A prevalent form of toxic pollutant NO_x formed in power-plant combustors is NO.

Question

A prevalent form of toxic pollutant NO_x formed in power-plant combustors is NO. Under favorable conditions, NO can be decomposed ("reduced") via the following reaction: 2NO rightarrow N_2 + O_2 At 1620 K, for a reaction rate expressed in gmol/L s and the amount of NO expressed in atmospheres, the reaction rate constant for this irreversible reaction is 0.0108 gmol/L s (atm)^2 (from reference 2, p. 813). a. Assuming that NO is the only reactant, use the units of the rate constant to determine the order of this reaction in terms of NO b. If a reactor is designed to reduce NO at a rate of 0.056 gmol/min L at 1620 K, what partial pressure of NO is needed in the reactor?

Explanation / Answer

Rate = k [reactant]n

Rate = [reactant]/time (how much reactants reacted per time)

[reactant]/time = k [reactant]n

k= [reactant](1-n) time-1

here k = gmol/L s

ie., gmol/L represent here concentration of reactant [(gmol/L)1]

so 1-n should be equal to 1

that means n=0

that indicates this is a zero order reaction.

they didnt depent on the concentration of NO

b) Kp=(PA)a(P)bB  / (PC)c(PD)d   ( A and B are product and C and D are reactants)

here kp = (PN2)(PO2) /(PNO)2

kp is inversily propotional to square of partial pressure of NO

kp1/kp2 = (P2NO)2/(P1NO)2

(P2NO)2 = 0.0108/0.056 (P1NO)2

   =0.1928(P1NO)2

(P2NO) = [0.1928(P1NO)2](1/2)

this much pressure want to reduce to make rate 0.056 gmol/L at 1620 K

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