A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

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A student puts 0.020 mol of methyl methanoate into an empty and 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.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+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.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K

Chemistry

1 answer:

KiRa [2.9K] · Answer 1 · 2026-02-25T15:27:12+03:00

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

= 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product $(CH3COOH)^{2}$ gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH)

= 0.005 mol $(CH3COOH)^{2}$

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

= 0.74 atm * (0.015 mol / 0.020 mol)

= 0.555 atm