Question

5. Consider the reaction below: CH COOC Hs(aq) + OH (aq CH Coo (aq)+CHsCH2OH(aq) The second-order rate constant for the reaction is 0.11 dmmol's1. What is the concentration of ester after (a) 10 s, (b) 10 min when ethyl acetate is added to sodium hydroxide so that the initial concentrations are [NaOH 0.050 mol dm3 and [CH COOC2Hs] 0.100 mol dm 3?

Answer 1

1/[A]_0 - [B]_0 ln [B]_0 {[A]_0 - x}/[A]_0 {[B]_0 - x} = k t where, [A]_0 = initial Concentration of ester = 0.100 mol dm^-3 [B]_0 = initial Concentration of NaOH = 0.050 mol dm^-3 x = change in the Concentration of ester or NaOH in time t k = rate constant = 0.11 dm^3 mol^-1 s^-1 t = time t = 10 s Substituting the values in equation (i) 1/(0.100 - 0.050) mol dm^-3 ln 0.050 times (0.100 - x)/0.100 times (0.050 - x) = 0.11 dm^3 mol^-1 s^-1 times 10 s ln (0.100 - x)/2 times (0.050 - x) = 0.11 times 10 times 0.05 0.100 - x/2 times (0.05 - x) = e^0.055 0.100 - x/0.05 - x = 2 times 1.056 0.100 - x = 2.112 times (0.05 - x) 0.100 - x = 0.105 - 2.112 x 1.112 x = 0.005 x = 0.005/1.112 = 4.50 times 10^-3 mol dm^-3 therefore Concentration of ester after 10 s = (0.100 - 4.50 times 10^-3) mol dm^-3 = 0.096 mol dm^-3 \r\n t =