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HOMEWORK-2 For the given stoichiometry of the oxidation reaction of NO, the rate expression is given as follows. 2NO + O$_2$ ? 2NO$_2$, r = k (C$_{NO}$)$^2$ (C$_{O_2}$), k = 1.4 x 10$^4$ (kmol/m$^3$)$^2$ sec$^{-1}$ The volumetric flow rate at the inlet to a plug flow reactor is 10m$^3$/sec and the feed composition is given as 9% NO, 1% NO$_2$, 8% O$_2$, 82% N$_2$. The reactor operates at 1 atm and 20$^\circ$C. Find the reactor volume necessary to obtain the C$_{NO_2}$/C$_{NO}$ = 5 at the outlet of the reactor.

          HOMEWORK-2
For the given stoichiometry of the oxidation reaction of NO, the rate expression is given as follows.
2NO + O$_2$ ? 2NO$_2$, r = k (C$_{NO}$)$^2$ (C$_{O_2}$), k = 1.4 x 10$^4$ (kmol/m$^3$)$^2$ sec$^{-1}$
The volumetric flow rate at the inlet to a plug flow reactor is 10m$^3$/sec and the feed composition is given as 9%
NO, 1% NO$_2$, 8% O$_2$, 82% N$_2$. The reactor operates at 1 atm and 20$^\circ$C. Find the reactor volume necessary to
obtain the C$_{NO_2}$/C$_{NO}$ = 5 at the outlet of the reactor.
        
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HOMEWORK-2
For the given stoichiometry of the oxidation reaction of NO, the rate expression is given as follows.
2NO + O2 ? 2NO2, r = k (CNO)^2 (CO2), k = 1.4 x 10^4 (kmol/m^3)^2 sec^-1
The volumetric flow rate at the inlet to a plug flow reactor is 10m^3/sec and the feed composition is given as 9%
NO, 1% NO$2$, 8% O$2$, 82% N$2$. The reactor operates at 1 atm and 20$^\circ$C. Find the reactor volume necessary to
obtain the CNO2/CNO = 5 at the outlet of the reactor.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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HOMEWORK-2 For the given stoichiometry of the oxidation reaction of NO, the rate expression is given as follows: 2NO + O2 → 2NO2, r = k(CNO)^2(CO2) k = 1.4x10^(-3) kmol/msec The volumetric flow rate at the inlet to a plug flow reactor is 10 m/sec and the feed composition is given as 9% NO, 1% NO2, 8% O2, and 82% N2. The reactor operates at 1 atm and 20°C. Find the reactor volume necessary to obtain the CNO2/CNO = 5 at the outlet of the reactor.
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00:01 So yeah in this question as per the question the recycle is very large hence therefore r is equals to infinity so clearly the w that is weight of catalyst is given as 3 kg then we have c of a o that is given as 2 mole per meter cube and similarly c of a is also given as that is 0 .5 mole per meter cube similarly v is given as 1 meter cube per hour now clearly we know that delta is equals to stoichiometric coefficient of product minus stoichiometric coefficient of reactant divided by stoichiometric coefficient of a so clearly it is equals to 1 minus 1 over 1 that is equals to 0 and y of a o comes out to be mole fraction of reactant a that is for pure a for pure a it is equals to 1 now that is volume expansion factor is equals to volume expansion factor that is equals to e a is equals to delta of multiplied by y of a o which is equals to 0 multiplied by 1 that is equals to 0 so volume expansion factor is equals to 0 now at constant temperature and pressure v is equals to v naught of 1 plus e a multiplied by x a where x a is the mole fraction so from here we can calculate the number of moles that is n a which is equals to n of a d 1 minus x a so we can rearrange this equation as well now clearly from…
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