The steam reforming of propane (R1) over a Ni catalyst is used to produce H2 gas

Question

The steam reforming of propane (R1) over a Ni catalyst is used to produce H2 gas. In the reactor, a simultaneous reaction occurs (R2), resulting in the formation of additional H2. The feed to the reactor contains steam and propane in the ratio 6:1 (molar) at a temperature of 125o C. Products are formed at a temperature of 800o C. Given the large excess of steam, we can assume that all the propane is consumed. The reactor is heated by passing a hot gas on the outside at a rate of 4.94 m3 /mol of propane entering the reactor. The heating gas enters the system at 1400o C and 1 atm. and is cooled to 900o C in the process. Given that the heat capacity of the heating gas is 0.040 kJ/mol o C, calculate the composition of the product gas from the steam reforming process. The reactions taking place in the reactor are as follows:

C3H8 (g) + 3H2O (g) ? 3CO (g) + 7H2 (g) (R1)

CO (g) + H2O (g) ? CO2 (g) + H2 (g) (R2)

Explanation / Answer

Number of mole of heating gas = 101300*4.94/(8.314*1673) = 35.5 mol/mol of propane.

Temperature of heating gas decreased from 1400C to 900C. The heat transferred to the reactor is

q= 0.040*(500)*(35.5)= 710.3 kJ/mol of propane

the temperature of product gas is increased from 125C to 800C. Heat required to increase the temperature can be calculated using E= nRT.

one mole of propane yields 13 mole of product gas. Therefore the energy released is

E = 13*8.314*675 = 72.955 kJ/mol

we more data such as heat of reaction, heat capacity of product gases to calculate the composition.

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