Question

Suppose a 500 ml flask is hilled with 0.30 mol of CO, 0.20 mol of H. and 1.1 mol of CO,. The following rection becomes possible CO(R) +HO(g)- CO ) +H, KB) The equilibrium constant for this reaction is 7.74 at the temperature of the flask Calculate the equilibrium molarity of CO2. Round your answer to two decimal places

Answer 1

concentration (M1 = moles Volumell) volume of flask = 500 mL = 0.5L ( 1000 mL FILL initial concentration of co= 0.30 mol = 0.5L 0.6M initial concentration of Hal = 0.20 mol = 05L 0.4M initial concentration of CO2 = 1.1 mol = 0.5L 2.2M ICE Table + CO2(g) concentration (M) H₂(g) cog) 0.6 + H₂Olg! 0.4 - 3 0.4.10 + 0.6 x 21 2+2 K- - > 7.74 10023 CH2 [co] CH₂O] (2:2+x) x (0.6-20) (0.4-30) → 6.74 r2 - 9.94 x + 1.8576 = 0 solving this quadratic equration, we get value Here, K. = equillibrum constant (CO₂] = equillibrium molarity of CO₂ CH2) = equillibnum mularity of H2 [Co]: eyer'llabrum melaw'ty of co (H2O) - equilibrium mularity of H2O x= -b Here & Joe yac lo a= 6.74 ba - 9.94 C = 1.8576 → Y 0.22 M rrol = 12.2 + x) M = (2-2 + 0.22 ) M = won 2.42 M. (CO2) = 2,42M libnum molarity of CO2 is 2.42M Hence, Please Rate