}^{150°C}{(C_{p})_{HCHO}  dT}\\ = (- 115.90+ 4.75)  kJ/mol = –  111.15  kJ/mol \\ CO_{2}(150°C):   \hat{H}_{8}= (\Delta \hat{H}^{°}_{f})_{CO_{2}} + \hat{H}_{CO_{2}} (150°C)\\=( – 393.5+ 4.75)  kJ/mol = – 388.6  kJ/mol\\ H_{2}O(v,150°C):   \hat{H}_{9}= (\Delta \hat{H}^{°}_{f})_{H_{2}O(v)} + \hat{H}_{H_{2}O(v)} (150°C)\\=( – 241.83+ 4.27)  kJ/mol = – 237.56  kJ/mol

As each of these values is calculated, it should be substituted in the inlet–outlet enthalpy table. The table finally appears as follows:

Evaluate ΔH
From Equation 9.5-2,

\Delta \dot{H}= \sum{\dot{n}_{out}\hat{H}_{out}} – \sum{ \dot{n}_{in} \hat{H}_{in}}               (9.5-2)

\Delta H= \sum{n_{out}\hat{H}_{out}} – \sum{ n_{in} \hat{H}_{in}} = –  15,300  kJ

If molecular species had been chosen as references for enthalpy calculations, the extents of each reaction (ξ_{1} and ξ_{2}) would have had to be calculated and Equation 9.5-1b used to determine ΔH. When more than one reaction occurs in a process, you are advised to choose elemental species as references and avoid these complications.
Energy Balance
Remember that we are dealing with a continuous process and hence an open system. [The reason we use n(mol) and not \dot{n}(mol/s) is that we took 100 mol CH_{4} as a basis of calculation.] With ΔE_{k}, ΔE_{p}, and W_{s} neglected, the open system energy balance yields

\boxed{Q= \Delta H=-  15,300  kJ}