Question

The vapor pressure of Y is 4.560 atm at 440K and 13.680 atm at 515K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(l) = Y(g).

G440o = -5.55 kJ/mol

G515o = -11.20 kJ/mol

Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values

delta S = J/(mol-K)

delta H = kJ/mol

What is the normal boiling point of Y in degrees Celcius?

t = oC

What is the vapor pressure of Y at 545 K?

Po = atm

Answer 1

ANSWER:

Y (1) =Y (9)

For this reaction, the equilibrium constant is

K, = Py

At the same time, the free energy is defined as

AGⓇ = -RT In (K)

where R = gas constant = 8.314 J/mol.K and T = temperature in K

Then, replacing

AGⓇ = -RT In (Py)

Now, we can calculate the standard free energy at each temperature:

• At 440 K

AG 440K = -8.314 = X 440 K x In (4.560)

k) AG440K = -5551 - = -5.55 – mol

• At 515 K

AG315K = -8.314 = x 515 K x In (13.680)

kᎫ AG$15K = -11201 =-11.20 mol mol

To find the standar enthalpy of vaporization we use the Clausius-Clapeyron equation:

$ln\left ( \frac{P_{2}}{P_{1}} \right )=-\frac{\Delta H^{\circ}_{vap}}{R}\left ( \frac{1}{T_{2}}- \frac{1}{T_{1}} \right )$

-Rin _9214J / 13.680 atm 4.560 atm AHap = 515 K - 440 K

kᎫ AH = 27596 - mol = 27.60 mol

And the standard entrophy is

AGT = AHap - TASAP

AH AS pap = - AGT

27596 -(-5551 mo) – 75.33_J Asap = 440 K mol.K

or

27596 Asap = - (-11201 mol) - 75.33_J 515 K mol.K

The nomral boiling point (at P = 1 atm) is calculated using the Clausius-Clapeyron equation:

$ln\left ( \frac{P_{2}}{P_{1}} \right )=-\frac{\Delta H^{\circ}_{vap}}{R}\left ( \frac{1}{T_{2}}- \frac{1}{T_{1}} \right )$

4.56 atm (1 atm ) 27596 mol 8.314 1440 K TO

xln(4.56) —8,314 27596 440 К то bg

-4.57x10-4 K-1 = 2.27x10-3K-1 -

= 2.27x10-3K-1 + 4.57.210-4K-1 = 2.727-10-3K-1

T = 366.70 K = 366.70 – 273.15 = 93.55 C

Finally, the vapor pressure of Y at 545 K is:

$ln\left ( \frac{P_{2}}{P_{1}} \right )=-\frac{\Delta H^{\circ}_{vap}}{R}\left ( \frac{1}{T_{2}}- \frac{1}{T_{1}} \right )$

P545K 27596 In Gam) = *8.314 T ( 545 K 366.70 K)

In (P) = 2.961

P545K = €2.961 = 19.35 atm