Question

Some CH.CH,OH is allowed to dissociate into CH CHO and H, at 523 K. At equilibrium, (CH.CH,OH) - 0.313 M. and [CH,CHO] - [H.] -6.48*10-M. Additional CH,CH,OH is added so that (CHCH,OH) -0.492 M and the system is allowed to once again reach equilibrium. CH,CH,OH(g) CH CHO(g) + H2(g) K-1.30x102 at 523 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [CH3CH2OH) = [C,H,CHO

Answer 1

du) Prútial contientration after adding mole Glo Chon, [Cons CM₂ OH] = 0.492M. [Cong Cho ] = 6.48810 2M [H₂ = 6.48810 ²M GASON (8) A Collo Cho (8) + H₂ CI) Q. Product of Concentration of Product Product of concentration of Reactant. = [CH CHO] [12] [Conson] Q = 6.48x1042 X 6.48X10-2 00492 = 85.35X20-4 . ke 7 hc Alo, Reaction is forward. Step I, ICE Table for Ron. GHOH (8) 2 CH CHO(g) + H₂ (8) 6.48X102 6.48X102 - x ta 0.492-8 6.48x10tn 6+48x67n 0.492 HUCH AK

K: t ConsChoT [th] 0.492 103X102 = (6.48X10 FW ( G. 48x1074) (00492-2) 1:3X102 - (6648X1024) 100492-21) 1. 3X102(00492 -_) = 30°+ 296X16ʻze + 411.99X16 0.6396X102 -3 xöze = seº+12.96X60*} +41.99x169 x? +14.26X10*4- 40.1194X10-4c0 10-2197 x = -140 26x10² + |(14.26 X10-2) 2 4X+06x10x1 2xl taking a vesign / Taking -ve sign x= -14.3 (not valid) (x) x= 0.015422)/ sco oftere equilibriumo [[Hg CH₂OH] = 0.492-000154= 0.4766 m [Colls Cho] = 000154M Hi] = 0.0154 M. Thank you, May this help you.