Question

The second-order rate constant for the reaction below is 0.11 dm³ mol⁻¹ s⁻¹.

CH₃COOC₂H₅(aq) + OH⁻(aq) → CH₃CO₂⁻(aq) + CH₃CH₂OH(aq)

What are the concentrations after the following times when ethyl acetate is added to sodium hydroxide so that the initial concentrations are [NaOH] = 0.050 mol dm⁻³ and [CH₃COOC₂H₅] = 0.100 mol dm⁻³?

(a) 10. s

CH3COOC2 Hs +OHPCH3 COz tCHsCteott iS o.jl Lmai.S shab is 6he concentraron OF ester curter los uhen ebhyl acebaute IS Cadsed 6oSodium hydroxide Ta ke COHJ O.OSmol/L cnd CCts COOC2 lts30 01mol A +BProduc6s KaOLLlmolS

Answer 1

solution we know reaction, the formula for second ordez * 2006, 7, C. (12 where, Karate constant = 0.ll Lmott git t = time = 10 sec. a = initial concentration of CH₃COOC2H5 b= initial concentration of [oH] So, here a= 0.100 mol/L b=0.050 molle x = Concentration of reactant reacted. NCC400W**TOPPPPPPP****&tf11111111 OOO OOO SO) r0.05010.100-)) 2.303 logro 10.100(0.050- de) 10x10.100 - 0,050) 0.11

2. 303 10x 0.050 10.100-10) 2 (0.050-x) 0.0239 = ley 100ml 12(0.050-x) 1057 = 0.100- 2 (0.050-x) 2.114 = 0.100-X 0.050-1 0.1057 - 2.114= 0.100-ve ? 22 0.0057 = 1.114 de 0.0057 = 0.0051 the given reaction, UUUUUUUU CH₃COOC2H5 + NaOH CH₃ Coonat C2 H₂ OH. at t=0 at t= 10 sec an b-x 2 っっっっっ so, after losec Concentration of CH₂COOH = (a-x) = {0.100- 0.0051) : 0.0949 mol/l Concentration of MaoH .050 - 0,0051 = 0.0449