Question

The oxidation of nitric oxide NO + 02 =NO2 takes place in an isothermal batch reactor. The reactor is charged with a mixture containing 20.0 volume percent NO and the balance air at an initial pressure of 380 kPa (absolute). (a) Assuming ideal-gas behavior, determine the composition of the mixture (component mole fractions) and the final pressure (kPa) if the conversion of NO is 90%. (b) Suppose the pressure in the reactor eventually equilibrates (levels out) at 360 kPa. What is the equilibrium percent conversion of NO? Calculate the reaction equilibrium constant at the prevailing temperature, Kyl(atm)-0.5), defined as (Pro) Кр where p,(atm) is the partial pressure of species i (NO, NO, 02) at equilibrium. (c) Assuming that K, depends only on temperature, estimate the final pressure and composition in the reactor if the feed ratio of NO to O, and the initial pressure are the same as in Part (a), but the feed to the reactor is pure 02 instead of air. (d) Replace the partial pressures in the expression for Kp, and use the result to explain how reactor pressure influences the conversion of NO to NO2. (Pro)(Po)03

Answer 1

NO + Į O NO2 Po 380 Kla & het initially 100 moles of air-Nos inixture is charged vol % of No 20 % 0.2 mole , Assuming ideal behaviour 7 Boro = P. Initical partial pressure NO 0.2 X 380 F6 lela pairs 380-76 = 304 kla air contains 21% (by volume) O bog 0.21 X 304 63.84 ula PON 304- 63.84 240.16 lela Nfinal x ДИ (final total no. of moles) 100 + (0.9) (1--1) 99.55 moles Not

Page! Date in = 1 + 1 = -0.5 .. lieleal gas o ) Po = NORT V P NRT 기 Po Р No N 100 99.55 380 P P 37.29 kle mixture final pressure of gas = 378.29 kPa - Pro Poro a (P-Po) where a = coeff of No in the reacth = 1 76 - 380) +(378.29 Pro tou) & Pro 72.58 kla PNOT Similarly, Po Pole b Δη ( p — P. when, b = coff of O aly the reacth = ㅗ - 6.5 2

po 63.84 0.5 ( 378.29 - 380) (در) -) por 62.13 KPa PNO + por + pre = 72.58 + 62.13 Pro 243.58 t prior P. 378.29 Kla Y No Pono 0 19 186 72.58 378.29 P You por 0.164239 62.13 378 29 Р PNow 0.64389 -243.58 378.29 Р b) NO + NO2 Tutal initial pressure 76 - 240.16 +380 = 139.84 63.84 Change - 20 13819-12 Equill presure 76-x 63.84 - 2 De -240.16+3 60 = 119.84 19.84 (76-x) + ( 63.84x) 기

here PN2 is substracted because Ne is not taking part in the reaction bence Por is not changing ban Date : Page : 40 76-40 = kPa 43.84 KPq 63.84 - 40 2 40 KCP a Galb. Y. Conversion of to 7636 -RNO - pro KtOD 76 Pro 52.63 15 r. & Equill. Constant PNow ou (tage (to) Кр 40 36 * (43.840.5 3 Кр 0.1678 mixture loo 380 het initially 100 moles of O- No fed to reacture No moles where voly. = moles of No 0.2 0.2 x 380 = 76 KRA 380 - 76 304 Baso - Por kPa

Date: Page: let, Conwersion of NO = gor (as in part (a) Nfind a X. An Not 100+ N = 0.9 (-05) 99.55 PE 379.29 kla 지 No Nf P Now PNo 72.58 Kla 76 ( 378-29 - 380) pe 309 - D.J (378.29-380) 302.29 kPa PNoz 3.42 kPa 378-29 (72.58+302.29) o.19186 Yono 72.8 378.29 Pro P por 0,799 302.29 378.29 P 3.42 378 29 3.42 ko PNow (No) (kos) . 0.37 (72.58). (302.29) 0.1 Kp 0.00271 therefore, as partial pressure changes, Kp. conversion changes accordingly