Answer the following questions about two isomers, methyl methanoate and ethanoic acid. The molecular — Equilibrium Chemistry Question
Problem Context
Answer the following questions about two isomers, methyl methanoate and ethanoic acid. The molecular formula of the compounds is C2H4O2.
A student puts 0.020 mol of methyl methanoate into a previously evacuated rigid 1.0 L 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:
2 CH3COOH(g) ⇌ (CH3COOH)2(g)
Model Answer
See diagram in scoring guidelines. (Note: A pair of electrons can be drawn as a pair of dots or a line segment.)
1 point is earned for a correct diagram showing all valence electrons, including single bonds, double bond to the carbonyl oxygen, and two lone pairs on each of the oxygen atoms.
Model Answer
Let n_initial be the number of moles of gas particles before any reaction occurs.
Because 50. percent of the molecules reacted, n_final = 0.75 n_initial.
P_final / P_initial = n_final / n_initial
P_final = P_initial * (n_final / n_initial) = (0.74 atm) * (0.015 mol gas particles) / (0.020 mol gas particles) = 0.56 atm
OR
PV = nRT
P_final = nRT / V = (0.015 mol)(0.08206 L atm mol−1 K−1)(450 K) / 1.0 L = 0.55 atm (or 0.56 atm)
1 point is earned for the correct pressure. || P_total = 0.56 atm
Mole fraction CH3COOH = 0.010 mol / 0.015 mol = 2/3
Mole fraction (CH3COOH)2 = 0.0050 mol / 0.015 mol = 1/3
P_CH3COOH = (2/3)(0.56 atm) = 0.37 atm
P_(CH3COOH)2 = (1/3)(0.56 atm) = 0.19 atm
Kp = P_(CH3COOH)2 / (P_CH3COOH)^2 = 0.19 / (0.37)^2 = 1.4
1 point is earned for the correct partial pressures.
1 point is earned for an answer that uses partial pressures correctly.