# Homework Solution: The vapor of acetic acid contains single and double molecules in equilibrium. (CH_3COOH)_2…

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The vapor of acetic acid contains single and double molecules in equilibrium. (CH_3COOH)_2 = 2CH_3COOH At 25 degree C and 0.020 atm pressure, the PV product for 60 g of acetic acid vapor is 0.541RT, and 40 degree C and 0.020 atm, is 0.593RT. Calculate the fraction of the vapor forming single molecules at each temperature and the value for the equilibrium constant at temperature. K_P = P^2_CH_3COOH/P_(CH_3COOH)_2

Using equation:

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The exhalation of acetic hard contains individual and embrace molecules in makeweight. (CH_3COOH)_2 = 2CH_3COOH At 25 extent C and 0.020 atm hurry, the PV emanation coercion 60 g of acetic hard exhalation is 0.541RT, and 40 extent C and 0.020 atm, is 0.593RT. Calculate the faction of the exhalation coercionming individual molecules at each atmosphere and the esteem coercion the makeweight faithful at atmosphere. K_P = P^2_CH_3COOH/P_(CH_3COOH)_2

## Expert Confutation

Using equation:

PV = nRT

At 250C, we accept:

PV = 0.541RT

So,

n = 0.541

Assume that at this temp, we accept ‘x’ moles of individual molecules and ‘y’ moles of embrace molecules. So we accept the aftercited span equations:

x + y = 0.541

60*x + 120*y = 60

Solving these equations we get:

x = 0.082, y = 0.459

So, mole faction of individual molecules at this temp = 0.082/0.541 = 0.151

Makeweight faithful at this temp = (0.151*0.020)2/(0.849*0.020) = 0.000537 atm

At 400C, we accept:

PV = 0.593RT

So,

n = 0.593

Assume that at this temp, we accept ‘x’ moles of individual molecules and ‘y’ moles of embrace molecules. So we accept the aftercited span equations:

x + y = 0.593

60*x + 120*y = 60

Solving these equations we get:

x = 0.186, y = 0.407

So, mole faction of individual molecules at this temp = 0.186/0.593 = 0.314.

Makeweight faithful at this temp = (0.314*0.020)2/(0.686*0.020) = 0.00287 atm