Show and explain each step (neatly in complete sentences), indicate units, and s
ID: 972115 • Letter: S
Question
Show and explain each step (neatly in complete sentences), indicate units, and specify the correct number of significant figures.
In the vapor phase, two acetic molecules may combine to form a hydrogen-bonded dimer:
We may determine the corresponding equilibrium constant by evaporating acetic acid in a container that is subsequently sealed and weighed. In one experiment, acetic acid was evaporated in a container of volume 21.45 cm^3 at 437 K. The pressure of the vapor was 101.9 kPa. The mass of acid in the sealed container was 0.0519 g. The experiment was repeated with the same container (same volume and pressure) at a temperature of 471 K; the mass of acid in the second container was 0.0380 g.
a. Compute the equilibrium constant for dimerization at each of these temperatures.
b. Determine the standard reaction enthalpy for this process.
c. Determine ?G° and ?S° for the dimerization reaction at 437 K.
Explanation / Answer
(a) Let us denote acetic acid as HA, so that the dimer can be represented as (HA)2.
2 HA ------> (HA)2
The pressure of the sealed container is 101.9 kPa (=1.006 atm) and the volume is 21.45 cm3 (= 0.02145 L) at 437 K. The container contains vapours of both the monomer and the dimer. As per Dalton’s law of partial pressures,
p = pM + pD where p is the total pressure and pM and pD are the partial pressures due to the monomer and the dimer.
Now, pM = nMRT/V where nM is the amount of the monomer in mole at equilibrium. Similarly,
pD = nDRT/V where nD is the molar concentration of the dimer at equilibrium. Molar mass of acetic acid is 60.05 gm/mol and we start with 0.0519 gm acetic acid monomer, so intial amount of the monomer is (0.0519 gm)/(60.05 gm/mol) = 8.643*10-4 mol and the initial molar concentration is (8.643*10-4 mol)/(0.02145 L) = 0.0403 M.
We construct the ICE chart as
2 HA --------------> (HA)2
initial 0.0403 0
change - 2x x
equilibrium (0.0403 – 2x) x
where x is the molar concentration of the dimer at equilibrium. Hence, amount of the monomer and the dimer at equilibrium are {0.02145(0.0403 -2x)} mole and 0.02145x mole.
Hence,
pM = {0.02145(0.0403 – 2x)}RT/V and
pD = 0.02145x.RT/V
Therefore,
p = {0.02145(0.0403 – 2x)}RT/V + 0.02145x.RT/V
or, (1.006 atm)(0.02145 L)/(0.082 L-atm/mol.K)(437 K) = 0.02145{(0.0403 – 2x) + x}
or, 0.0281 mol = 0.0403 – x
or x = 0.0122 mol
Therefore, the equilibrium concentration of the dimer is 0.0122 M and that of the monomer is (0.0403 – 2*0.0122) = 0.0159 M.
The equilibrium constant, KC = [(HA)2]/[HA]2 = (0.0122 M)/(0.0159 M)2 = 48.257 M-1 = 48.256 L/mol.
To find the equilibrium constant at 471 K, we should have proceeded in the same way. The initial concentration of the monomer in this case will be 0.0380 gm/(60.05 gm/mol) = 6.328*10-4 mol and the molar concentration is 0.0295 M.
Since, the pressure and volume remain same, we have
0.0281 = 0.0295 – x
or, x = 1.4*10-3
Solving, we will get KC = 1.773 L/mol
(b) From Vant Hoff equation, we know,
ln (K1/K2) = H0/R(1/T1 – 1/T2)
or, 3.304 = H0/R( 2.288*10-3 – 2.123*10-3)K-1
or, (3.304 * 8.314 J/mol.K)/1.65*10-4/K =H0
or, H0 = 166.481 kJ/mol (ans)