Cinetica

Homework 2 - Reaction Kinetics

Equations
Find the equilibrium state for a mixture at temperature of 1000K and 25 atm of pressure an compare both results. The mixture is compound by : 1 mole of CO3 mole of H2 We have 2 non-dependant reactions in our system: 2CO + 2H2 ←→CH4 + CO2 ( 1 ) CO + 3H2 ←→ CH4 + H2O ( 2 ) And we are going to have just one equilibrium, but it will depend of the reactionyield of each one reaction. This defines the total mole in equilibrium molT = 4 − 2 · ξ1 − 2 · ξ2 And we define the mole of each component, again in the equilibrium of the system molCO = 1 − 2 · ξ1 −ξ2 molH2 = 3 − 2 · ξ1 − 3 · ξ2 molCH4 = ξ1 + ξ2 molCO2 = ξ1 molH2O = ξ2 Now this represents the fraction by component yCO = molCO /molT yH2 = molH2 /molT yCH4 = molCH4 /molT yCO2 = molCO2 /molT yH2O =molH2O /molT The problem give us the Kp values for the constant temperature Kp1 = 0.046 atm-2 Kp2 = 0.034 atm-2 (12) (13) (7) (8) (9) (10) (11) (2) (3) (4) (5) (6) (1)

As we know the Kp value isdefinied by de continuos product of the partial pressure of each component in the system, powered by the stoichiometric factor in the reaction. Making some algebraic work we can simplify to this 2expressions: Kp1 = Kp2 = And here we check the composition summatory yac = yCO + yH2 + yCH4 + yCO2 + yH2O (16) yCH4 · yCO2 2 2 yCO · yH2 yCH4 · yH2O 3 yCO · yH2 · · 1 P2 1 P2 (14) (15)

1

ParametricTable: T able 1
Run 1 2 3 4 5 6 7 8 9 10 11 P [atm] 1 5 10 15 20 25 30 35 40 45 50 ξ1 0.05833 0.1177 0.1294 0.1305 0.129 0.1267 0.1242 0.1217 0.1193 0.117 0.1149 ξ2 0.1429 0.3627 0.4687 0.5276 0.5670.5962 0.619 0.6376 0.6531 0.6664 0.6779 yCH4 0.05594 0.1581 0.2133 0.2452 0.2669 0.283 0.2957 0.306 0.3146 0.322 0.3284 yH2O 0.03973 0.1194 0.1672 0.1966 0.2174 0.2334 0.2463 0.2569 0.266 0.2739 0.2808yCO 0.2058 0.1322 0.09717 0.07878 0.0671 0.05891 0.05277 0.04798 0.04411 0.04091 0.03822 yCO2 0.01621 0.03872 0.04616 0.04863 0.04946 0.04959 0.0494 0.04904 0.04859 0.0481 0.0476 yH2 0.6823 0.5516...