Question

Consider a chemical reaction CO(g)+2H, (g) = CH,OH (g). When 2 moles of CO were reacting with 4 moles of H, at temperature 500K and pressure p 100 bar, and chemical equilibrium was reached, 0.366 moles of COwas consumed. a) (10 pts) Based on the given data determine the value of the chemical equilibrium constant K(T) for T= 500K b) (10 pts) Consider now the same reaction, when 3 moles of CO are reacting with 3 moles of H at the temperature T 500K and the pressure p 100bar. Find the mass m of the product, i.e., CH OH, obtained in the reaction.

Answer 1

Co(g) +242 (3) = CH30H18) Preparing ICE Table for it I= initial = change to equilibrium Initial Total mol = 2molcot u mol 42 - 6 mol Temprature - 5ook pressure = loo box I bus = 0.987 am 10obar = 98.7 ahm vo PV = nRT T= Temprature P= Pressure v = volume n= number of mol R= gas constant V = nRT = - 6 moi * 008 6 mol x 0.0821 Laton xsook Molok E 98.7 apon so CO (9) + 242419) 4. - 2x Change - 4-2X 2-2 AF quilibrium At equilibrium 0.366 mol of co consumed. so na x=0.366 At equilibrium mol - 2-0.366 - 1634 mol 4-0.732= 3.268 mal mol J H₂ = moly chzon = 0.366 mol

Volume = 2:54 0.36 mol - 0:1464M [CH₂ou] = mol 25 volume = 1.634 mol 2.5 L 0.6536 M [4 ] 3.266 mol = = 1.3072 M 3.266 mol D.SL k= [CHzou] [co] [42] (0.1464) = = 0.131 (0.6536) (1 3072)2 K= 0.131 equilibrium constant = 0.13) ² CH₂on19) colgt 2420 (9) initial 3 chя 8 Chay - 3-2 3-2x 9- A+ epuili bolum neol Ko 3-2 A+ беше bo'qm Conco [Chjon) [co] [127² (xlys) xac 3 nglezte in 480 in 2.5

013! = 1044 0.188 = x mo At equilibrium a= 0.18 9 mol = 0.189 mass = mol x molar mass = 0'18 9hx 32.04 g = 6.06 g mor massy alizon formed = 6:06g Answer