Question

The vapor pressure of Y is 4.050 atm at 290K and 13.365 atm at 373K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). «J/mol AG290º = AG373° = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values J/(mol-K) AS° = AH° = kJ/mol What is the normal boiling point of Y in degrees Celcius? 1 t= t = ° What is the vapor pressure of Y at 403 K? po = am atm Submit Answer

Answer 1

Vapour pressure of Y is 4.050(p1) atm at 290K(T1) and 13.365(p2) atm at 373 K(T2)

We know the equation,

  
AG = -RTINK
where $\small \Delta G$ - gibbs free energy.

R - Universal gas constant (atm) = 8.314J/(mol.K)​​​​​​, T - temperature at Kelvin, K - equilibrium constant at pressure. We can substitute pressure itself.

At T = 290K, K = 4.050 atm,

  

AG = -8.314J/(K.mol) x 290K x In(4.050)

AG290 = -3372.39J/mol = -3.372K J/mol

At T = 373K K = 13.365 atm

AG373 = -8.314J/K.mol x 373K x In(13.365)

AG373 = -8040J/mol = -8.04K J/mol

​​To find the standard Enthalpy of vaporization, we use the equation,

$\small ln(\frac{p1}{p2}) =\frac{\Delta H}{R}*( \frac{1}{T2}-\frac{1}{T1})$

$\small ln(\frac{4.050}{13.365}) =\frac{\Delta H}{8.314 }*( \frac{1}{373}-\frac{1}{290})$

AH = 1.29 * 104J/mol = 12.9K J/mol

To find standard entropy, we use the equation,

ΔG = ΔΗ – TAS
where delta S is the entropy and T is the temperature.

At 290K,  
AG = -3372J/mol
,  
ΔΗ = 12900J/mol

-3372 = 12900 + 290 * AS

AS = 56.11J/(mol – K)

Boiling point = the ratio of standard Enthalpy to standard Entropy.

AS

12900J/mol 56.11J/mol - K) = 229.9K

  

= (229.9 + 273) C = 502.9°C

At 403 K,
AG =?

ΔG = ΔΗ – TAS

  
AG = 12900(J/mol) – 403T * 56.11(J/(mol – K))

  
AG = -9712.33J/mol = -9.712K J/mol

From this we calculate vapour pressure using the first equation we used above.

AG = -RTINK

RT -= yut
, where R = 8.314 J/K.mol , T = 403 K ,
AG = -9712J/mol

  

-9712.33 InK = - 8.314 * 403

= - 2.898

  

K = antiln(–2.898)

= 0.05513

Therefore vapour pressure at 403K is

P = 0.05513 atm