Question

5. Answer the follwing questions about two isomers, methyl methanoate and ethanoic acid. The molecular formula of the compounds is CH402 (a) Complete the Lewis electron-dot diagram of methyl methanoate in the box below. Show all valence electrons. A student puts 0.020 mol of methyl methanoate into a previously evacuated rigid 1.0L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction. CH,COOH)+ CH,COOHg) (CH,COOH)28) (b) Assume that when equilibrium has been reached, 50. percent of the ethanoic acid molecules have reacted. (i) Calculate the total pressure in the vessel at equilibrium at 450 K. (i) Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K.

Answer 1

The lewis dot structure Initial pressure p = 0.74 atm % association = 50% alpha = % association/100 alpha = 50/100 alpha = 0.5 consider a reaction 2 ch_3 cooh(g) rightarrow (ch_3cooh)_2 (g) I 0.74 atm 0 atm C 0.74 - 0.74 alpha 0.74/2 alpha E 0.74 (1-alpha) 0.74/2 alpha at equlibrium, p_ch_3cooh = 0.74 (1-alpha) = 0.74 (1-0.5) = 0.37 atm p_(ch_3cooh__2 = 0.74 alpha/2 = 0.185 atm total pressure at equilibrium p_tot = p_ch_3cooh + p(ch_3cooh)_2 = 0.37 + 0.185 P_tot = 0.555 atm k_p = p(ch_3cooh)_2/p_ch_3cooh^2 k_p = 0.185 atm/[0.37 atm]^2 k_p = 1.3 atm^-1