The kinetics of the catalytic reaction SO_2 + 2H_2S rightarrow 3S + 2H_2O are be

Question

The kinetics of the catalytic reaction SO_2 + 2H_2S rightarrow 3S + 2H_2O are being studied in an ideal CSTR. Hydrogen sulphide (H_2S) is fed to the reactor at a rate of 1000 mol/h. The rate at which H_2S leaves the reactor is measured and found to be 115 mol/h. The feed to the reactor is a mixture of SO_2, H_2S, and N_2 in the molar ratio 1/2/7.5. The total pressure and temperature in the reactor are 1.1 atm and 250 degree C. The reactor contains 3.5 kg of catalyst. What is the rate of disappearance of H_2S? Don't forget to include units. What are the corresponding concentrations of H_2S, SO_2, S and H_2O? You might want to revisit the concept of extent of reaction (xi) from CHE 320 or use the newly introduced concept of conversion (X).

Explanation / Answer

Feed to the reactor contains 1000 moles/hr of H2S . SO2=1000/2= 500 moles/hr, N2= 7.5*500= 3750 moles/hr. There is no S and H2S in the reactants.

From the definition of rate law, -rH2S = (1/W)*dNA/dt

Where w= weight of catalyst and -rH2S = (1000-115)/3.5= 252.85 moles/h.kg catalyst

From the stoichiometry, -dH2S/2dt = -dSO2/dt

-dSO2/dt= (252.85/2)*3.5= -379. 275   moles/kg catalyst

SO2 at t=t – SO2 at t=0= -379.275, SO2 at t=t = 500-379.275= 120.725 moles/hr

-rH2S/2 =rS/3 ( since S is formed)

rS= 1.5*(-rH2S)= 1.5*252.825 moles/kg catalsyst= 379.2375

S at t=t- S at t=0 = 379.2375*3.5= 1327 and H2O= (2/3)* 1327

S at t=0 = 1327 moles/hr and H2O= 885 moles/hr N2 remains the same at 3750 moles/hr

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