Question

The thermal decomosition of dimethyl ether (CH3)20(g) - CH4() + H2(g) + CO(9) is to be carried out in an isothermal, 4.00-liter laboratory reactor at 600.0 °C. The reactor is charged with pure dimethyl ether at a pressure of 350.0 torr. After about two hours, the reactor pressure is 725.0 torr.

Answer 1

4L (constant) 600 C (600 + 273.15) K = 873.15 K Volume of reactor (V) Initial Temperature (T1) Since reactor is isothermal, Final Temperature (T2) T 873.15 K (350 Initial Pressure (P1) 350 torr atm 760 Final Pressure (P2) 725 torr atm 760 Assume the gas behaves as ideal gas Using Ideal gas law for initial state, Initial no of moles (ni) P1V RT: 350 *4 7604 0.08206873.15 moles 0.02571 moles Since reactor is charged with pure dimethyl ether, Initial moles of dimethyl ether 0.02571 moles Since Using Ideal gas law for final state, Final no of moles (na) P2V RT2 725 moles 0.08206+ 873.15 0.053256 moles Reaction is: (CH3)20(g) → CH4 (g) + H2(g) + CO(g) Let moles of (CH3)20 reacted be 'x' moles By stoichiometry, *x' moles of each product will be formed.

CH4 Thus, final no of moles of each component is: (CH3)20 0.02571 - x 0+ x H2 0+ x CO 0 + x Total 0.02571 + 2x -0.02571 + 2x 0.053256 On solving, X 0.013773 Thus, final no of moles of each component is: (CH3)20 0.02571 -0.013773 CH4 0+ 0.013773 H2 0+ 0.013773 CO 0+ 0.013773 Total 0.02571 + 2 * 0.013773 0.011937 0.013773 0.013773 0.013773 0.053256 moles moles moles moles moles We need to find Specific Enthalpy of each component in each of the stream at their respective temperature. Here, Reference Temperature is 25C Tref= 25 C Heat of Formation for the components are: Component (CH3)20 (g) CH4 (g) H (9) CO (g) Std. Heat of Formation (KJ/mol) -180.16 - 74.85 0 -110.52

Specific Heat Capacity of the components are: Cp.(CH3)20) = 26.86 +0.1659T - 4.179 * 10-5 * T where Cp is in J/mol.K, T in Kelvin C,CH, (B) = 34.31 + 10-3 + 5.469 + 10-5T + 0.3661 + 10-8 + T2 – 11 + 10-12 + T3 where Cp is in kJ/mol.K, T is in C Note: Units of Cp and T are different for both components Specific Enthalpy of component at given temperature is given as: h + Cp dT AH,250 We can calculate specific enthalpy of H2 and CO directly from standard temperature at given temperature with respect to 25 C as reference temperature. For (CH3)20 and CH4, we need to follow the following procedure given below: For (CH3)20, At T = 600 C = 873.15 K h(CH3)20 AH(CH3)20.25 + Store CP (CH3)20 (8) dT Here, AH (CH3)20.250 -180.16 kJ/mol -180160 J/mol h(CH3)20 -180160 + 1973,15 (26.86 +0.1659T - 4.179 * 10-5 * T?)dt -180160 + 62407 J/mol - 117753 J/mol -117.753 kJ/mol Same for inlet and outlet since temperature is same for both For CH4, At T = 600 C hch AH CH4,25C + SCPCHAdT hch = -74.85 +3.50 (34.31. 10-3 + 5.469 - 10-5T +0.3661 - 10-8 - T2 – 11 * 10-12 . T?) dT -74.85 +29.463 kJ/mol = -45.387 kJ/mol

Now, we need to find it for CO and H2 We can directly find value of sea Cp dT for CO and Hz from table For H2: S25°Cph, dt 16.81 kJ/mol For CO: Sesso Cp.co dT 17.57 kJ/mol Thus, hh2 0 + 16.81 16.81 kJ/mol hco -110.52 + 17.57 -92.95 kJ/mol Thus, we found specific heat enthalpies of all the components. Molar flowrate and Specific Enthalpy table is: + Component Dimethyl ether Carbon Monoxide Methane Hydrogen nin * 10% (mol) Hin (kJ/mol) nout * 103 (mol) 25.71 -117.753 11.937 0 13.773 0 13.773 0 13.773 Hout (kJ/mol) -117.753 -92.95 -45.387 16.81