A gas mixtures containing 85 mole % of methane and the balance of oxygen is to b

Question

A gas mixtures containing 85 mole % of methane and the balance of oxygen is to be charged into
an evacuated well-insulated 10-liter reaction CLOSED (a hint for your calculations of heat
management!) vessel at 25°C and 200 kPa. An electrical coil in the reactor, which delivers heat
at a rate of 100 W, will be turned on for 85 seconds and then turned off. As a reference state take
all molecules as gases at 25°C. Formaldehyde will be produced in the reaction:
CH4(g) + O2(g) ® HCHO(g) + H2O(g)

a) Calculate the maximum pressure that the reactor is likely to have to withstand, assuming
there are no side reactions. If you were ordering the reactor, why would you specify an
even greater pressure in your order? Give at least two reasons.
b) Why would heat be added to the feed mixture rather than running the reactor
adiabatically?

c) Suppose the reaction is run as planned, the reaction products are analyzed
chromatographically, and some CO2 is detected. Where did it come from? If you had
taken this CO2 into account, would your calculated pressure in part a) have been larger,
or smaller, or you cannot tell without doing the detailed calculations?

Explanation / Answer

Initial moles of gas mixture in the reactor = PV/RT

P= 200 Kpa= 200/101.3 atm= 1.974 atm, V= 10 Liters, R= 0.0821 L.atm/mole,K, T= 25+273= 298K

Total nNumber of moles initially= 1.974*10/(0.0821*298)= 0.81 moles

Moles of methane = 0.81*0.85= 0,69, moles of O2= 0.81-0.69=0.12 moles

As per the reaction CH4(g)+ O2(g) --à HCHO(g) + H2O(g)

Molar ratio of CH4:O2= 1:1 molar ratio given : 0.69:0.12 = 5.75:1. So formaldehyde is the excess reactant. All the oxygen is consumed. So moles of HCHO=0.12, moles of H2O= 0.12, moles of CH4 remaining =0.69-0.12 =0.57

Heat supplied = 100J/s* 85 seconds= 8500 Joules.

For maximum pressures the reactor withstands, the heat supplied is expected to rise the temperature of the gas without change in temperature.

From 1st law of thermodynamics applied to closed system, deltaU= Q+W

Q= -W= -PdV

V= nRT/P, dV= (-nRT/P2) dP

Q= -PdV= P*nRT/P2= nRT ln(P2/P1), P1= 200Kpa, T= 298K

8500 = 0.81*8.314*298*ln (P2/200)

P2= 200*69.1 Kpa = 13820 Kpa.

So the reaction is gas phase reaction and the design pressure is normally fixed at 1.25 times the operating pressure. Since the hydraulic testing of the equipment is done at design pressure.

The standard suggest the design pressure to be higher than the operating pressure. The design pressure also takes care of other loads like wind load etc.

3. When the heat is added adiabatically, it increases the temperature of reaction and may leas to side reaction of combustion of CH4.

4. This is due to combustion of CH4 according to the reaction, CH4+2O2--àCO2+ 2H2O

Here the number of moles of products is more than the moles of reactants. Hence the Calculated pressure will be more than the pressure calculated earlier. For that detailed calculation are required.

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