Question

5) Carbon dioxide (CO2) and oxygen (O2) in a 1:2 molar ratio enter a reactor operating at steady state in separate streams at 1 atm, 127°C and 1 atm, 277°C, respectively. An equilibrium mixture of CO2, CO, and O2 exits at 1 atm. If the mole fraction of CO in the exiting mixture is 0.1, determine the rate of heat transfer from the reactor, in kJ per kmol of CO2 entering. Ignore kinetic and potential energy effects. (Ans: 425239 kJ/kmol)

Answer 1

Given nco2 in formation, molay ratio of coz and oz is hoz = 2 cog initial pressure p= 1 atm coz inibad Temperature 5 = 127°C Og inital pressure palatm. Os initial temperantare ==#°c Equilibrium Mixture : Com, Co and on Equilibrium pressure palatma The mole fraction of coin the exiting mixture is go=0-1 Fov cl....ня та офія uCoron pw Scpa. The reaction takes the form Hop 700x **C0+ (1-x) (02 + ( ** Jo, The total number of rolex of mixture et nastal = x +(1-x) ( het me t from the given trole en hele fraction of coz in the exciting mintare geo - no more) For Calculating on nroto From 0.) nio 0.1 X x = 0.316

2 2 1. Substituting x wallee in the quation © 1 Coz + 202 +260 + (1-x) CO2 + ( 4+2) 1602.4202 – 0.316 Co + (1-0.316) (oz +(4+0.316)22 Then the equation becomes Ilog tao, 0.316 Co. +0.684 CO2 +2.15802 Then the total no. of roles of the mixture is motal = 642 .670.816 = 6.316 :3.158 Total = 3.155) quilibrium Co, o cot £02 The equilibrium Coretant i experession for an equilibrum 197rture of ideal gases is given by a 'ka na noga ( P/ Pred the Da n icom ( nitotal) K: (0:31 (2.16) 3 ( 3.198]" 6-684 K-10.316 70 2.1587 Y - 10.654 ] [ 3.158 ) K : 0-3819 logio k= -0.418 From The the logarithms logarithe to the base lo of The quil Table to the corresponding logiok toke temperature Iz By Pokopolation in table we will get T3 • 3000 + (3100-3000) (08.418 70.485 ) IÖ.33L 40.985), K: 1 3.158

i Tg = 3044K The Temperature TB = 3044 K the energy fake balance quation at Steady Stake is given you os n moment hay (1) ta for (-focus hco(@) 0 48 5The Le +2.176h(io)] Substituting the valles in the equation cv : : ( +10%) - (29]. + 64 [+ + (15) - (%) +2156[hep th (79) - 7 (26) Jocos2 [ üç (5) 46 (1o) - (218) Jez 2[ū(12) - T (098) Jaz Take enthalpy values from the ideal gos tables and this Subalitating the values we will get a BEK = 0.3166-110530 +103849 -8669] +0.684[-0935207 162964-9364.3 +2:158[108538' 8682] -[(-393520) +13 372 -9364 J -2(14 4to -8682] Qev =425239 xTxemal ricor 2 ) ñicos te mit o lat teme as young kmal transtor WOZ